This deck reverses differentiation into antiderivatives, including the constant of integration you must never drop. It then builds area from left, right, and midpoint Riemann sums, defines the definite integral as signed area, and proves out both parts of the Fundamental Theorem. It targets the classic errors: losing the plus C, using the power rule at an exponent of negative one, calling net area total area, dropping the chain factor in Part 1 of the FTC, and reversing the evaluation bar.
Subject: Calculus I · 149 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 17
Running the derivative backwards, building area out of rectangles, and the theorem that ties them together.
Objectives
This is the deck where derivatives and areas turn out to be the same subject. By the end you can:
Warm-up
Discussion prompt
Before we open Antiderivatives, Riemann Sums, and the FTC: without looking back, what was the main idea of Applied Optimization, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
The full applied-optimization workflow: name the objective and the constraint, reduce to one variable, state the realistic domain, find critical points, and justify the max or min. Worked classics include fences, a cut-corner box, a minimum-metal can, a poster with margins, closest points, revenue and profit, and a least-cost pipeline.
Section
Section 1
Concept
Every rule you have learned so far runs one direction: function in, rate out. Now we run the machine backwards. You are handed the rate and asked to recover the function.
antiderivative — A function F is an antiderivative of f on an interval when F'(x) = f(x) for every x in that interval. You check an antiderivative by differentiating it.
\[ \frac{d}{dx}\left[x^3\right] = 3x^2 \qquad \Longrightarrow \qquad x^3 \ \text{is an antiderivative of}\ 3x^2 \]
Counterexample
Discussion prompt
Every rule you have learned so far runs one direction: function in, rate out. Now we run the machine backwards. You are handed the rate and asked to recover the function.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Picture it
Animation
Shows: Why the two parts fit together — a rendered Manim animation.
Rendered with Manim.
Takeaway: Part two is a corollary of part one, not a separate miracle.
Picture it
Animation
Shows: An antiderivative undoes a derivative — a rendered Manim animation.
Rendered with Manim.
Takeaway: The plus C is not decoration; it is the whole family.
Intuition
A derivative takes a record of where you were and produces a record of how fast you were going.
Antidifferentiation asks the reverse question: here is the speedometer tape for the whole drive. Reconstruct the trip.
You can do it, but only up to one missing fact: where you started. Two cars that leave from mile marker 10 and mile marker 60 and then drive identically record the exact same speedometer tape.
That missing starting value is the constant of integration. It is not a formality. It is the one piece of information the rate genuinely does not contain.
Analogy
Discussion prompt
Explain The speedometer and the trip by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A derivative takes a record of where you were and produces a record of how fast you were going.
Ranking
Put in order
Put the moves of Worked example: guess, differentiate, adjust into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Differentiating knocks the exponent down by one.
Worked example
Find an antiderivative of this function.
\[ f(x) = 8x^3 \]
Guess a power one higher
Why: Differentiating knocks the exponent down by one. To land on exponent 3 we must start at exponent 4.
\[ \text{guess: } x^4 \]
Differentiate the guess and compare
Why: The shape is right but the coefficient is wrong: we produced a 4 in front and we wanted an 8.
\[ \frac{d}{dx}\left[x^4\right] = 4x^3 \]
Scale the guess by 2
Why: Constants ride along through differentiation, so doubling the guess doubles the derivative from 4 to 8.
\[ F(x) = 2x^4 \]
Verify by differentiating the answer
Why: The derivative of two x to the fourth is eight x cubed, which is exactly the function we started with. The antiderivative checks out.
\[ F'(x) = 2 \cdot 4x^3 = 8x^3 = f(x) \ \checkmark \]
Concept
Add any constant to an antiderivative and it is still an antiderivative, because the derivative of a constant is zero.
\[ \frac{d}{dx}\left[2x^4 + 7\right] = 8x^3, \qquad \frac{d}{dx}\left[2x^4 - 100\right] = 8x^3 \]
So the answer is not a function. It is a family of functions, all the same shape, stacked vertically.
\[ F(x) = 2x^4 + C, \qquad C \ \text{any real constant} \]
Picture it
Animation
Shows: A family of parallel curves differing only by a vertical shift.
Same slope everywhere, different heights.
Takeaway: Every curve here has the same slope at every x, which is why the constant of integration can never be recovered from the derivative alone.
Concept
Figure (svg): Three identically shaped upward parabolas stacked vertically, each a vertical shift of the others
Every curve in the family has the same slope at every x. Sliding a graph straight up does not change any of its slopes.
That is why the rate cannot possibly tell you which curve you are on. It only tells you the shape.
One extra fact - a single point the curve passes through - picks out exactly one member of the family.
Explain it
Discussion prompt
Explain The family, drawn to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Every curve in the family has the same slope at every x. Sliding a graph straight up does not change any of its slopes.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The problem: a particle has velocity given below and sits at position 5 when time is zero. Find the position function.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The derivative of two t cubed is six t squared, so this looks finished.
Keep the constant. It is the slot where the starting information goes.
Why: The derivative of two t cubed is six t squared, so this looks finished. The constant was never written down.
Trap
The problem: a particle has velocity given below and sits at position 5 when time is zero. Find the position function.
\[ v(t) = 6t^2 \]
Antidifferentiate and stop
Why: The derivative of two t cubed is six t squared, so this looks finished. The constant was never written down.
\[ s(t) = 2t^3 \]
Now test the given fact
Why: The problem said the position at time zero was 5, but this answer gives 0. There is no knob left to turn - the initial condition cannot be satisfied.
\[ s(0) = 2(0)^3 = 0 \neq 5 \quad \text{(contradiction)} \]
Keep the constant. It is the slot where the starting information goes.
\[ v(t) = 6t^2 \]
Antidifferentiate with the constant attached
Why: Every antiderivative of six t squared has this form; the constant is not decoration, it is the unknown starting height.
\[ s(t) = 2t^3 + C \]
Use the initial condition to solve for C
Why: Substituting time zero collapses the whole family down to the one curve that actually passes through the given point.
\[ s(0) = 0 + C = 5 \ \Longrightarrow \ C = 5, \qquad s(t) = 2t^3 + 5 \]
Notation
Annotate
From Trap: dropping the constant of integration — read this one piece at a time. What is each part doing?
On: \( s(0) = 2(0)^3 = 0 \neq 5 \quad \text{(contradiction)} \)
Concept
The whole family gets its own symbol. Read it as the general antiderivative of the function with respect to the named variable.
\[ \int f(x)\,dx = F(x) + C \qquad \text{means} \qquad F'(x) = f(x) \]
indefinite integral — The complete family of antiderivatives of f, written with an integral sign, no limits, and a plus C. Indefinite because no specific member is selected.
The piece after the integral sign is the integrand, and the differential names the variable you are undoing the derivative in. Both matter.
Definition probe
Sort into buckets
Every line below is part of the definition of antiderivative or of indefinite integral — one or the other, never both. Put each where it belongs.
Picture it
Animation
Shows: Notation: the indefinite integral — a rendered Manim animation.
Rendered with Manim.
Takeaway: The definite integral is a number. This one is not.
Concept
To reverse the power rule, raise the exponent by one and divide by the new exponent.
\[ \int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \neq -1) \]
Check it the only way that counts: differentiate the right side. The new exponent comes down, cancels the denominator, and the exponent drops back.
\[ \frac{d}{dx}\left[\frac{x^{n+1}}{n+1}\right] = \frac{(n+1)x^{n}}{n+1} = x^n \]
The exclusion is not a footnote. At that one exponent the formula divides by zero, and we handle it separately in two slides.
Step zero
Discussion prompt
Worked example: a polynomial integrand — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Handle each term separately
Answer:
Worked example
\[ \int \left(6x^2 - 4x + 5\right)dx \]
Handle each term separately
Why: Antidifferentiation is linear: the antiderivative of a sum is the sum of the antiderivatives, and constant factors come along for the ride.
