This deck treats substitution as the chain rule run backwards: choosing u, converting dx into du, back-substituting, and changing the limits on a definite integral. It then covers the area between two curves - top minus bottom, and left minus right in y - and net change, distinguishing displacement from total distance. It targets the traps of leaving a stray x in the integrand, losing a constant factor from du, keeping the old x-limits, subtracting in the wrong order, and computing distance without splitting at the sign changes.
Subject: Calculus I · 138 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 18
The chain rule run backwards, then adding up the gap between two graphs.
Objectives
By the end of this deck you can:
Warm-up
Discussion prompt
Before we open u-Substitution and Area Between Curves: without looking back, what was the main idea of Antiderivatives, Riemann Sums, and the FTC, 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:
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.
Section
Part 1
Concept
When you differentiate a composition, the chain rule multiplies by the derivative of the inside function. That extra factor is still sitting there in the answer.
\[ \frac{d}{dx}\left[\sin\!\left(x^{2}\right)\right] = \cos\!\left(x^{2}\right)\cdot 2x \]
So if you ever meet that product as an integrand, you already know the antiderivative - you just read the line above from right to left.
\[ \int 2x\cos\!\left(x^{2}\right)dx = \sin\!\left(x^{2}\right) + C \]
Counterexample
Discussion prompt
When you differentiate a composition, the chain rule multiplies by the derivative of the inside function. That extra factor is still sitting there in the answer.
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: Every chain-rule derivative leaves a fingerprint — a rendered Manim animation.
Rendered with Manim.
Takeaway: Spot the leftover factor and you have spotted the substitution.
Intuition
A chain-rule integrand always looks like two things stuck together: some complicated inside expression, and a copy of that inside's derivative hanging around as a factor.
Substitution just gives the inside a short name so you can see the simple integral hiding underneath.
\[ \underbrace{\cos\!\left(x^{2}\right)}_{\text{outer, evaluated at the inside}}\;\cdot\;\underbrace{2x}_{\text{derivative of the inside}} \]
Rename the inside and the whole thing collapses into an integral you can do in one line.
Analogy
Discussion prompt
Explain Look for a package and its wrapper 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 chain-rule integrand always looks like two things stuck together: some complicated inside expression, and a copy of that inside's derivative hanging around as a factor.
Concept
The rule is the chain rule stated for integrals.
\[ \int f\!\left(g(x)\right)g'(x)\,dx = \int f(u)\,du \quad\text{where}\quad u = g(x) \]
substitution (u-substitution) — Renaming an inner expression as u, replacing its derivative times dx by du, and integrating in the new variable. It converts a composite integrand into a basic one.
The differential is part of the substitution. If you change the variable and leave the differential alone, the new integral is not equal to the old one.
Explain it
Discussion prompt
Explain The substitution rule 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:
The rule is the chain rule stated for integrals.
Picture it
Animation
Shows: The substitution rule with the change of variable highlighted.
Look for an inside function whose derivative is already present.
Takeaway: Substitution is the chain rule run backwards. The signal is an inside function whose derivative is already sitting in the integrand.
Ranking
Put in order
Put the moves of Worked example: cosine of x squared, times 2x 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 cosine is applied to x squared, so x squared is the inner function - and its derivative is present as a factor.
Worked example
Find the antiderivative.
\[ \int 2x\cos\!\left(x^{2}\right)dx \]
Name the inside expression u
Why: The cosine is applied to x squared, so x squared is the inner function - and its derivative is present as a factor.
\[ u = x^{2} \]
Differentiate to get du
Why: Differentiating u with respect to x gives 2x, and multiplying both sides by dx packages the whole factor 2x dx into du.
\[ \frac{du}{dx} = 2x \quad\Longrightarrow\quad du = 2x\,dx \]
Swap every piece into the new variable
Why: The cosine's argument becomes u, and the entire factor 2x dx becomes du. No x survives, which is the sign the substitution was the right one.
\[ \int 2x\cos\!\left(x^{2}\right)dx = \int \cos u \, du \]
Integrate in u, then put x back
Why: The antiderivative of cosine is sine; back-substituting restores the original variable so the answer is a function of x.
\[ \int\cos u\,du = \sin u + C = \sin\!\left(x^{2}\right)+C \]
Verify by differentiating the answer
Why: The chain rule on the answer must reproduce the original integrand exactly - that is the only proof an antiderivative needs.
\[ \frac{d}{dx}\left[\sin\!\left(x^{2}\right)+C\right] = \cos\!\left(x^{2}\right)\cdot 2x \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "cosine of x squared, times 2x", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The chain rule on the answer must reproduce the original integrand exactly - that is the only proof an antiderivative needs.
Pattern
Concept
Look for something wrapped up: an expression under a root, inside parentheses raised to a power, in a denominator, in an exponent, or inside a trig function. That expression is your first candidate for u.
Then ask the deciding question: is the derivative of that candidate sitting in the integrand, at least up to a constant factor?
| Integrand contains | Try u equal to |
|---|---|
| A quantity raised to a power | the quantity inside the parentheses |
| A root | the expression under the root |
| An exponential | the exponent |
| A fraction | the denominator |
| A trig function | its argument |
Comparison
Comparison matrix
From How to choose u: refill the Try u equal to column from what you know. The rest of the table is as it appeared.
| Integrand contains | Try u equal to |
|---|---|
| A quantity raised to a power | the quantity inside the parentheses |
| A root | the expression under the root |
| An exponential | the exponent |
| A fraction | the denominator |
| A trig function | its argument |
Picture it
Animation
Shows: How to choose u — a rendered Manim animation.
Rendered with Manim.
Takeaway: A constant multiple is fine. A missing variable factor is not.
Step zero
Discussion prompt
Worked example: a leftover constant factor — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Let u be the expression inside the parentheses
Answer:
Worked example
Find the antiderivative.
\[ \int x\left(x^{2}+1\right)^{5}dx \]
Let u be the expression inside the parentheses
Why: It is the inner function of the fifth power, and its derivative 2x is a constant multiple of the x already in the integrand.
\[ u = x^{2}+1 \quad\Longrightarrow\quad du = 2x\,dx \]
Solve for the piece you actually have
Why: The integrand offers x dx, not 2x dx, so divide the du equation by 2 to get an exact match.
\[ x\,dx = \tfrac{1}{2}\,du \]
Substitute and pull the constant out front
Why: Constants slide through an integral freely, so the one half rides outside while the power rule handles the rest.
\[ \int x\left(x^{2}+1\right)^{5}dx = \frac{1}{2}\int u^{5}\,du \]
Apply the power rule and back-substitute
Why: Raise the exponent by one and divide by the new exponent, then restore the original variable.
\[ \frac{1}{2}\cdot\frac{u^{6}}{6}+C = \frac{\left(x^{2}+1\right)^{6}}{12}+C \]
Verify by differentiating the answer
Why: The chain rule gives six times the fifth power times the inner derivative 2x, and twelve divides that back down to the original integrand.
\[ \frac{d}{dx}\!\left[\frac{\left(x^{2}+1\right)^{6}}{12}\right] = \frac{6\left(x^{2}+1\right)^{5}\cdot 2x}{12} = x\left(x^{2}+1\right)^{5}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a leftover constant factor", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The chain rule gives six times the fifth power times the inner derivative 2x, and twelve divides that back down to the original integrand.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The integrand looks close enough, so the student replaces the parentheses by u and writes du where dx was.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The power rule gives u to the sixth over six, so the reported antiderivative divides by 6.
Convert the differential honestly: work out du first, then trade it for exactly what the integrand contains.
Why: The power rule gives u to the sixth over six, so the reported antiderivative divides by 6.