Apply the reversed power rule term by term
Why: Exponent 2 becomes 3 with a divisor of 3; exponent 1 becomes 2 with a divisor of 2; the constant 5 is five times x to the zero, which becomes five x.
\[ 6 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 5x \]
Simplify the coefficients and write one constant
Why: Three separate constants would just add up to one arbitrary constant, so we write a single C for the whole family.
\[ \int \left(6x^2 - 4x + 5\right)dx = 2x^3 - 2x^2 + 5x + C \]
Verify by differentiating the answer
Why: Six x squared minus four x plus five is exactly the integrand, and the constant dies. The answer is confirmed.
\[ \frac{d}{dx}\left[2x^3 - 2x^2 + 5x + C\right] = 6x^2 - 4x + 5 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a polynomial integrand", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Six x squared minus four x plus five is exactly the integrand, and the constant dies. The answer is confirmed.
Concept
Ask which function has a reciprocal as its derivative. The reversed power rule cannot answer, because raising the exponent by one gives zero in the denominator.
\[ \int x^{-1}\,dx \ \neq \ \frac{x^{0}}{0} \qquad \text{(division by zero)} \]
But we already met the function whose derivative is the reciprocal: the natural logarithm. The absolute value lets it cover negative inputs too.
\[ \int \frac{1}{x}\,dx = \ln|x| + C \]
Anomaly
Predict first
A student writes this, and it looks reasonable:
Rewriting the reciprocal as a power and reaching for the usual rule.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Mechanically the rule says exponent negative one becomes zero, and we divide by that new exponent - which is zero.
Recognize the exception and reach for the logarithm instead.
Why: Mechanically the rule says exponent negative one becomes zero, and we divide by that new exponent - which is zero. The expression is meaningless.
Trap
Rewriting the reciprocal as a power and reaching for the usual rule.
\[ \int \frac{1}{x}\,dx = \int x^{-1}\,dx \]
Raise the exponent and divide
Why: Mechanically the rule says exponent negative one becomes zero, and we divide by that new exponent - which is zero. The expression is meaningless.
\[ \frac{x^{-1+1}}{-1+1} = \frac{x^0}{0} \quad \text{(undefined)} \]
Recognize the exception and reach for the logarithm instead.
\[ \int \frac{1}{x}\,dx = \ln|x| + C \]
Confirm by differentiating
Why: The derivative of the natural log of the absolute value of x is one over x on both sides of zero, so this really is the antiderivative.
\[ \frac{d}{dx}\left[\ln|x|\right] = \frac{1}{x} \ \checkmark \]
Concept
Every entry here is a derivative rule you already know, read from right to left. Nothing new is being asserted.
| Integrand | Antiderivative | Because the derivative of |
|---|---|---|
| x to the n, n not negative one | x to the n+1 over n+1 | that quotient is x to the n |
| one over x | natural log of absolute x | that log is one over x |
| e to the x | e to the x | e to the x is itself |
| cosine x | sine x | sine is cosine |
| sine x | negative cosine x | negative cosine is sine |
| secant squared x | tangent x | tangent is secant squared |
\[ \int e^x dx = e^x + C \qquad \int \cos x\,dx = \sin x + C \qquad \int \sin x\,dx = -\cos x + C \]
Watch the minus sign on the sine row. It is the single most miscopied entry in the table.
Comparison
Comparison matrix
From The basic antiderivative table: refill the Antiderivative column from what you know. The rest of the table is as it appeared.
| Integrand | Antiderivative | Because the derivative of |
|---|---|---|
| x to the n, n not negative one | x to the n+1 over n+1 | that quotient is x to the n |
| one over x | natural log of absolute x | that log is one over x |
| e to the x | e to the x | e to the x is itself |
| cosine x | sine x | sine is cosine |
| sine x | negative cosine x | negative cosine is sine |
| secant squared x | tangent x | tangent is secant squared |
Picture it
Animation
Shows: The basic table, read backwards — a rendered Manim animation.
Rendered with Manim.
Takeaway: The exception at n equals minus one is why the logarithm is on the list.
Concept
Sums split and constants factor out, exactly as they do for derivatives.
\[ \int \left[af(x) + bg(x)\right]dx = a\int f(x)\,dx + b\int g(x)\,dx \]
What does not split: products and quotients. There is no product rule for antiderivatives. If you see a product, your move is to rewrite it into a sum first.
Worked example
\[ \int \left(\sqrt{x} + \frac{1}{x^2}\right)dx \]
Rewrite both terms as plain powers of x
Why: The reversed power rule only recognizes exponents. A radical is a one-half power and a reciprocal power moves upstairs with a negative exponent.
\[ = \int \left(x^{1/2} + x^{-2}\right)dx \]
Apply the rule to the first term
Why: One half plus one is three halves, and dividing by three halves means multiplying by two thirds.
\[ \int x^{1/2}dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \]
Apply the rule to the second term
Why: Negative two plus one is negative one, and dividing by negative one flips the sign. The result is a negative reciprocal.
\[ \int x^{-2}dx = \frac{x^{-1}}{-1} = -\frac{1}{x} \]
Assemble the answer
Why: One constant covers the whole family.
\[ \frac{2}{3}x^{3/2} - \frac{1}{x} + C \]
Verify by differentiating
Why: Two thirds times three halves is one, leaving the one-half power, which is the square root; and the derivative of negative x inverse is positive x to the negative two. Both terms match the integrand.
\[ \frac{d}{dx}\left[\frac{2}{3}x^{3/2} - x^{-1}\right] = x^{1/2} + x^{-2} = \sqrt{x} + \frac{1}{x^2} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "rewrite before you integrate", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two thirds times three halves is one, leaving the one-half power, which is the square root; and the derivative of negative x inverse is positive x to the negative two. Both terms match the integrand.
Fill the middle
Fill in the blanks
From Worked example: a trig and exponential mix — finish the line. Write what belongs on the right of the equals sign before you look.
4\int \frac4\ln|x|___dx = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Linearity lets each term be antidifferentiated on its own, with its coefficient parked out front.
Worked example
\[ \int \left(\frac{4}{x} + 3e^{x} - 2\cos x\right)dx \]
Split the integral and pull out the constants
Why: Linearity lets each term be antidifferentiated on its own, with its coefficient parked out front.
Take the first term with the logarithm rule
Why: The reciprocal is the power-rule exception, so this term becomes a natural log with the absolute value.
\[ 4\int \frac{1}{x}dx = 4\ln|x| \]
Take the exponential and cosine terms from the table
Why: The natural exponential is its own antiderivative, and the antiderivative of cosine is sine - no sign flip on this one.
\[ 3\int e^x dx = 3e^x, \qquad -2\int \cos x\,dx = -2\sin x \]
Write the family
Why: Combine the three pieces and attach a single constant.
\[ 4\ln|x| + 3e^{x} - 2\sin x + C \]
Verify by differentiating
Why: Four over x, plus three e to the x, minus two cosine x - term for term this is the integrand we started with.
\[ \frac{d}{dx}\left[4\ln|x| + 3e^x - 2\sin x\right] = \frac{4}{x} + 3e^x - 2\cos x \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a trig and exponential mix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Four over x, plus three e to the x, minus two cosine x - term for term this is the integrand we started with.
Concept
An initial-value problem hands you a rate and one point on the graph, and asks for the single function that fits both.
\[ f'(x) = 6x^2 - 4, \qquad f(1) = 3 \]
The rate gives you the family. The point picks the member. Two facts, two jobs.
Worked example
\[ f'(x) = 6x^2 - 4, \qquad f(1) = 3 \]
Antidifferentiate to get the whole family
Why: Reversed power rule on each term, and the constant stays symbolic because we do not know it yet.
\[ f(x) = 2x^3 - 4x + C \]
Substitute the given point
Why: The condition says the output at input one is three. Plugging in turns the unknown constant into a solvable equation.
\[ f(1) = 2(1)^3 - 4(1) + C = -2 + C \]
Solve for the constant
Why: Setting the expression equal to the required value 3 and adding two to both sides isolates C.
\[ -2 + C = 3 \ \Longrightarrow \ C = 5 \]
State the specific antiderivative
Why: This is the one curve in the family passing through the given point.
\[ f(x) = 2x^3 - 4x + 5 \]
Verify both conditions
Why: The derivative comes back as six x squared minus four, and the value at one is two minus four plus five, which is three. Both requirements hold.
\[ f'(x) = 6x^2 - 4 \ \checkmark \qquad f(1) = 2 - 4 + 5 = 3 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "solving an initial-value problem", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The derivative comes back as six x squared minus four, and the value at one is two minus four plus five, which is three. Both requirements hold.