Trap
The integrand looks close enough, so the student replaces the parentheses by u and writes du where dx was.
\[ \int x\left(x^{2}+1\right)^{5}dx \;\longrightarrow\; \int u^{5}\,du \]
Integrate and report the answer
Why: The power rule gives u to the sixth over six, so the reported antiderivative divides by 6.
\[ \frac{\left(x^{2}+1\right)^{6}}{6}+C \]
Differentiating shows it is twice too big
Why: The check fails: the derivative comes out as 2x times the fifth power, but the integrand only had one x.
\[ \frac{d}{dx}\!\left[\frac{\left(x^{2}+1\right)^{6}}{6}\right] = 2x\left(x^{2}+1\right)^{5} \neq x\left(x^{2}+1\right)^{5} \]
Convert the differential honestly: work out du first, then trade it for exactly what the integrand contains.
\[ du = 2x\,dx \quad\Longrightarrow\quad x\,dx = \tfrac{1}{2}\,du \]
Carry the one half through
Why: The integrand supplies only half of du, so the new integral carries a factor of one half.
\[ \frac{1}{2}\int u^{5}\,du = \frac{u^{6}}{12}+C = \frac{\left(x^{2}+1\right)^{6}}{12}+C \]
Differentiating returns the integrand
Why: Six times two divided by twelve is one, so the check closes exactly.
\[ \frac{6\left(x^{2}+1\right)^{5}\cdot 2x}{12} = x\left(x^{2}+1\right)^{5}\;\checkmark \]
Notation
Annotate
From Trap: swapping the variable but not the differential — read this one piece at a time. What is each part doing?
On: \( \int x\left(x^{2}+1\right)^{5}dx \;\longrightarrow\; \int u^{5}\,du \)
Elimination
Eliminate the wrong options
Which substitution (with its correct differential) turns this into an integral in u alone?
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 inside of the seventh power is x cubed plus 5, and its derivative is 3x squared, which matches the x squared already in the integrand up to the constant 3. So x squared dx equals one third du, and the integral becomes one third times the integral of u to the seventh.
Check
Decide which substitution clears the original variable out of the integrand completely.
\[ \int x^{2}\left(x^{3}+5\right)^{7}dx \]
Check your understanding
Which substitution (with its correct differential) turns this into an integral in u alone?
Answer: A
Why: The inside of the seventh power is x cubed plus 5, and its derivative is 3x squared, which matches the x squared already in the integrand up to the constant 3. So x squared dx equals one third du, and the integral becomes one third times the integral of u to the seventh.
Concept
After substituting, the integral must be written entirely in the new variable. A single leftover x makes the expression meaningless: you cannot antidifferentiate with respect to u while an x is still floating in there.
If an x is left over, you have exactly two options.
What you may not do is treat the leftover x as a constant and pull it outside the integral. It is not constant - it is the very thing you are integrating over.
Picture it
Animation
Shows: No original variable may survive — a rendered Manim animation.
Rendered with Manim.
Takeaway: If an x is left behind, the substitution was the wrong one.
Fill the middle
Fill in the blanks
From Worked example: sine cubed times cosine — finish the line. Write what belongs on the right of the equals sign before you look.
\frac\frac{\sin^{4}x}{4}+C}___+C = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Sine cubed means the cube of sine, so sine is the inner function and the cube is the outer one.
Worked example
Find the antiderivative.
\[ \int \sin^{3}x\,\cos x\,dx \]
Rewrite the integrand to see the composition
Why: Sine cubed means the cube of sine, so sine is the inner function and the cube is the outer one.
\[ \int \left(\sin x\right)^{3}\cos x\,dx \]
Let u be sine of x
Why: The derivative of sine is cosine, and a cosine factor is already sitting in the integrand - an exact match, no constant needed.
\[ u = \sin x \quad\Longrightarrow\quad du = \cos x\,dx \]
Substitute the whole integrand
Why: The cube becomes u cubed and the factor cosine x dx becomes du, so nothing in the original variable remains.
\[ \int u^{3}\,du \]
Integrate and back-substitute
Why: The power rule gives u to the fourth over four; replacing u by sine x returns the answer in the original variable.
\[ \frac{u^{4}}{4}+C = \frac{\sin^{4}x}{4}+C \]
Verify by differentiating the answer
Why: The chain rule gives four times sine cubed times cosine, and the four in the denominator cancels it, returning the integrand.
\[ \frac{d}{dx}\!\left[\frac{\sin^{4}x}{4}\right] = \frac{4\sin^{3}x\cos x}{4} = \sin^{3}x\cos x\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "sine cubed times cosine", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The chain rule gives four times sine cubed times cosine, and the four in the denominator cancels it, returning the integrand.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The root is the obvious thing to rename, so the student substitutes and then treats the lonely x as if it were a constant.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This is the fatal move: x is the variable of integration in disguise, not a constant that may be factored out.
Use the u-equation a second time: it also tells you what the stray x equals.
Why: This is the fatal move: x is the variable of integration in disguise, not a constant that may be factored out.
Trap
The root is the obvious thing to rename, so the student substitutes and then treats the lonely x as if it were a constant.
\[ \int x\sqrt{x+1}\,dx,\qquad u = x+1,\quad du = dx \]
Pull the x outside the integral
Why: This is the fatal move: x is the variable of integration in disguise, not a constant that may be factored out.
\[ x\int u^{1/2}\,du = \frac{2}{3}x\left(x+1\right)^{3/2}+C \]
Differentiating exposes the error
Why: The product rule produces an extra term that was never in the original integrand, so the answer is wrong.
\[ \frac{d}{dx}\!\left[\frac{2}{3}x\left(x+1\right)^{3/2}\right] = \frac{2}{3}\left(x+1\right)^{3/2} + x\left(x+1\right)^{1/2} \neq x\sqrt{x+1} \]
Use the u-equation a second time: it also tells you what the stray x equals.
\[ u = x+1 \quad\Longrightarrow\quad x = u-1,\quad du = dx \]
Replace the stray x too, then expand
Why: Now every symbol is in the new variable, and distributing the root turns the product into a sum of plain powers.
\[ \int (u-1)\sqrt{u}\,du = \int \left(u^{3/2}-u^{1/2}\right)du \]
Integrate term by term and restore x
Why: Two ordinary power-rule integrals; back-substituting gives an answer whose derivative really is the integrand.
\[ \frac{2}{5}\left(x+1\right)^{5/2}-\frac{2}{3}\left(x+1\right)^{3/2}+C \]
Estimation
Predict first
Find the antiderivative. The derivative of the inside is 1, so there is no matching factor to cancel - the leftover x must be converted.
Commit before you compute: what does Worked example: solving the u-equation for x 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 by differentiating and factoring
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. Differentiate each power, then factor the common square root: the bracket collapses to x, which is exactly the missing factor.