Pattern
Check
Antidifferentiate first, then use the point. Do it on paper before you click.
\[ f'(x) = 4x - 6, \qquad f(2) = 5 \]
Check your understanding
Which function is f?
Answer: A
Why: Antidifferentiating gives 2x^2 - 6x + C. Substituting x = 2 gives 8 - 12 + C = -4 + C, and setting that equal to 5 gives C = 9. Verify: f(2) = 8 - 12 + 9 = 5, and f'(x) = 4x - 6.
Concept
Differentiating position twice gives velocity and then acceleration. Antidifferentiating walks back up the same ladder.
\[ a(t) \ \xrightarrow{\ \int\ } \ v(t) \ \xrightarrow{\ \int\ } \ s(t) \]
Each step up needs its own initial condition: the starting velocity for the first, the starting position for the second. Two integrations, two constants, two facts.
Estimation
Predict first
A ball is thrown straight up from a 64 foot roof at 48 feet per second. Gravity gives constant acceleration downward. Find the position function and the time it lands.
Commit before you compute: what does Worked example: a ball thrown upward come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the answer at the landing time and at launch
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At four seconds the height is negative two hundred fifty-six plus one hundred ninety-two plus sixty-four, which is zero, so the ball is on the ground; and at time zero the height is the roof height of 64 feet.
Worked example
A ball is thrown straight up from a 64 foot roof at 48 feet per second. Gravity gives constant acceleration downward. Find the position function and the time it lands.
\[ a(t) = -32, \qquad v(0) = 48, \qquad s(0) = 64 \]
Antidifferentiate acceleration to get velocity
Why: The antiderivative of the constant negative thirty-two is negative thirty-two t, plus a constant that represents the launch velocity.
\[ v(t) = -32t + C_1 \]
Use the starting velocity
Why: At time zero the velocity term vanishes, so the constant is exactly the given launch speed of 48.
\[ v(0) = C_1 = 48 \ \Longrightarrow \ v(t) = -32t + 48 \]
Antidifferentiate velocity to get position
Why: Reversed power rule on each term: negative thirty-two t becomes negative sixteen t squared, and forty-eight becomes forty-eight t.
\[ s(t) = -16t^2 + 48t + C_2 \]
Use the starting height
Why: At time zero both variable terms vanish, so the second constant is the roof height of 64 feet.
\[ s(t) = -16t^2 + 48t + 64 \]
Set the height to zero to find the landing time
Why: Dividing by negative sixteen gives a clean quadratic that factors; only the positive root is physically meaningful.
\[ -16t^2 + 48t + 64 = 0 \ \Longrightarrow \ t^2 - 3t - 4 = 0 \ \Longrightarrow \ (t-4)(t+1) = 0 \]
Verify the answer at the landing time and at launch
Why: At four seconds the height is negative two hundred fifty-six plus one hundred ninety-two plus sixty-four, which is zero, so the ball is on the ground; and at time zero the height is the roof height of 64 feet. Both check.
\[ s(4) = -256 + 192 + 64 = 0 \ \checkmark \qquad s(0) = 64 \ \checkmark \qquad t = 4\ \text{seconds} \]
Picture it
Animation
Shows: Each line of the worked example "a ball thrown upward", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At four seconds the height is negative two hundred fifty-six plus one hundred ninety-two plus sixty-four, which is zero, so the ball is on the ground; and at time zero the height is the roof height of 64 feet. Both check.
Section
Section 2
Concept
Geometry hands you the area of a rectangle, a triangle, a circle. It has nothing to say about the region under a curved graph.
That region is not a decoration. It is total distance from a velocity graph, total cost from a marginal-cost graph, total charge from a current graph. Whenever a rate is graphed, the area under it is the accumulated amount.
So we need a way to measure area under a curve, and the only shapes we can measure are the straight ones.
Intuition
Figure (svg): A curve rising from the origin with four rectangles beneath it, each rectangle's height set by the curve's value at the left edge of that strip
Cut the interval into thin vertical strips. Over one thin strip the curve barely changes, so a rectangle is almost the right shape.
Add up the rectangles and you get an estimate. Use thinner strips and the estimate gets better, because the curve has less room to wander across each one.
The area is what those estimates settle down to as the strips get infinitely thin. That limit is the whole idea; the rest is bookkeeping.
Concept
Chop the interval into a chosen number of equal-width strips. The width is the length of the interval divided by the number of strips.
\[ \Delta x = \frac{b-a}{n}, \qquad x_i = a + i\,\Delta x \quad (i = 0, 1, 2, \ldots, n) \]
Notice the count: there are n strips but n plus one grid points, because the endpoints are shared. Miscounting these is the number-one arithmetic error in this topic.
The only thing left to decide is where inside each strip you sample the height.
Concept
Sample the height at the left edge of each strip, and add up width times height.
\[ L_n = \sum_{i=0}^{n-1} f(x_i)\,\Delta x = \left[f(x_0) + f(x_1) + \cdots + f(x_{n-1})\right]\Delta x \]
The left sum uses the first n grid points and never touches the right endpoint of the interval.
Picture it
Animation
Shows: The left Riemann sum — a rendered Manim animation.
Rendered with Manim.
Takeaway: Sample at the left edge of every strip. On a rising curve that under-estimates.
Picture it
Animation
Shows: The gap between left and right Riemann sums shrinking to zero.
The gap is squeezed out by the limit.
Takeaway: The left and right sums bracket the true value, and the distance between them shrinks to nothing — so in the limit the choice of sample point is irrelevant.
Step zero
Discussion prompt
Worked example: a left sum with four rectangles — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the strip width
Answer:
Worked example
Estimate the area under this curve using four equal strips and left endpoints.
\[ f(x) = x^2 \ \text{on}\ [0, 2], \quad n = 4 \]
Find the strip width
Why: Interval length two divided by four strips gives width one half.
\[ \Delta x = \frac{2-0}{4} = 0.5 \]
List the grid points and the left endpoints
Why: Five grid points bound four strips; the left endpoints are the first four of them.
\[ 0,\ 0.5,\ 1,\ 1.5,\ 2 \qquad \text{left endpoints: } 0,\ 0.5,\ 1,\ 1.5 \]
Evaluate the function at each left endpoint
Why: Squaring each left endpoint gives the height of that rectangle.
| strip | left endpoint | height | area (height times 0.5) |
|---|---|---|---|
| 1 | 0 | 0 | 0 |
| 2 | 0.5 | 0.25 | 0.125 |
| 3 | 1 | 1 | 0.5 |
| 4 | 1.5 | 2.25 | 1.125 |
Add the four rectangle areas
Why: Factor the common width out of the sum of heights, then multiply once.
\[ L_4 = (0 + 0.25 + 1 + 2.25)(0.5) = (3.5)(0.5) = 1.75 \]
Verify the estimate is plausible and on the low side
Why: The exact area turns out to be eight thirds, about 2.667. Since the curve rises, left edges are the shortest point of each strip, so an underestimate of 1.75 is exactly what we should see.
\[ L_4 = 1.75 \ < \ \frac{8}{3} \approx 2.667 \ \checkmark \]
Concept
Same strips, same widths, but now sample the height at the right edge of each strip.
\[ R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x = \left[f(x_1) + f(x_2) + \cdots + f(x_n)\right]\Delta x \]
The right sum skips the left endpoint of the interval and does use the far right one. Left and right sums share every interior height and differ in exactly two terms.