Worked example
Find the antiderivative. The derivative of the inside is 1, so there is no matching factor to cancel - the leftover x must be converted.
\[ \int x\sqrt{x+1}\,dx \]
Let u be the expression under the root
Why: The root is the hard part; renaming what is under it is the only way to reach a basic power integral.
\[ u = x+1,\qquad du = dx,\qquad x = u-1 \]
Substitute all three pieces
Why: The factor x becomes u minus 1, the root becomes the one-half power of u, and dx becomes du. Nothing in x is left.
\[ \int (u-1)\,u^{1/2}\,du \]
Distribute before integrating
Why: You cannot integrate a product directly, but you can integrate each power separately once the parentheses are expanded.
\[ \int \left(u^{3/2}-u^{1/2}\right)du = \frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}+C \]
Back-substitute
Why: Replacing u by x plus 1 puts the answer back in the original variable, which is the form a test expects.
\[ \frac{2}{5}\left(x+1\right)^{5/2}-\frac{2}{3}\left(x+1\right)^{3/2}+C \]
Verify by differentiating and factoring
Why: Differentiate each power, then factor the common square root: the bracket collapses to x, which is exactly the missing factor.
\[ \left(x+1\right)^{3/2}-\left(x+1\right)^{1/2} = \left(x+1\right)^{1/2}\left[(x+1)-1\right] = x\sqrt{x+1}\;\checkmark \]
Concept
If the inner function is a straight line, its derivative is a constant. That makes the substitution almost automatic: antidifferentiate as usual and divide by the coefficient of the variable.
\[ \int f(ax+b)\,dx = \frac{1}{a}F(ax+b)+C \quad\text{where } F' = f \]
This single case covers a huge share of homework problems. Notice that it fails the moment the inside stops being linear - then the derivative is not constant and cannot be pulled out.
Picture it
Animation
Shows: When the inside is linear, the fix is one constant — a rendered Manim animation.
Rendered with Manim.
Takeaway: Only a constant is missing, and a constant can always be supplied.
Hypothesis
Predict first
Worked example: a linear inside function 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: Let u be the linear inside expression
Why: It is what the seventh power is applied to, and its derivative is the constant 2.
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
Find the antiderivative.
\[ \int \left(2x+5\right)^{7}dx \]
Let u be the linear inside expression
Why: It is what the seventh power is applied to, and its derivative is the constant 2.
\[ u = 2x+5 \quad\Longrightarrow\quad du = 2\,dx \quad\Longrightarrow\quad dx = \tfrac{1}{2}\,du \]
Substitute and factor the one half out
Why: The integrand has no 2 to spare, so the 2 is manufactured by writing dx as half of du; that constant then rides outside.
\[ \frac{1}{2}\int u^{7}\,du \]
Apply the power rule and back-substitute
Why: One half times u to the eighth over eight gives a denominator of sixteen.
\[ \frac{1}{2}\cdot\frac{u^{8}}{8}+C = \frac{\left(2x+5\right)^{8}}{16}+C \]
Verify by differentiating the answer
Why: Eight from the power and 2 from the chain rule make sixteen, which cancels the denominator exactly.
\[ \frac{d}{dx}\!\left[\frac{\left(2x+5\right)^{8}}{16}\right] = \frac{8\left(2x+5\right)^{7}\cdot 2}{16} = \left(2x+5\right)^{7}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a linear inside function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Eight from the power and 2 from the chain rule make sixteen, which cancels the denominator exactly.
Concept
The natural exponential is its own derivative, so the only thing the chain rule ever adds is the derivative of the exponent. That tells you exactly what to call u.
\[ \frac{d}{dx}\left[e^{g(x)}\right] = e^{g(x)}g'(x) \quad\Longrightarrow\quad \int e^{g(x)}g'(x)\,dx = e^{g(x)}+C \]
So when you see an exponential, look at its exponent and ask whether the exponent's derivative is available as a factor.
Missing information
Discussion prompt
Find the antiderivative. This integrand shows up constantly in probability.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The exponent is the inner function; its derivative is negative 2x, which is a constant multiple of the x in the integrand.
Worked example
Find the antiderivative. This integrand shows up constantly in probability.
\[ \int x\,e^{-x^{2}}dx \]
Let u be the exponent
Why: The exponent is the inner function; its derivative is negative 2x, which is a constant multiple of the x in the integrand.
\[ u = -x^{2} \quad\Longrightarrow\quad du = -2x\,dx \]
Solve for the piece the integrand actually has
Why: Only x dx is available, so divide by negative 2 - and keep the sign, which is where most errors happen.
\[ x\,dx = -\tfrac{1}{2}\,du \]
Substitute and integrate
Why: The exponential of u is its own antiderivative, so only the constant out front needs handling.
\[ -\frac{1}{2}\int e^{u}\,du = -\frac{1}{2}e^{u}+C \]
Back-substitute
Why: Restore the exponent in terms of x to finish.
\[ -\frac{1}{2}e^{-x^{2}}+C \]
Verify by differentiating the answer
Why: The chain rule brings down negative 2x; times the negative one half out front gives a positive x, matching the integrand.
\[ \frac{d}{dx}\!\left[-\frac{1}{2}e^{-x^{2}}\right] = -\frac{1}{2}e^{-x^{2}}\cdot(-2x) = x\,e^{-x^{2}}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "x times a Gaussian exponential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The chain rule brings down negative 2x; times the negative one half out front gives a positive x, matching the integrand.
Concept
There is one antiderivative the power rule cannot produce: the reciprocal. Whenever a fraction has the derivative of its own denominator sitting in the numerator, the answer is a natural log.
\[ \int \frac{g'(x)}{g(x)}\,dx = \ln\left|g(x)\right| + C \]
The absolute value matters: the natural log is only defined for positive inputs, but the formula must work on both sides of a zero of the denominator.
Step zero
Discussion prompt
Worked example: x over a quadratic — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Let u be the denominator
Answer:
Worked example
Find the antiderivative.
\[ \int \frac{x}{x^{2}+3}\,dx \]
Let u be the denominator
Why: For a fraction, the denominator is the natural candidate: its derivative 2x is a constant multiple of the numerator.
\[ u = x^{2}+3 \quad\Longrightarrow\quad du = 2x\,dx \quad\Longrightarrow\quad x\,dx = \tfrac{1}{2}\,du \]
Substitute
Why: The numerator and the differential together become half of du, leaving the reciprocal of u.
\[ \frac{1}{2}\int \frac{du}{u} \]
Integrate to a natural log and back-substitute
Why: The reciprocal integrates to the natural log of the absolute value; here the quadratic is always positive, so the bars can be dropped.
\[ \frac{1}{2}\ln\left|u\right| + C = \frac{1}{2}\ln\!\left(x^{2}+3\right)+C \]
Verify by differentiating the answer
Why: The derivative of the log is one over the inside times the inside's derivative, and the one half cancels the 2.
\[ \frac{1}{2}\cdot\frac{2x}{x^{2}+3} = \frac{x}{x^{2}+3}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "x over a quadratic", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The derivative of the log is one over the inside times the inside's derivative, and the one half cancels the 2.
Worked example
This one is worth memorizing, but it is really just a substitution.
\[ \int \tan x\,dx \]
Rewrite tangent as a quotient
Why: Substitution needs a visible inner function; writing tangent as sine over cosine reveals one.
\[ \int \frac{\sin x}{\cos x}\,dx \]
Let u be the cosine in the denominator
Why: Its derivative is negative sine, which is exactly the numerator up to a minus sign.
\[ u = \cos x \quad\Longrightarrow\quad du = -\sin x\,dx \quad\Longrightarrow\quad \sin x\,dx = -du \]
Substitute and integrate
Why: The minus sign travels outside the integral and stays in the final answer.
\[ -\int\frac{du}{u} = -\ln\left|u\right| + C = -\ln\left|\cos x\right| + C \]
Verify by differentiating the answer
Why: The derivative of the log gives one over cosine, the chain rule supplies negative sine, and the two minus signs cancel to leave tangent.
\[ \frac{d}{dx}\left[-\ln\left|\cos x\right|\right] = -\frac{1}{\cos x}\cdot(-\sin x) = \tan x\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the integral of tangent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The derivative of the log gives one over cosine, the chain rule supplies negative sine, and the two minus signs cancel to leave tangent.
Check
Work it out on paper first, then differentiate your answer before you look at the choices.
\[ \int \cos(5x)\,dx \]
Check your understanding
Which is the antiderivative of cosine of 5x?
Answer: A
Why: With u equal to 5x, du equals 5 dx, so dx is one fifth du and the integral becomes one fifth times the integral of cosine u. Differentiating one fifth of sine of 5x gives one fifth times cosine of 5x times 5, which is cosine of 5x.
Concept
Substitution only works when the derivative of the inner function is genuinely present, up to a constant. If it is not, no amount of algebra will force it.
\[ \int x^{2}e^{x}\,dx \qquad\text{and}\qquad \int e^{x^{2}}\,dx \]
In the first one, the exponent is x, whose derivative is 1 - the extra x squared is not a constant multiple of that, so substitution stalls. It needs integration by parts, a Calculus II tool.
The second has no elementary antiderivative at all. Recognizing a dead end quickly is part of the skill: a missing constant is fine, a missing variable factor is not.
Picture it
Animation
Shows: A definite integral with its limits converted to the new variable.
The commonest slip in this section.
Takeaway: For a definite integral either convert the limits to the new variable or substitute back before evaluating. Forgetting is the commonest error here.
Section
Part 2
Concept
Once limits are attached, there are two correct ways to finish a substitution.
Route 2 is shorter and much safer, because the messy back-substitution step - where sign and algebra errors breed - is skipped entirely.
Intuition
The numbers on an integral sign are not decorations: they are values of the variable written next to the differential. Change the variable and those numbers are now measured in the wrong units.
Think of switching from miles to kilometres. The trip is the same trip, but every marker on the route has to be relabelled - you cannot keep the mile markers and call the distance kilometres.
\[ \int_{x=a}^{x=b} f(g(x))g'(x)\,dx = \int_{u=g(a)}^{u=g(b)} f(u)\,du \]
Worked example
Evaluate the definite integral.
\[ \int_{0}^{2} x\left(x^{2}+1\right)^{3}dx \]
Choose u and find du
Why: The inside of the cube is the candidate, and its derivative is twice the x that is already present.
\[ u = x^{2}+1 \quad\Longrightarrow\quad du = 2x\,dx \quad\Longrightarrow\quad x\,dx = \tfrac{1}{2}\,du \]
Push both limits through the substitution
Why: Each limit is an x-value; feeding it into the u-equation gives the matching u-value.
| x | u equals x squared plus 1 |
|---|---|
| 0 | 1 |
| 2 | 5 |
Rewrite the whole integral in u
Why: Integrand, differential, and limits all change together, so the value of the integral is unchanged.
\[ \frac{1}{2}\int_{1}^{5} u^{3}\,du \]
Antidifferentiate and evaluate
Why: The power rule gives u to the fourth over four; evaluate top minus bottom with the new limits - no back-substitution needed.
\[ \frac{1}{2}\left[\frac{u^{4}}{4}\right]_{1}^{5} = \frac{1}{8}\left(625-1\right) = \frac{624}{8} = 78 \]
Verify by the other route
Why: Back-substituting gives the antiderivative in x; evaluating it at the original limits must give the same number, and it does.
\[ \left[\frac{\left(x^{2}+1\right)^{4}}{8}\right]_{0}^{2} = \frac{625}{8}-\frac{1}{8} = 78\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "converting the limits", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Back-substituting gives the antiderivative in x; evaluating it at the original limits must give the same number, and it does.
Trap
The substitution is done correctly, but the numbers on the integral sign are left untouched.
\[ \int_{0}^{2} x\left(x^{2}+1\right)^{3}dx \;\longrightarrow\; \frac{1}{2}\int_{0}^{2} u^{3}\,du \]
Evaluate with the x-limits still attached
Why: The bar now says: run u from 0 to 2, which describes a completely different region than the original problem.
\[ \frac{1}{2}\left[\frac{u^{4}}{4}\right]_{0}^{2} = \frac{1}{8}(16-0) = 2 \]
The answer is off by a factor of 39
Why: The true value is 78, so this is not a small slip - mismatched limits change the answer completely.
Convert each limit through the same equation you used for the integrand.
\[ x=0 \Rightarrow u=1, \qquad x=2 \Rightarrow u=5 \]
Evaluate with the u-limits
Why: Integrand, differential, and limits are now all in the same variable, so the bar means what it says.
\[ \frac{1}{2}\left[\frac{u^{4}}{4}\right]_{1}^{5} = \frac{625-1}{8} = 78 \]
A habit that prevents this
Why: Write the limits as x equals 0 and x equals 2 on the original integral. Then a bare 0 and 2 next to a du can never look right.
Translation
\( \frac{1}{2}\left[\frac{u^{4}}{4}\right]_{0}^{2} = \frac{1}{8}(16-0) = 2 \)
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.
Pattern
If you would rather back-substitute, that is legal too - but then you must strip the limits off while you work and only reattach the original ones at the very end.
Ranking
Put in order
Put the moves of Worked example: a logarithm inside the integrand 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. Its derivative is one over x, and a factor of one over x is exactly what the denominator provides.
Worked example
Evaluate the definite integral.
\[ \int_{1}^{e} \frac{\ln x}{x}\,dx \]
Let u be the logarithm
Why: Its derivative is one over x, and a factor of one over x is exactly what the denominator provides.
\[ u = \ln x \quad\Longrightarrow\quad du = \frac{1}{x}\,dx \]
Convert the limits
Why: Run each x-limit through the natural log: the log of 1 is 0 and the log of e is 1.
| x | u equals the natural log of x |
|---|---|
| 1 | 0 |
| e | 1 |
Rewrite and integrate
Why: The integrand collapses to just u, a basic power integral in the new variable.
\[ \int_{0}^{1} u\,du = \left[\frac{u^{2}}{2}\right]_{0}^{1} = \frac{1}{2} \]
Verify with the antiderivative in x
Why: Back-substituting gives half the square of the natural log; evaluating at e and 1 reproduces the same value.
\[ \left[\frac{\left(\ln x\right)^{2}}{2}\right]_{1}^{e} = \frac{1^{2}}{2}-\frac{0^{2}}{2} = \frac{1}{2}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a logarithm inside the integrand", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Back-substituting gives half the square of the natural log; evaluating at e and 1 reproduces the same value.
Worked example
Evaluate the definite integral.
\[ \int_{0}^{\pi/4} \tan x\,\sec^{2}x\,dx \]
Let u be the tangent
Why: The derivative of tangent is secant squared, and that factor is already present with nothing to spare or borrow.
\[ u = \tan x \quad\Longrightarrow\quad du = \sec^{2}x\,dx \]
Convert the limits with exact values
Why: The tangent of 0 is 0 and the tangent of a quarter pi is 1 - both come straight off the unit circle.
| x | u equals the tangent of x |
|---|---|
| 0 | 0 |
| a quarter of pi | 1 |
Integrate in the new variable
Why: What remains is the integral of u from 0 to 1, one of the simplest integrals there is.
\[ \int_{0}^{1} u\,du = \left[\frac{u^{2}}{2}\right]_{0}^{1} = \frac{1}{2} \]
Verify by differentiating the antiderivative in x
Why: Half of tangent squared differentiates to tangent times secant squared, which is the original integrand; its values at the two limits give one half minus zero.
\[ \frac{d}{dx}\!\left[\frac{\tan^{2}x}{2}\right] = \tan x\sec^{2}x, \qquad \left[\frac{\tan^{2}x}{2}\right]_{0}^{\pi/4} = \frac{1}{2}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "tangent times secant squared", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Half of tangent squared differentiates to tangent times secant squared, which is the original integrand; its values at the two limits give one half minus zero.
Check
Substitute, convert the limits, and evaluate before you look at the choices.
\[ \int_{0}^{3} \frac{x}{\sqrt{x^{2}+16}}\,dx \]
Check your understanding
What is the value of this definite integral?