Picture it
Animation
Shows: The right Riemann sum — a rendered Manim animation.
Rendered with Manim.
Takeaway: Sample at the right edge. On a rising curve that over-estimates.
Pattern
Predict first
The table runs: 1 | 0.5 | 0.25 | 0.125 · 2 | 1 | 1 | 0.5 · 3 | 1.5 | 2.25 | 1.125
In Worked example: the right sum on the same curve, given the rows so far: what is the next one — the row where strip is 4?
Correct: 4 | 2 | 4 | 2
| strip | right endpoint | height | area (height times 0.5) |
|---|---|---|---|
| 1 | 0.5 | 0.25 | 0.125 |
| 2 | 1 | 1 | 0.5 |
| 3 | 1.5 | 2.25 | 1.125 |
| 4 | 2 | 4 | 2 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Drop the leftmost grid point and pick up the rightmost one; everything in between is shared with the left sum.
Worked example
\[ f(x) = x^2 \ \text{on}\ [0, 2], \quad n = 4, \quad \Delta x = 0.5 \]
List the right endpoints
Why: Drop the leftmost grid point and pick up the rightmost one; everything in between is shared with the left sum.
\[ \text{right endpoints: } 0.5,\ 1,\ 1.5,\ 2 \]
Evaluate the heights
Why: Square each right endpoint. Only the first and last entries differ from the left-sum table.
| strip | right endpoint | height | area (height times 0.5) |
|---|---|---|---|
| 1 | 0.5 | 0.25 | 0.125 |
| 2 | 1 | 1 | 0.5 |
| 3 | 1.5 | 2.25 | 1.125 |
| 4 | 2 | 4 | 2 |
Add them up
Why: Sum the heights, then multiply by the common width one half.
\[ R_4 = (0.25 + 1 + 2.25 + 4)(0.5) = (7.5)(0.5) = 3.75 \]
Check the difference against the shortcut
Why: The two sums differ only in the swapped end heights, so the gap should be the last height minus the first, times the width: four minus zero, times one half, equals two. And 3.75 minus 1.75 is indeed 2.
\[ R_4 - L_4 = \left[f(2) - f(0)\right]\Delta x = (4-0)(0.5) = 2 \ \checkmark \]
Concept
Sample the height at the center of each strip instead of an edge.
\[ M_n = \sum_{i=1}^{n} f\!\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x \]
The midpoint sum usually beats both edge sums, because on each strip the rectangle cuts off about as much as it adds. It is the cheapest big accuracy upgrade in the whole topic.
Picture it
Animation
Shows: The midpoint rule does better for free — a rendered Manim animation.
Rendered with Manim.
Takeaway: Over- and under-shoot cancel within each strip, so the error falls faster.
Picture it
Animation
Shows: Riemann rectangles refining from four to forty-eight under a curve.
The region never changed. The estimate did.
Takeaway: Nothing about the region changes as the rectangles refine — only the crudeness of the estimate. The limit is the area itself.
Pattern
Predict first
The table runs: left sum | 1.75 | 0.917 low · midpoint sum | 2.625 | 0.042 low
In Worked example: the midpoint sum, given the rows so far: what is the next one — the row where estimate is right sum?
Correct: right sum | 3.75 | 1.083 high
| estimate | value | error from 8/3 |
|---|---|---|
| left sum | 1.75 | 0.917 low |
| midpoint sum | 2.625 | 0.042 low |
| right sum | 3.75 | 1.083 high |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Average the two edges of each strip; equivalently, start a quarter-unit in and step by the strip width.
Worked example
\[ f(x) = x^2 \ \text{on}\ [0, 2], \quad n = 4, \quad \Delta x = 0.5 \]
Find the midpoint of each strip
Why: Average the two edges of each strip; equivalently, start a quarter-unit in and step by the strip width.
\[ \text{midpoints: } 0.25,\ 0.75,\ 1.25,\ 1.75 \]
Square each midpoint to get the heights
Why: These heights are neither the smallest nor the largest value on their strips, which is why the estimate lands so much closer.
| strip | midpoint | height |
|---|---|---|
| 1 | 0.25 | 0.0625 |
| 2 | 0.75 | 0.5625 |
| 3 | 1.25 | 1.5625 |
| 4 | 1.75 | 3.0625 |
Add and multiply by the width
Why: The four heights total 5.25; multiplying by one half gives the estimate.
\[ M_4 = (5.25)(0.5) = 2.625 \]
Verify against the other two estimates and the exact value
Why: The exact area is eight thirds, about 2.667. The midpoint estimate misses by about 0.042 while the left and right sums miss by about 0.917 and 1.083. The midpoint sum is the clear winner, exactly as predicted.
| estimate | value | error from 8/3 |
|---|---|---|
| left sum | 1.75 | 0.917 low |
| midpoint sum | 2.625 | 0.042 low |
| right sum | 3.75 | 1.083 high |
Picture it
Animation
Shows: Each line of the worked example "the midpoint sum", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The exact area is eight thirds, about 2.667. The midpoint estimate misses by about 0.042 while the left and right sums miss by about 0.917 and 1.083. The midpoint sum is the clear winner, exactly as predicted.
Concept
For a function that only rises across the interval, the left edge is the shortest point of every strip and the right edge is the tallest.
| Function on the interval | Left sum | Right sum |
|---|---|---|
| increasing | underestimate | overestimate |
| decreasing | overestimate | underestimate |
There is no rule to memorize here beyond one sentence: the sum that samples the taller side overestimates. Sketch the curve, look at which edge is higher, and read the answer off the picture.
If the function rises and then falls, no rule applies at all. Split the interval or say nothing.
Trade off
Comparison matrix
From Which sums over- and under-estimate: every row here is a choice with a cost. Fill the Left sum column, then say which row you would actually pick and what you give up for it.
| Function on the interval | Left sum | Right sum |
|---|---|---|
| increasing | underestimate | overestimate |
| decreasing | overestimate | underestimate |
Trap
The estimate: a left sum with four strips, and the reflex that left sums come in low.
\[ f(x) = 4 - x^2 \ \text{on}\ [0,2], \quad n = 4, \quad \Delta x = 0.5 \]
Compute the left sum and declare it an underestimate
Why: The heights at 0, 0.5, 1, and 1.5 are 4, 3.75, 3, and 1.75, totalling 12.5, so the left sum is 6.25 - but calling it low is the mistake.
\[ L_4 = (4 + 3.75 + 3 + 1.75)(0.5) = 6.25 \quad \text{claimed: an underestimate} \]
Compare with the true area
Why: The exact area is sixteen thirds, about 5.333. The left sum is nearly a whole unit too big, not too small.
\[ 6.25 \ > \ \frac{16}{3} \approx 5.333 \]
Look at the direction of the curve first. This one is falling on the whole interval.
\[ f(x) = 4 - x^2 \ \text{on}\ [0,2], \quad f'(x) = -2x < 0 \ \text{for}\ x > 0 \]
Because the function decreases, the left edge is the tallest point of each strip
Why: Sampling the tallest point means every rectangle pokes out above the curve, so the left sum must come in high.
\[ L_4 = 6.25 \ \text{is an overestimate} \ \checkmark \]
Confirm the right sum goes the other way
Why: Right edges are the shortest points here: heights 3.75, 3, 1.75, and 0 total 8.5, giving 4.25, which is below the true 5.333 as expected.
\[ R_4 = (3.75 + 3 + 1.75 + 0)(0.5) = 4.25 \ < \ \frac{16}{3} \ \checkmark \]
Translation
\( f(x) = 4 - x^2 \ \text{on}\ [0,2], \quad f'(x) = -2x < 0 \ \text{for}\ x > 0 \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Writing out twenty rectangles is impossible and writing out n of them is worse. Sigma notation is just an instruction to loop.
\[ \sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \cdots + a_n \]
index of summation — The counter under the sigma. It starts at the lower value, steps up by one, and stops at the value on top. It is a placeholder and never appears in the final answer.
Three standard formulas turn a sum into a closed expression in n, which is what makes the limit computable.
\[ \sum_{i=1}^{n} 1 = n, \qquad \sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
Picture it
Animation
Shows: Sigma notation, read once — a rendered Manim animation.
Rendered with Manim.
Takeaway: The symbol is an instruction to add, nothing more.
Ranking
Put in order
Put the moves of Worked example: a sum two ways into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Substitute the index values one through five into the expression and list what you get.
Worked example
\[ \sum_{i=1}^{5} (2i - 1) \]
Expand the sum term by term
Why: Substitute the index values one through five into the expression and list what you get.
\[ = 1 + 3 + 5 + 7 + 9 = 25 \]
Now redo it with the formulas
Why: Sums split across addition and constants pull out front, exactly like integrals. The second piece is a constant summed five times.
\[ \sum_{i=1}^{5}(2i-1) = 2\sum_{i=1}^{5} i - \sum_{i=1}^{5} 1 \]
Apply the closed forms
Why: Five times six over two is fifteen, doubled is thirty; the constant sum is simply five.
\[ = 2\left(\frac{5 \cdot 6}{2}\right) - 5 = 30 - 5 = 25 \]
Verify the two routes agree
Why: Direct expansion gave 25 and the formula route gave 25. The formulas are trustworthy, which matters because for a general n we cannot expand at all.
\[ 25 = 25 \ \checkmark \]
Concept
Now let the number of strips grow without bound. If the estimates converge to a single number no matter where inside each strip you sampled, that number is the area.
\[ \int_{a}^{b} f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^{*})\,\Delta x \]
Read the notation as a leftover from the sum: the integral sign is a stretched S for sum, the integrand is the height, and the differential is the vanishing width.
Every continuous function on a closed interval is integrable, so for everything in this course the limit exists.
Picture it
Animation
Shows: Three properties worth knowing by heart — a rendered Manim animation.
Rendered with Manim.
Takeaway: The last one is what lets you split at a crossing.
Picture it
Animation
Shows: The definite integral written as a limit of Riemann sums.
A limit of sums — not an antiderivative.
Takeaway: The definite integral is defined as a limit of sums. Its connection to antiderivatives is a theorem to be proved, not part of the definition.
Hypothesis
Predict first
Worked example: the limit of the right sums is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Write the width and the right endpoints in terms of n
Why: With n equal strips on an interval of length two, each is two over n wide, and the i-th right endpoint is that width taken i times.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Compute the exact area under the parabola from zero to two, using only the definition.
\[ \int_{0}^{2} x^2\,dx = \lim_{n\to\infty} R_n \]
Write the width and the right endpoints in terms of n
Why: With n equal strips on an interval of length two, each is two over n wide, and the i-th right endpoint is that width taken i times.
\[ \Delta x = \frac{2}{n}, \qquad x_i = \frac{2i}{n} \]
Build the sum
Why: Square the endpoint for the height, multiply by the width, and pull everything that does not depend on the index out front.
\[ R_n = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2\frac{2}{n} = \frac{8}{n^3}\sum_{i=1}^{n} i^2 \]
Apply the sum-of-squares formula
Why: This is the step that turns an n-term sum into a single algebraic expression in n.
\[ = \frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{4(n+1)(2n+1)}{3n^2} \]
Take the limit as the strip count grows without bound
Why: Expanding the numerator gives a quadratic over a quadratic, so the limit is the ratio of the leading coefficients: eight over three.
\[ \lim_{n\to\infty}\frac{4(2n^2+3n+1)}{3n^2} = \frac{8}{3} \]
Verify the formula reproduces the earlier estimate
Why: Putting n equal to four into the closed form gives four times five times nine over forty-eight, which is 3.75 - exactly the right sum we computed by hand. The algebra is sound.
\[ \frac{4(4+1)(8+1)}{3(16)} = \frac{180}{48} = 3.75 = R_4 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the limit of the right sums", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Putting n equal to four into the closed form gives four times five times nine over forty-eight, which is 3.75 - exactly the right sum we computed by hand. The algebra is sound.
Pattern
Edge cases
Discussion prompt
Pattern: computing a Riemann sum works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Elimination
Eliminate the wrong options
What is the left Riemann sum with four strips?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The width is 0.5 and the left endpoints are 0, 0.5, 1, 1.5, giving heights 4, 3.75, 3, 1.75. Those heights total 12.5, and 12.5 times 0.5 is 6.25.
Check
Four equal strips, left endpoints. Build the table before you choose.
\[ f(x) = 4 - x^2 \ \text{on}\ [0, 2], \quad n = 4 \]
Check your understanding
What is the left Riemann sum with four strips?
Answer: A
Why: The width is 0.5 and the left endpoints are 0, 0.5, 1, 1.5, giving heights 4, 3.75, 3, 1.75. Those heights total 12.5, and 12.5 times 0.5 is 6.25.
Prediction
Predict first
For this increasing function, how do the left and right sums compare with the true area?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Left sum underestimates, right sum overestimates
Why: On a rising curve the left edge of each strip is its lowest point, so those rectangles sit under the curve; the right edge is the highest point, so those rectangles poke out above it.
Check
No arithmetic needed. Picture one strip and ask which edge is taller.
\[ g \ \text{is positive and increasing on the whole interval} \ [a,b] \]
Check your understanding
For this increasing function, how do the left and right sums compare with the true area?
Answer: A
Why: On a rising curve the left edge of each strip is its lowest point, so those rectangles sit under the curve; the right edge is the highest point, so those rectangles poke out above it.
Section
Section 3
Concept
Every piece of the symbol came from the sum it replaced.
\[ \int_{a}^{b} f(x)\,dx \]
| Piece | Name | What it was in the sum |
|---|---|---|
| the elongated S | integral sign | the sigma that added the rectangles |
| lower and upper numbers | limits of integration | the left and right ends of the interval |
| the function | integrand | the height of a rectangle |
| the differential | variable of integration | the vanishing width of a strip |
The variable is a dummy. Renaming it changes nothing, because the answer is a number, not a function.
\[ \int_{0}^{2} x^2\,dx = \int_{0}^{2} t^2\,dt = \frac{8}{3} \]
Comparison
Comparison matrix
From Reading the notation: refill the Name column from what you know. The rest of the table is as it appeared.
| Piece | Name | What it was in the sum |
|---|---|---|
| the elongated S | integral sign | the sigma that added the rectangles |
| lower and upper numbers | limits of integration | the left and right ends of the interval |
| the function | integrand | the height of a rectangle |
| the differential | variable of integration | the vanishing width of a strip |
Concept
Figure (svg): A rising straight line crossing the horizontal axis, with the region below the axis shaded on the left and the region above the axis shaded on the right
In a Riemann sum the width is always positive, but the height carries the sign of the function.
So a strip where the curve sits below the axis contributes a negative amount. The integral adds up area above the axis and subtracts area below it.
net area — Area above the axis minus area below it. This is what a definite integral computes. It can be positive, negative, or zero.
This is a feature, not a bug. Integrating a velocity that goes negative gives displacement, which is exactly what you want when the object turns around.
Picture it
Animation
Shows: One full period of sine shaded above and below the axis.
Above counts positive, below counts negative.
Takeaway: Area above the axis counts positive and below counts negative. Over a full period of sine they cancel exactly, giving zero.
Estimation
Predict first
The integrand is a straight line, so we can get the exact answer with triangles and no calculus at all.
Commit before you compute: what does Worked example: signed area by geometry come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with an antiderivative
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. An antiderivative of two x minus four is x squared minus four x.