Answer: A
Why: With u equal to x squared plus 16, du is 2x dx so x dx is half of du, and the limits become 16 and 25. Half the integral of u to the negative one half is the square root of u, evaluated from 16 to 25, which is 5 minus 4, or 1.
Section
Part 3
Concept
A definite integral of a positive function measures the area between that curve and the horizontal axis. But the axis is nothing special - it is just the line at height zero.
Replace that line by a second curve and the same integral measures the area trapped between the two graphs.
\[ \int_{a}^{b} f(x)\,dx = \int_{a}^{b}\left[f(x)-0\right]dx \quad\longrightarrow\quad \int_{a}^{b}\left[f(x)-g(x)\right]dx \]
Picture it
Animation
Shows: The region between a line and a parabola, shaded.
The line is above here, so the line goes first.
Takeaway: The integrand is always the upper curve minus the lower one. Deciding which is which is the step that determines the sign.
Intuition
Slice the region with thin vertical strips. Each strip has a tiny width and a height equal to how far the upper curve sits above the lower one at that place.
The height of a strip is always top minus bottom, and that subtraction is what makes the height a positive number. Adding up all the strips is exactly what the integral does.
\[ \text{strip area} \approx \left[y_{\text{top}}-y_{\text{bottom}}\right]\Delta x \]
This picture also explains why the region can sit below the axis and the formula still works: both curves shift down together, so the gap between them is unchanged.
Picture it
Animation
Shows: Stack up thin vertical strips — a rendered Manim animation.
Rendered with Manim.
Takeaway: Each strip has height given by the function and width dx.
Concept
If one curve stays above the other across the whole interval, the area of the region between them is a single integral.
\[ A = \int_{a}^{b}\left[f(x)-g(x)\right]dx \quad\text{whenever } f(x)\ge g(x) \text{ on } [a,b] \]
top minus bottom — The integrand for an area between curves. Which function is the top one is decided by testing a point inside the interval, never by which one was written first.
Definition probe
Sort into buckets
Every line below is part of the definition of substitution (u-substitution) or of top minus bottom — one or the other, never both. Put each where it belongs.
Step zero
Discussion prompt
Worked example: the line and the parabola — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set the curves equal to find where they meet
Answer:
Worked example
Figure (svg): The line y equals x running above the parabola y equals x squared between x equals 0 and x equals 1, with the region between them shaded.
Find the area of the region enclosed by the two curves.
\[ y = x \qquad\text{and}\qquad y = x^{2} \]
Set the curves equal to find where they meet
Why: The enclosed region starts and ends at the intersection points, so those x-values are the limits of integration.
\[ x = x^{2} \;\Longrightarrow\; x(1-x)=0 \;\Longrightarrow\; x=0,\; x=1 \]
Test an interior point to see which curve is on top
Why: At one half, the line is at one half and the parabola is at one quarter, so the line is the upper curve throughout the interval.
\[ \text{at } x=\tfrac{1}{2}: \quad x=\tfrac{1}{2} \;>\; x^{2}=\tfrac{1}{4} \]
Integrate top minus bottom
Why: Every strip has height equal to the line minus the parabola, and the strips run from 0 to 1.
\[ A = \int_{0}^{1}\left(x-x^{2}\right)dx \]
Antidifferentiate and evaluate
Why: The power rule on each term, then substitute the limits: at 0 everything vanishes.
\[ \left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} = \frac{1}{2}-\frac{1}{3} = \frac{1}{6} \]
Check the size against the picture
Why: The region fits inside a unit square, and its widest gap is one quarter at the middle, so an area near one sixth (about 0.17) is exactly the right order of size - and it is positive, as an area must be.
Picture it
Animation
Shows: Each line of the worked example "the line and the parabola", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The region fits inside a unit square, and its widest gap is one quarter at the middle, so an area near one sixth (about 0.17) is exactly the right order of size - and it is positive, as an area must be.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The parabola was written first in the problem, so the student subtracts in that order.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Every strip height comes out negative, so the integral does too - and a region cannot have negative area.
Decide the order by testing a point inside the interval, not by reading order.
Why: Every strip height comes out negative, so the integral does too - and a region cannot have negative area.
Trap
The parabola was written first in the problem, so the student subtracts in that order.
\[ \int_{0}^{1}\left(x^{2}-x\right)dx \]
Evaluate and report a negative area
Why: Every strip height comes out negative, so the integral does too - and a region cannot have negative area.
\[ \left[\frac{x^{3}}{3}-\frac{x^{2}}{2}\right]_{0}^{1} = \frac{1}{3}-\frac{1}{2} = -\frac{1}{6} \]
Why this is a real error, not a sign slip
Why: The order of subtraction is not a convention you may choose - it encodes which curve is physically above the other.
Decide the order by testing a point inside the interval, not by reading order.
\[ \text{at } x=\tfrac{1}{2}: \quad \tfrac{1}{2} > \tfrac{1}{4} \;\Longrightarrow\; y=x \text{ is on top} \]
Integrate the upper curve minus the lower one
Why: Now every strip height is positive, so the sum of the strips is a genuine area.
\[ \int_{0}^{1}\left(x-x^{2}\right)dx = \frac{1}{6} \]
A safety net
Why: If an area comes out negative, you have not made an arithmetic mistake - you subtracted backwards. Flip the order and the magnitude is already correct.
Break the constraint
Discussion prompt
The rule this trap just fixed:
If an area comes out negative, you have not made an arithmetic mistake - you subtracted backwards. Flip the order and the magnitude is already correct.
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:
Every strip height comes out negative, so the integral does too - and a region cannot have negative area.
Concept
When a problem says the region is enclosed by two curves and gives you no interval, the curves supply their own limits: solve for where they meet.
\[ f(x) = g(x) \quad\Longrightarrow\quad x = a,\; x = b \]
Solve that equation completely. A missed solution means a missed piece of the region, and an extra one means integrating over territory that is not enclosed at all.
Explain it
Discussion prompt
Explain The limits come from the intersections 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:
When a problem says the region is enclosed by two curves and gives you no interval, the curves supply their own limits: solve for where they meet.
Picture it
Animation
Shows: Two curves solved for their intersection points, which become the limits.
Solve for the crossings first.
Takeaway: Setting the two functions equal gives the crossings, and those crossings are exactly the limits of integration.
Estimation
Predict first
Find the area of the region enclosed by the two curves.
Commit before you compute: what does Worked example: two parabolas 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 using the symmetry of the region
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 integrand is even and the interval is symmetric, so the area must be twice the right half - and that route gives the same number.
Worked example
Find the area of the region enclosed by the two curves.
\[ y = x^{2} \qquad\text{and}\qquad y = 8-x^{2} \]
Find the intersections
Why: Set the two expressions equal and solve; both solutions are needed because they are the two edges of the region.
\[ x^{2} = 8-x^{2} \;\Longrightarrow\; 2x^{2}=8 \;\Longrightarrow\; x = \pm 2 \]
Identify the upper curve at an interior point
Why: At zero the downward parabola is at 8 and the upward one is at 0, so the downward parabola is on top across the whole interval.
\[ \text{at } x=0: \quad 8-x^{2}=8 \;>\; x^{2}=0 \]
Write and simplify the integral
Why: Top minus bottom collapses the two squares into a single quadratic, which is easier to antidifferentiate.
\[ A = \int_{-2}^{2}\left[\left(8-x^{2}\right)-x^{2}\right]dx = \int_{-2}^{2}\left(8-2x^{2}\right)dx \]
Antidifferentiate and evaluate
Why: Evaluate at 2 and at negative 2 and subtract; the odd-power term contributes twice.
\[ \left[8x-\frac{2x^{3}}{3}\right]_{-2}^{2} = \left(16-\frac{16}{3}\right)-\left(-16+\frac{16}{3}\right) = 32-\frac{32}{3} = \frac{64}{3} \]
Verify using the symmetry of the region
Why: The integrand is even and the interval is symmetric, so the area must be twice the right half - and that route gives the same number.
\[ 2\int_{0}^{2}\left(8-2x^{2}\right)dx = 2\left(16-\frac{16}{3}\right) = \frac{64}{3}\;\checkmark \]
Concept
If the two graphs cross somewhere inside the interval, no single subtraction works: the curve that was on top becomes the one underneath.