Worked example
The integrand is a straight line, so we can get the exact answer with triangles and no calculus at all.
\[ \int_{0}^{3} (2x - 4)\,dx \]
Find where the line crosses the axis
Why: The sign of the integrand changes there, so that point splits the region into a below-axis piece and an above-axis piece.
\[ 2x - 4 = 0 \ \Longrightarrow \ x = 2 \]
Measure the triangle from zero to two
Why: It has base two and height four, since the line is at negative four when x is zero. That is area four, but it lies below the axis, so it counts negative.
\[ -\tfrac{1}{2}(2)(4) = -4 \]
Measure the triangle from two to three
Why: Base one and height two, since the line reaches two when x is three. This piece is above the axis, so it counts positive.
\[ +\tfrac{1}{2}(1)(2) = +1 \]
Add the signed pieces
Why: Net area is the sum of the signed contributions, and here the negative piece dominates.
\[ \int_{0}^{3}(2x-4)\,dx = -4 + 1 = -3 \]
Verify with an antiderivative
Why: An antiderivative of two x minus four is x squared minus four x. At three that is nine minus twelve, or negative three; at zero it is zero. The difference is negative three, matching the geometry exactly.
\[ \left[x^2 - 4x\right]_{0}^{3} = (9 - 12) - (0) = -3 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "signed area by geometry", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: An antiderivative of two x minus four is x squared minus four x. At three that is nine minus twelve, or negative three; at zero it is zero. The difference is negative three, matching the geometry exactly.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The question: how much total area lies between this line and the axis from zero to three?
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Area is a size and can never be negative, so an answer of negative three is not an area at all.
Split at the crossing point and make each piece positive before adding.
Why: Area is a size and can never be negative, so an answer of negative three is not an area at all. The integral answered a different question than the one asked.
Trap
The question: how much total area lies between this line and the axis from zero to three?
\[ \int_{0}^{3}(2x-4)\,dx = -3 \]
Report the integral as the total area
Why: Area is a size and can never be negative, so an answer of negative three is not an area at all. The integral answered a different question than the one asked.
\[ \text{total area} = -3 \quad \text{(impossible)} \]
Split at the crossing point and make each piece positive before adding.
\[ 2x - 4 = 0 \ \text{at}\ x = 2 \]
Integrate the absolute value, piece by piece
Why: On the first piece the function is negative, so flipping its sign makes that contribution positive; on the second piece it is already positive.
\[ \int_{0}^{3}\left|2x-4\right|dx = -\!\int_{0}^{2}(2x-4)\,dx + \int_{2}^{3}(2x-4)\,dx \]
Evaluate and add
Why: The first piece has size four and the second has size one, so the total area between the line and the axis is five - while the net area stays negative three. Two different questions, two different answers.
\[ = 4 + 1 = 5 \ \checkmark \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
On the first piece the function is negative, so flipping its sign makes that contribution positive; on the second piece it is already positive.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Area is a size and can never be negative, so an answer of negative three is not an area at all. The integral answered a different question than the one asked.
Concept
Sums split apart and constants slide out front, inherited straight from the sums the integral is built out of.
\[ \int_{a}^{b}\left[f(x) \pm g(x)\right]dx = \int_{a}^{b}f(x)\,dx \pm \int_{a}^{b}g(x)\,dx \]
\[ \int_{a}^{b} k\,f(x)\,dx = k\int_{a}^{b} f(x)\,dx \]
One more worth memorizing: the integral of a constant is that constant times the width of the interval, because the region is a rectangle.
\[ \int_{a}^{b} k\,dx = k(b-a) \]
Concept
Running the interval backwards makes every strip width negative, so every term in the sum changes sign.
\[ \int_{b}^{a} f(x)\,dx = -\int_{a}^{b} f(x)\,dx \]
And an interval of zero width has no strips at all, so it accumulates nothing - regardless of how big the function is there.
\[ \int_{a}^{a} f(x)\,dx = 0 \]
Concept
Accumulating from the start to a middle point, then from that point onward, is the same as accumulating the whole way.
\[ \int_{a}^{c} f(x)\,dx = \int_{a}^{b} f(x)\,dx + \int_{b}^{c} f(x)\,dx \]
Combined with the sign-flip rule, this holds even when the middle point sits outside the interval, which is what makes it so useful for solving for a missing piece.
Step zero
Discussion prompt
Worked example: working with given integrals — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Join the adjacent intervals for the first answer
Answer:
Worked example
You are told two values and asked for three more. No formula for the function is given, and none is needed.
\[ \int_{1}^{4} f(x)\,dx = 10, \qquad \int_{4}^{7} f(x)\,dx = -3 \]
Join the adjacent intervals for the first answer
Why: The two given intervals meet at four and cover one through seven with no gap or overlap, so their values simply add.
\[ \int_{1}^{7} f(x)\,dx = 10 + (-3) = 7 \]
Flip the limits for the second answer
Why: Reversing the direction of travel negates the value; nothing else changes.
\[ \int_{7}^{4} f(x)\,dx = -(-3) = 3 \]
Use linearity for the third answer
Why: Split the sum, pull the two out of the first piece, and use the constant rule on the second piece: the interval from one to four has width three.
\[ \int_{1}^{4}\left[2f(x) + 5\right]dx = 2(10) + 5(4-1) = 20 + 15 = 35 \]
Verify the joining result runs backwards too
Why: If the whole integral from one to seven is seven and the first stretch is ten, then the remaining stretch must be seven minus ten, which is negative three - exactly the value we were given. The answers are internally consistent.
\[ \int_{4}^{7} f = \int_{1}^{7} f - \int_{1}^{4} f = 7 - 10 = -3 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "working with given integrals", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: If the whole integral from one to seven is seven and the first stretch is ten, then the remaining stretch must be seven minus ten, which is negative three - exactly the value we were given. The answers are internally consistent.
Check
Split the interval first, then handle the constant factor.
\[ \int_{0}^{5} f(x)\,dx = 12, \qquad \int_{0}^{2} f(x)\,dx = 5 \]
Check your understanding
What is the integral of 3f(x) from 2 to 5?
Answer: A
Why: Additivity gives the integral of f from 2 to 5 as 12 minus 5, which is 7. Pulling the constant 3 out front multiplies that by 3, giving 21.
Section
Section 4
Concept
Freeze the left limit and let the right limit move. What you get is no longer a number - it is a function of where you stopped.
\[ A(x) = \int_{a}^{x} f(t)\,dt \]
The variable inside is renamed to keep it distinct from the limit outside. Using the same letter for both is legal but genuinely confusing, so do not.
accumulation function — The running total of a rate from a fixed starting point up to a moving endpoint. Its input is the stopping place; its output is how much has piled up by then.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of antiderivative, indefinite integral, index of summation, net area, accumulation function as Antiderivatives, Riemann Sums, and the FTC uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Your speedometer is the rate. Your odometer is the running total of that rate since the trip began.
Now ask: how fast is the odometer climbing right now? Obviously, at whatever the speedometer currently reads. Nothing else could be the answer.
That sentence is the entire content of the Fundamental Theorem, Part 1. The rate of growth of a running total is the thing being totalled.
It is startling only because one side is built from areas and the other from slopes, and no one expected those two to be the same subject.
Concept
If the integrand is continuous, then differentiating the accumulation function simply hands the integrand back.
\[ \frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x) \]
Two consequences worth saying out loud. First, every continuous function has an antiderivative, even ones with no formula. Second, differentiation and integration undo each other.
The lower limit is irrelevant to the derivative. Changing it shifts the accumulation function by a constant, and constants have zero derivative - the plus C, showing up again from a new direction.
Explain it
Discussion prompt
Explain Fundamental Theorem, Part 1 to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
If the integrand is continuous, then differentiating the accumulation function simply hands the integrand back.
Picture it
Animation
Shows: The first part of the Fundamental Theorem written out.
The two operations undo each other.
Takeaway: The area accumulated up to x grows at exactly the height at x. Accumulation and differentiation are inverse operations.
Worked example
\[ g(x) = \int_{1}^{x} \left(t^2 + 1\right)dt, \qquad \text{find } g'(x) \]
Check the shape matches the theorem
Why: Constant on the bottom, a bare x on top, continuous integrand. All three hypotheses are met, so Part 1 applies directly.