Split the interval at every crossing, and on each piece integrate whichever function is on top there.
\[ A = \int_{a}^{c}\left[f-g\right]dx + \int_{c}^{b}\left[g-f\right]dx \]
This is the same idea as splitting for total distance later on: an integral adds signed quantities, so you must keep every piece positive by hand.
Picture it
Animation
Shows: The rule for splitting an area integral where two curves cross.
One integral would let the regions cancel.
Takeaway: If the curves cross inside the interval, split the integral at the crossing and swap which one goes first. A single integral would let the two regions cancel instead of add.
Worked example
Find the total area of the regions enclosed by the two curves.
\[ y = x^{3} \qquad\text{and}\qquad y = x \]
Find every intersection
Why: Factor completely: three solutions means the curves cross in the middle, which is the warning sign that a split is coming.
\[ x^{3}=x \;\Longrightarrow\; x\left(x^{2}-1\right)=0 \;\Longrightarrow\; x=-1,\,0,\,1 \]
Test one point in each subinterval
Why: At negative one half the cubic is higher; at positive one half the line is higher. The curves really do swap at the origin.
| test point | value of the cubic | value of the line | on top |
|---|---|---|---|
| negative one half | negative 0.125 | negative 0.5 | the cubic |
| positive one half | 0.125 | 0.5 | the line |
Write one integral per piece
Why: Each integrand is that piece's top minus that piece's bottom, so both integrands are positive on their own interval.
\[ A = \int_{-1}^{0}\left(x^{3}-x\right)dx + \int_{0}^{1}\left(x-x^{3}\right)dx \]
Evaluate the left piece
Why: Antidifferentiate, then subtract the value at negative 1 from the value at 0.
\[ \left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0} = 0-\left(\frac{1}{4}-\frac{1}{2}\right) = \frac{1}{4} \]
Evaluate the right piece and add
Why: The second piece gives the same value, which the symmetry of the picture predicts.
\[ \left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1} = \frac{1}{4}, \qquad A = \frac{1}{4}+\frac{1}{4} = \frac{1}{2} \]
Verify with the odd symmetry of both curves
Why: Both functions are odd, so the left lobe is a point-reflection of the right lobe and the two areas must be equal. Each came out as one quarter, and the total one half is positive, as an area must be.
Picture it
Animation
Shows: Each line of the worked example "a cubic crossing a line", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both functions are odd, so the left lobe is a point-reflection of the right lobe and the two areas must be equal. Each came out as one quarter, and the total one half is positive, as an area must be.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The interval runs from negative 1 to 1 and the line looked like the top curve, so the student writes a single integral.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The two lobes have opposite signs in this single integral, so they cancel each other exactly.
Find the crossing first, then split the interval there and take top minus bottom on each side.
Why: The two lobes have opposite signs in this single integral, so they cancel each other exactly.
Trap
The interval runs from negative 1 to 1 and the line looked like the top curve, so the student writes a single integral.
\[ \int_{-1}^{1}\left(x-x^{3}\right)dx \]
Evaluate it
Why: The two lobes have opposite signs in this single integral, so they cancel each other exactly.
\[ \left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{-1}^{1} = \frac{1}{4}-\frac{1}{4} = 0 \]
An area of zero for a visible region
Why: The picture clearly encloses two lobes, so zero cannot be right. The integral computed net signed area, not area.
Find the crossing first, then split the interval there and take top minus bottom on each side.
\[ x^{3}=x \;\Longrightarrow\; x=-1,\,0,\,1 \]
Two integrals, each with its own order
Why: On the left the cubic is on top; on the right the line is. Both integrands are then positive.
\[ \int_{-1}^{0}\left(x^{3}-x\right)dx + \int_{0}^{1}\left(x-x^{3}\right)dx = \frac{1}{4}+\frac{1}{4} = \frac{1}{2} \]
The tell to watch for
Why: Solving the intersection equation gave three roots. A root strictly between the outer two always means the curves swap and the integral must be split.
Notation
Annotate
From Trap: one integral across a crossing point — read this one piece at a time. What is each part doing?
On: \( \int_{-1}^{0}\left(x^{3}-x\right)dx + \int_{0}^{1}\left(x-x^{3}\right)dx = \frac{1}{4}+\frac{1}{4} = \frac{1}{2} \)
Pattern
Edge cases
Discussion prompt
The area-between-curves recipe 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:
Concept
If a region's left and right edges are each a single curve, but its top or bottom is made of two different pieces, vertical strips force you to split. Horizontal strips often do not.
Then integrate with respect to the vertical variable, and each strip's length is right minus left.
\[ A = \int_{c}^{d}\left[x_{\text{right}}(y)-x_{\text{left}}(y)\right]dy \]
Everything else is the same: the limits are now the smallest and largest heights of the region, and both curves must be solved for the horizontal variable.
Analogy
Discussion prompt
Explain Sometimes the strips should be horizontal 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 a region's left and right edges are each a single curve, but its top or bottom is made of two different pieces, vertical strips force you to split. Horizontal strips often do not.
Picture it
Animation
Shows: Sometimes the strips should be horizontal — a rendered Manim animation.
Rendered with Manim.
Takeaway: Turn your head sideways and the awkward region becomes routine.
Intuition
A sideways parabola is not a function of the horizontal variable - one input gives two outputs - so vertical strips hit it twice. But it is a perfectly good function of the vertical variable.
Choosing which way to slice is the same decision a bricklayer makes about which way to lay a course: pick the direction where every strip runs from exactly one curve to exactly one other.
| Slice direction | Strip length | Integrate over |
|---|---|---|
| Vertical strips | top minus bottom | the horizontal variable |
| Horizontal strips | right minus left | the vertical variable |
Trade off
Comparison matrix
From Turn your head sideways: every row here is a choice with a cost. Fill the Strip length column, then say which row you would actually pick and what you give up for it.
| Slice direction | Strip length | Integrate over |
|---|---|---|
| Vertical strips | top minus bottom | the horizontal variable |
| Horizontal strips | right minus left | the vertical variable |
Missing information
Discussion prompt
Find the area of the region enclosed by the sideways parabola and the line.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Both curves are already solved for the horizontal variable, and each horizontal strip runs from the parabola to the line - no splitting needed.
Worked example
Find the area of the region enclosed by the sideways parabola and the line.
\[ x = y^{2} \qquad\text{and}\qquad x = y+2 \]
Slice horizontally
Why: Both curves are already solved for the horizontal variable, and each horizontal strip runs from the parabola to the line - no splitting needed.