Replace the dummy variable with the upper limit
Why: That is the whole operation: the derivative of the running total is the integrand evaluated at the stopping point.
\[ g'(x) = x^2 + 1 \]
Verify by evaluating the integral the long way first
Why: Antidifferentiating gives t cubed over three plus t; evaluating from one to x gives x cubed over three plus x minus four thirds. Differentiating that gives x squared plus one, and the constant four thirds dies - matching the shortcut exactly.
\[ g(x) = \frac{x^3}{3} + x - \frac{4}{3} \ \Longrightarrow \ g'(x) = x^2 + 1 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Part 1, straight up", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Constant on the bottom, a bare x on top, continuous integrand. All three hypotheses are met, so Part 1 applies directly.
Concept
If the top of the integral is not a bare variable but something built out of it, the accumulation function is a composition - and compositions need the chain rule.
\[ \frac{d}{dx}\int_{a}^{u(x)} f(t)\,dt = f\!\left(u(x)\right)\cdot u'(x) \]
Think of it as the outer accumulation machine composed with the inner formula for where you stopped. Substitute into the integrand, then multiply by the derivative of the upper limit.
A variable on the bottom is handled by flipping the limits first, which costs a minus sign out front.
Analogy
Discussion prompt
Explain When the upper limit is a function by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
If the top of the integral is not a bare variable but something built out of it, the accumulation function is a composition - and compositions need the chain rule.
Fill the middle
Fill in the blanks
From Worked example: Part 1 with a chain — finish the line. Write what belongs on the right of the equals sign before you look.
F(x) = \int_{0}^{x^2} \sin t\,dt, \qquad \text{find } F'(x)
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The top is x squared, whose derivative is two x.
Worked example
\[ F(x) = \int_{0}^{x^2} \sin t\,dt, \qquad \text{find } F'(x) \]
Name the upper limit and its derivative
Why: The top is x squared, whose derivative is two x. This is the factor the plain version of Part 1 does not include.
\[ u(x) = x^2, \qquad u'(x) = 2x \]
Substitute the upper limit into the integrand
Why: The integrand is sine of the dummy variable, so it becomes sine of whatever the stopping point is.
\[ f(u(x)) = \sin\!\left(x^2\right) \]
Multiply by the derivative of the upper limit
Why: The chain rule factor accounts for how fast the stopping point itself is moving as x changes.
\[ F'(x) = 2x\sin\!\left(x^2\right) \]
Verify by evaluating the integral explicitly
Why: The antiderivative of sine is negative cosine, so the integral equals one minus cosine of x squared. Differentiating that with the chain rule gives sine of x squared times two x - identical to the shortcut answer.
\[ F(x) = \left[-\cos t\right]_{0}^{x^2} = 1 - \cos\!\left(x^2\right) \ \Longrightarrow \ F'(x) = 2x\sin\!\left(x^2\right) \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Part 1 with a chain", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The antiderivative of sine is negative cosine, so the integral equals one minus cosine of x squared. Differentiating that with the chain rule gives sine of x squared times two x - identical to the shortcut answer.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The problem, with the upper limit built out of the variable.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The pattern from the simple version of Part 1 is applied on autopilot: swap the dummy variable for the upper limit and call it done.
Treat it as a composition and bring the chain rule along.
Why: The pattern from the simple version of Part 1 is applied on autopilot: swap the dummy variable for the upper limit and call it done.
Trap
The problem, with the upper limit built out of the variable.
\[ G(x) = \int_{1}^{x^3} e^{t}\,dt \]
Substitute the top into the integrand and stop
Why: The pattern from the simple version of Part 1 is applied on autopilot: swap the dummy variable for the upper limit and call it done.
\[ G'(x) = e^{x^3} \quad \text{(incomplete)} \]
Test it against the explicit answer
Why: The integral really equals e to the x cubed minus e, whose derivative carries an extra factor of three x squared. The autopilot answer is off by that whole factor.
\[ G(x) = e^{x^3} - e \ \Longrightarrow \ G'(x) \neq e^{x^3} \]
Treat it as a composition and bring the chain rule along.
\[ G(x) = \int_{1}^{x^3} e^{t}\,dt, \qquad u(x) = x^3 \]
Substitute the top, then multiply by its derivative
Why: The derivative of the upper limit is three x squared, and that factor is exactly what the chain rule contributes.
\[ G'(x) = e^{x^3}\cdot 3x^2 \]
Confirm against the explicit evaluation
Why: Differentiating e to the x cubed minus e by the ordinary chain rule gives three x squared times e to the x cubed. The two routes agree.
\[ \frac{d}{dx}\left[e^{x^3} - e\right] = 3x^2 e^{x^3} \ \checkmark \]
Notation
Annotate
From Trap: dropping the chain factor in Part 1 — read this one piece at a time. What is each part doing?
On: \( \frac{d}{dx}\left[e^{x^3} - e\right] = 3x^2 e^{x^3} \ \checkmark \)
Fill the middle
Fill in the blanks
From Worked example: the variable on the bottom — finish the line. Write what belongs on the right of the equals sign before you look.
h'(x) = -\frac{1}{1+x^2}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Part 1 is only stated for a moving upper limit, and swapping the limits of a definite integral costs exactly one minus sign.
Worked example
\[ h(x) = \int_{x}^{5} \frac{1}{1+t^2}\,dt, \qquad \text{find } h'(x) \]
Flip the limits to put the variable on top
Why: Part 1 is only stated for a moving upper limit, and swapping the limits of a definite integral costs exactly one minus sign.
\[ h(x) = -\int_{5}^{x} \frac{1}{1+t^2}\,dt \]
Apply Part 1 to the flipped integral
Why: Now the top is a bare x, so the derivative is just the integrand at x - and the minus sign out front rides along.
\[ h'(x) = -\frac{1}{1+x^2} \]
Sanity-check the sign
Why: The integrand is always positive, so pushing the lower limit to the right eats away at the accumulated area. A shrinking total must have a negative derivative, which is what we got.
Verify with the explicit antiderivative
Why: The antiderivative of one over one plus t squared is arctangent, so the integral is arctangent of five minus arctangent of x. Differentiating gives negative one over one plus x squared, matching.
\[ h(x) = \arctan 5 - \arctan x \ \Longrightarrow \ h'(x) = -\frac{1}{1+x^2} \ \checkmark \]
Pattern
Picture it
Animation
Shows: Differentiating an accumulation with a chain — a rendered Manim animation.
Rendered with Manim.
Takeaway: The chain rule follows the accumulation function upstairs.
Concept
Part 1 says differentiation undoes integration. Part 2 turns that around and gives us a way to actually compute a definite integral without any limits of sums.
\[ \int_{a}^{b} f(x)\,dx = F(b) - F(a) \qquad \text{where } F' = f \]
Any antiderivative works. Add a constant to it and the same constant appears in both terms and cancels, which is why no plus C ever appears on a definite integral.
\[ \left[F(x) + C\right]_{a}^{b} = \left(F(b)+C\right) - \left(F(a)+C\right) = F(b)-F(a) \]
Counterexample
Discussion prompt
Part 1 says differentiation undoes integration. Part 2 turns that around and gives us a way to actually compute a definite integral without any limits of sums.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Any antiderivative works. Add a constant to it and the same constant appears in both terms and cancels, which is why no plus C ever appears on a definite integral.
Picture it
Animation
Shows: The evaluation form of the Fundamental Theorem.
This is why calculus is a subject and not a chore.
Takeaway: A limit of infinitely many sums is computed by evaluating one antiderivative at two points and subtracting. That shortcut is the whole payoff.
Estimation
Predict first
This is the same area we ground out with sigma notation and a limit. Watch how long it takes now.
Commit before you compute: what does Worked example: the parabola, finally the easy way come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the Riemann work from Section 2
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The limit of the right sums gave eight thirds, and the midpoint sum with only four rectangles gave 2.625 against the true 2.667.
Worked example
This is the same area we ground out with sigma notation and a limit. Watch how long it takes now.
\[ \int_{0}^{2} x^2\,dx \]
Find any antiderivative of the integrand
Why: The reversed power rule gives x cubed over three. No constant is needed because it would cancel.
\[ F(x) = \frac{x^3}{3} \]
Evaluate at the top limit, then subtract the value at the bottom
Why: The evaluation bar is shorthand for exactly this subtraction, always top minus bottom.
\[ \left[\frac{x^3}{3}\right]_{0}^{2} = \frac{8}{3} - \frac{0}{3} = \frac{8}{3} \]
Verify against the Riemann work from Section 2
Why: The limit of the right sums gave eight thirds, and the midpoint sum with only four rectangles gave 2.625 against the true 2.667. Three independent routes, one answer.
\[ \lim_{n\to\infty} R_n = \frac{8}{3} \approx 2.667, \qquad M_4 = 2.625 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the parabola, finally the easy way", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The limit of the right sums gave eight thirds, and the midpoint sum with only four rectangles gave 2.625 against the true 2.667. Three independent routes, one answer.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The integrand is positive on the whole interval, so the answer has to be positive. Watch what happens anyway.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Reading the bar as bottom minus top - usually because the lower limit was written first and substituted first - reverses the whole subtraction.
Always top minus bottom, in that order, every time.
Why: Reading the bar as bottom minus top - usually because the lower limit was written first and substituted first - reverses the whole subtraction.
Trap
The integrand is positive on the whole interval, so the answer has to be positive. Watch what happens anyway.
\[ \int_{1}^{3} x^2\,dx, \qquad F(x) = \frac{x^3}{3} \]
Subtract the top value from the bottom value
Why: Reading the bar as bottom minus top - usually because the lower limit was written first and substituted first - reverses the whole subtraction.
\[ \frac{1}{3} - \frac{27}{3} = -\frac{26}{3} \]
Notice the answer is impossible
Why: The region sits entirely above the axis between one and three, so its signed area cannot be negative. The sign alone convicts the answer.
Always top minus bottom, in that order, every time.
\[ \left[F(x)\right]_{a}^{b} = F(b) - F(a) \]
Evaluate at three first, then subtract the value at one
Why: Twenty-seven thirds minus one third is twenty-six thirds, a positive number, as the picture demands.
\[ \left[\frac{x^3}{3}\right]_{1}^{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \]
Bound-check the size
Why: On this interval the integrand runs between one and nine over a width of two, so the area must lie between two and eighteen. Twenty-six thirds is about 8.67, comfortably inside.
\[ 2 \ \le \ \frac{26}{3} \approx 8.67 \ \le \ 18 \ \checkmark \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Step zero
Discussion prompt
Worked example: rewrite, then evaluate — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Rewrite the radical as a power
Answer:
Worked example
\[ \int_{1}^{4} 6\sqrt{x}\,dx \]
Rewrite the radical as a power
Why: The reversed power rule needs an exponent, and the constant six stays out front.
\[ = 6\int_{1}^{4} x^{1/2}\,dx \]
Antidifferentiate
Why: One half plus one is three halves, and dividing by three halves multiplies by two thirds; six times two thirds is four.
\[ = \left[6\cdot\frac{2}{3}x^{3/2}\right]_{1}^{4} = \left[4x^{3/2}\right]_{1}^{4} \]
Evaluate top minus bottom
Why: Four to the three-halves is the cube of the square root of four, which is eight; one to any power is one.
\[ = 4(8) - 4(1) = 32 - 4 = 28 \]
Verify the antiderivative and bound the answer
Why: Differentiating four x to the three halves gives six x to the one half, the original integrand. And the integrand runs from six to twelve across a width of three, so the answer must sit between eighteen and thirty-six - and twenty-eight does.
\[ \frac{d}{dx}\left[4x^{3/2}\right] = 6x^{1/2} \ \checkmark \qquad 18 \le 28 \le 36 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "rewrite, then evaluate", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating four x to the three halves gives six x to the one half, the original integrand. And the integrand runs from six to twelve across a width of three, so the answer must sit between eighteen and thirty-six - and twenty-eight does.
Ranking
Put in order
Put the moves of Worked example: one arch of the sine curve into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The derivative of cosine is negative sine, so the antiderivative of sine carries the minus sign.
Worked example
\[ \int_{0}^{\pi} \sin x\,dx \]
Recall the antiderivative of sine
Why: The derivative of cosine is negative sine, so the antiderivative of sine carries the minus sign. This is the most-dropped sign in the table.
\[ F(x) = -\cos x \]
Evaluate top minus bottom
Why: Cosine of pi is negative one and cosine of zero is one, so the two minus signs on the first term cancel into a plus.
\[ \left[-\cos x\right]_{0}^{\pi} = -(-1) - (-1) = 1 + 1 = 2 \]
Verify the size against the picture
Why: The arch fits inside a rectangle of height one and width about 3.14, so the area must be less than that; and it clearly exceeds the inscribed triangle of area about 1.57. The value two sits right between, as it should.
\[ 1.57 \ < \ 2 \ < \ \pi \approx 3.14 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "one arch of the sine curve", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The arch fits inside a rectangle of height one and width about 3.14, so the area must be less than that; and it clearly exceeds the inscribed triangle of area about 1.57. The value two sits right between, as it should.
Pattern
Real world
Discussion prompt
Outside this lesson: where does Antiderivatives, Riemann Sums, and the FTC actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: evaluating a definite integral is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck reverses differentiation into antiderivatives, including the constant of integration you must never drop. It then builds area from left, right, and midpoint Riemann sums, defines the definite integral as signed area, and proves out both parts of the Fundamental Theorem. It targets the classic errors: losing the plus C, using the power rule at an exponent of negative one, calling net area total area, dropping the chain factor in Part 1 of the FTC, and reversing the evaluation bar.
Concept
Two problems, developed for two thousand years by different people for different reasons: finding tangent slopes, and finding areas.
| Part | What it says | What it gives you |
|---|---|---|
| Part 1 | the derivative of a running total is the rate being totalled | every continuous function has an antiderivative |
| Part 2 | a definite integral is a difference of antiderivative values | a way to compute areas exactly, with no limits of sums |
The theorem says those two problems are the same problem, read in opposite directions. That is why every integral you will ever evaluate starts by asking what function has this as its derivative.
Comparison
Comparison matrix
From Why this theorem gets the name: refill the What it gives you column from what you know. The rest of the table is as it appeared.
| Part | What it says | What it gives you |
|---|---|---|
| Part 1 | the derivative of a running total is the rate being totalled | every continuous function has an antiderivative |
| Part 2 | a definite integral is a difference of antiderivative values | a way to compute areas exactly, with no limits of sums |
Check
Identify the upper limit and its derivative before you write anything down.
\[ \frac{d}{dx}\int_{0}^{x^2} \sqrt{1+t^3}\,dt \]
Check your understanding
What is this derivative?
Answer: A
Why: The upper limit is x squared, so substituting it for t gives the square root of 1 plus x squared cubed, which is 1 plus x to the sixth; then multiply by the derivative of the upper limit, which is 2x.
Elimination
Eliminate the wrong options
What is the value of this definite integral?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: An antiderivative is x cubed minus 2x squared. At 3 that is 27 minus 18, which is 9; at 1 it is 1 minus 2, which is -1. Then 9 minus negative 1 gives 10.
Check
Antidifferentiate, then evaluate top minus bottom. Mind the value at the lower limit.
\[ \int_{1}^{3}\left(3x^2 - 4x\right)dx \]
Check your understanding
What is the value of this definite integral?
Answer: A
Why: An antiderivative is x cubed minus 2x squared. At 3 that is 27 minus 18, which is 9; at 1 it is 1 minus 2, which is -1. Then 9 minus negative 1 gives 10.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Reversing the Derivative · Area, Built From Rectangles · The Definite Integral · The Fundamental Theorem of Calculus. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You started this deck able to differentiate. You finish it able to run that machine backwards, and able to compute areas that geometry has no formula for.
Next up is the technique that makes most of these integrals actually doable: substitution, which is the chain rule run backwards - and then using integrals to measure the area trapped between two curves.
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