Find the intersections in terms of the vertical variable
Why: Set the two expressions for the horizontal variable equal and solve the resulting quadratic; those heights are the limits.
\[ y^{2}=y+2 \;\Longrightarrow\; y^{2}-y-2=0 \;\Longrightarrow\; (y-2)(y+1)=0 \;\Longrightarrow\; y=-1,\,2 \]
Decide which curve is on the right
Why: At height zero the line is at 2 and the parabola is at 0, so the line is the right-hand boundary all the way up the region.
\[ \text{at } y=0: \quad y+2 = 2 \;>\; y^{2}=0 \]
Integrate right minus left
Why: Each strip has length equal to the line minus the parabola, and the strips stack from the lower intersection to the upper one.
\[ A = \int_{-1}^{2}\left[(y+2)-y^{2}\right]dy \]
Antidifferentiate and evaluate
Why: Power rule term by term in the vertical variable, then top limit minus bottom limit.
\[ \left[\frac{y^{2}}{2}+2y-\frac{y^{3}}{3}\right]_{-1}^{2} = \frac{10}{3}-\left(-\frac{7}{6}\right) = \frac{27}{6} = \frac{9}{2} \]
Check the size against a bounding box
Why: The region sits inside a box three units tall and four units wide, so its area must be well under 12; four and a half is a believable fraction of that box, and it is positive.
Picture it
Animation
Shows: Each line of the worked example "integrating with respect to y", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The region sits inside a box three units tall and four units wide, so its area must be well under 12; four and a half is a believable fraction of that box, and it is positive.
Prediction
Predict first
Which integral gives the area of the region enclosed by these two curves?
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: the integral from negative 1 to 2 of the quantity x plus 2 minus x squared, dx
Why: Setting x squared equal to x plus 2 gives x squared minus x minus 2 equal to zero, which factors to give x equal to negative 1 and x equal to 2. Testing x equal to 0 shows the line is at 2 and the parabola at 0, so the line is on top and the integrand is the line minus the parabola. The value works out to nine halves.
Check
Find where the curves meet and which one is on top, then choose the correct set-up.
\[ y = x^{2} \qquad\text{and}\qquad y = x+2 \]
Check your understanding
Which integral gives the area of the region enclosed by these two curves?
Answer: A
Why: Setting x squared equal to x plus 2 gives x squared minus x minus 2 equal to zero, which factors to give x equal to negative 1 and x equal to 2. Testing x equal to 0 shows the line is at 2 and the parabola at 0, so the line is on top and the integrand is the line minus the parabola. The value works out to nine halves.
Commit first
Predict first
What is the area of the region enclosed by these two curves?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: one third
Why: The curves meet at 0 and 1, and testing one quarter shows the square root is on top. The integral of the square root minus the square from 0 to 1 gives two thirds minus one third, which equals one third.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Both curves pass through the origin and through the point where both coordinates are 1. Find the area they enclose.
\[ y = \sqrt{x} \qquad\text{and}\qquad y = x^{2} \]
Check your understanding
What is the area of the region enclosed by these two curves?
Answer: A
Why: The curves meet at 0 and 1, and testing one quarter shows the square root is on top. The integral of the square root minus the square from 0 to 1 gives two thirds minus one third, which equals one third.
Section
Part 4
Concept
The Fundamental Theorem, read as a sentence about the real world, says this: if you integrate how fast something is changing, you get how much it changed in total.
\[ \int_{a}^{b} F'(t)\,dt = F(b)-F(a) \]
net change — The end value minus the start value of a quantity. It is what the definite integral of that quantity's rate always computes - never the total amount of movement.
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 substitution (u-substitution), top minus bottom, net change as u-Substitution and Area Between Curves uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: The integral of a rate is a net change — a rendered Manim animation.
Rendered with Manim.
Takeaway: The FTC, restated in the language of the application.
Intuition
Drive ten miles east, then ten miles back west. Your net change in position is zero - you are home. Your odometer says twenty miles.
Integrating velocity gives the first number, because velocity is negative on the way home and those negative contributions cancel the positive ones.
To get the odometer reading you must throw away the signs first, by integrating the speed - the absolute value of the velocity - instead.
Counterexample
Discussion prompt
Drive ten miles east, then ten miles back west. Your net change in position is zero - you are home. Your odometer says twenty miles.
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:
Integrating velocity gives the first number, because velocity is negative on the way home and those negative contributions cancel the positive ones.
Estimation
Predict first
A particle moves along a line with the velocity below, measured in metres per second. Find its displacement over the first three seconds.
Commit before you compute: what does Worked example: displacement of a particle 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 sign against the motion
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 velocity is negative until 2 seconds and positive after, so the particle moves backwards first and only partly returns.
Worked example
A particle moves along a line with the velocity below, measured in metres per second. Find its displacement over the first three seconds.
\[ v(t) = t^{2}-4, \qquad 0 \le t \le 3 \]
Recognize displacement as the integral of velocity
Why: Velocity is the rate of change of position, so integrating it over the interval gives the net change in position.
\[ s(3)-s(0) = \int_{0}^{3}\left(t^{2}-4\right)dt \]
Antidifferentiate
Why: Power rule on the squared term, and the constant term integrates to a linear one.
\[ \left[\frac{t^{3}}{3}-4t\right]_{0}^{3} \]
Evaluate at the limits
Why: At 3 the bracket is 9 minus 12; at 0 it is zero, so the whole value is negative 3.
\[ (9-12)-(0) = -3 \ \text{metres} \]
Verify the sign against the motion
Why: The velocity is negative until 2 seconds and positive after, so the particle moves backwards first and only partly returns. Ending 3 metres behind the start is exactly what a negative displacement should mean, and differentiating the bracket returns the velocity.
\[ \frac{d}{dt}\!\left[\frac{t^{3}}{3}-4t\right] = t^{2}-4\;\checkmark \]
Concept
Total distance travelled is the odometer reading: every metre counts, forwards or backwards. That means integrating the speed, not the velocity.
\[ \text{distance} = \int_{a}^{b}\left|v(t)\right|dt \]
You cannot antidifferentiate an absolute value directly. Instead, find where the velocity is zero, split the interval there, and integrate the velocity or its negative on each piece so that every piece comes out positive.
Picture it
Animation
Shows: Total distance strips the signs off first — a rendered Manim animation.
Rendered with Manim.
Takeaway: Going out and coming back gives zero displacement and plenty of distance.
Step zero
Discussion prompt
Worked example: total distance for the same particle — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find where the velocity changes sign
Answer:
Worked example
Same particle, same three seconds. Now find the total distance travelled.
\[ v(t) = t^{2}-4, \qquad 0 \le t \le 3 \]
Find where the velocity changes sign
Why: The direction can only reverse where the velocity is zero, and only the positive root lies in the interval.
\[ t^{2}-4 = 0 \;\Longrightarrow\; t = 2 \ \text{(in the interval)} \]
Determine the sign on each piece
Why: Test a point in each subinterval: at 1 the velocity is negative 3, at 2.5 it is positive 2.25.
| interval | test value of the velocity | direction |
|---|---|---|
| 0 to 2 | negative 3 at t equal to 1 | backwards |
| 2 to 3 | positive 2.25 at t equal to 2.5 | forwards |
Flip the sign on the backwards piece
Why: Integrating the negative of the velocity where the velocity is negative makes that contribution positive, which is what an odometer records.
\[ \text{distance} = \int_{0}^{2}\left(4-t^{2}\right)dt + \int_{2}^{3}\left(t^{2}-4\right)dt \]
Evaluate both pieces
Why: Each bracket is evaluated top limit minus bottom limit; both results come out positive, as they must.
\[ \left[4t-\frac{t^{3}}{3}\right]_{0}^{2} = \frac{16}{3}, \qquad \left[\frac{t^{3}}{3}-4t\right]_{2}^{3} = -3-\left(-\frac{16}{3}\right) = \frac{7}{3} \]
Add them
Why: Sixteen thirds backwards plus seven thirds forwards is the full path length.
\[ \text{distance} = \frac{16}{3}+\frac{7}{3} = \frac{23}{3} \approx 7.67 \ \text{metres} \]
Check it against the displacement
Why: Going seven thirds forwards and sixteen thirds backwards must give a net of seven thirds minus sixteen thirds, which is negative 3 - exactly the displacement found earlier. And the distance is larger than the size of the displacement, as it always must be.
\[ \frac{7}{3}-\frac{16}{3} = -3\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "total distance for the same particle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Going seven thirds forwards and sixteen thirds backwards must give a net of seven thirds minus sixteen thirds, which is negative 3 - exactly the displacement found earlier. And the distance is larger than the size of the displacement, as it always must be.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The student integrates the velocity across the whole interval and then makes the answer positive at the end.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Inside the integral the backwards metres subtracted from the forwards metres.
Take the absolute value inside the integral, which in practice means splitting at each sign change.
Why: Inside the integral the backwards metres subtracted from the forwards metres. Taking an absolute value afterwards cannot put back what was cancelled.
Trap
The student integrates the velocity across the whole interval and then makes the answer positive at the end.
\[ \left|\int_{0}^{3}\left(t^{2}-4\right)dt\right| = \left|-3\right| = 3 \ \text{metres} \]
The cancellation already happened
Why: Inside the integral the backwards metres subtracted from the forwards metres. Taking an absolute value afterwards cannot put back what was cancelled.
The reported distance is too small
Why: The true path is about 7.67 metres, so this answer is off by more than a factor of two.
Take the absolute value inside the integral, which in practice means splitting at each sign change.
\[ \int_{0}^{3}\left|t^{2}-4\right|dt = \int_{0}^{2}\left(4-t^{2}\right)dt + \int_{2}^{3}\left(t^{2}-4\right)dt \]
Each piece is positive before it is added
Why: No cancellation can occur once both integrands are positive on their own intervals.
\[ \frac{16}{3}+\frac{7}{3} = \frac{23}{3} \]
The reliable order of operations
Why: Absolute value first, then integrate. It is the same rule as area between curves: fix the signs before you add, never after.
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.
Pattern
| Question asked | What to compute | Sign of the answer |
|---|---|---|
| How far from the start does it end up | the integral of the velocity | any sign |
| How far did it travel | the integral of the speed | always positive |
| Which is larger | the distance, unless the motion never reverses | distance is at least the size of the displacement |
If the velocity never changes sign on the interval, the two answers agree in size - and that is the only case where they do.
Comparison
Comparison matrix
From Displacement versus total distance: refill the Sign of the answer column from what you know. The rest of the table is as it appeared.
| Question asked | What to compute | Sign of the answer |
|---|---|---|
| How far from the start does it end up | the integral of the velocity | any sign |
| How far did it travel | the integral of the speed | always positive |
| Which is larger | the distance, unless the motion never reverses | distance is at least the size of the displacement |
Prediction
Predict first
What is the total distance travelled, in metres?
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: thirteen halves
Why: The velocity is zero at 3 seconds, negative before it and positive after. The integral of 3 minus t from 0 to 3 is nine halves, and the integral of t minus 3 from 3 to 5 is 2. Adding them gives thirteen halves, or 6.5 metres.
Check
A particle moves with the velocity below, in metres per second. Find the total distance it travels over the five seconds.
\[ v(t) = t-3, \qquad 0 \le t \le 5 \]
Check your understanding
What is the total distance travelled, in metres?
Answer: A
Why: The velocity is zero at 3 seconds, negative before it and positive after. The integral of 3 minus t from 0 to 3 is nine halves, and the integral of t minus 3 from 3 to 5 is 2. Adding them gives thirteen halves, or 6.5 metres.
Concept
Any time a problem hands you a rate, the integral of that rate is the accumulated change in the underlying quantity.
| If the rate is | Then the integral gives |
|---|---|
| velocity in metres per second | change in position, in metres |
| litres per minute flowing in | litres added to the tank |
| marginal cost in dollars per unit | extra cost of producing those units |
| births per year minus deaths per year | change in the population |
Reading the units of the rate tells you the units of the answer: the units of the rate multiplied by the units of the variable you integrate over.
Comparison
Comparison matrix
From Net change is not only about motion: refill the Then the integral gives column from what you know. The rest of the table is as it appeared.
| If the rate is | Then the integral gives |
|---|---|
| velocity in metres per second | change in position, in metres |
| litres per minute flowing in | litres added to the tank |
| marginal cost in dollars per unit | extra cost of producing those units |
| births per year minus deaths per year | change in the population |
Ranking
Put in order
Put the moves of Worked example: filling a tank, with a substitution 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 amount added is the integral of the inflow rate over the time interval.
Worked example
Water flows into a tank at the rate below, in litres per minute. How much water enters during the first four minutes?
\[ r(t) = 4t\sqrt{t^{2}+9} \]
Set up the net change integral
Why: The amount added is the integral of the inflow rate over the time interval.
\[ V = \int_{0}^{4} 4t\sqrt{t^{2}+9}\,dt \]
Substitute the expression under the root
Why: Its derivative is 2t, and the integrand supplies 4t - a constant multiple, which is all substitution needs.
\[ u = t^{2}+9 \;\Longrightarrow\; du = 2t\,dt \;\Longrightarrow\; 4t\,dt = 2\,du \]
Convert the limits
Why: Feed each time value through the u-equation: at 0 minutes u is 9, and at 4 minutes u is 25.
| t in minutes | u equals t squared plus 9 |
|---|---|
| 0 | 9 |
| 4 | 25 |
Integrate in the new variable
Why: The square root is the one-half power, so the power rule gives two thirds of the three-halves power.
\[ 2\int_{9}^{25} u^{1/2}\,du = \frac{4}{3}\left[u^{3/2}\right]_{9}^{25} \]
Evaluate with exact values
Why: The three-halves power of 25 is 125 and of 9 is 27, since both bases are perfect squares.
\[ \frac{4}{3}\left(125-27\right) = \frac{4}{3}\cdot 98 = \frac{392}{3} \approx 130.7 \ \text{litres} \]
Verify by differentiating the antiderivative
Why: Differentiating four thirds of the three-halves power returns the original rate, and the answer is positive with units of litres, which is what a filling tank requires.
\[ \frac{d}{dt}\!\left[\frac{4}{3}\left(t^{2}+9\right)^{3/2}\right] = \frac{4}{3}\cdot\frac{3}{2}\left(t^{2}+9\right)^{1/2}\cdot 2t = 4t\sqrt{t^{2}+9}\;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "filling a tank, with a substitution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating four thirds of the three-halves power returns the original rate, and the answer is positive with units of litres, which is what a filling tank requires.
Pattern
Every one of these ends the same way: differentiate your antiderivative, or check the sign and size against the picture. The check is part of the method, not an optional extra.
Real world
Discussion prompt
Outside this lesson: where does u-Substitution and Area Between Curves 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 Which tool does this problem want? 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:
Substitution as the chain rule run backwards: choosing u, converting dx into du, back-substituting, and changing the limits on a definite integral. Then area between two curves (top minus bottom, and left minus right in y) and net change - displacement versus total distance.
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: With u equal to 2x plus 1, dx is half of du and the limits run from 1 to 3. Half the integral of u to the fourth is u to the fifth over ten, and 243 minus 1 over ten is 24.2, which is one hundred twenty-one fifths.
Check
The inside is linear, so this one is quick - but the limits still have to move.
\[ \int_{0}^{1}\left(2x+1\right)^{4}dx \]
Check your understanding
What is the value of this definite integral?
Answer: A
Why: With u equal to 2x plus 1, dx is half of du and the limits run from 1 to 3. Half the integral of u to the fourth is u to the fifth over ten, and 243 minus 1 over ten is 24.2, which is one hundred twenty-one fifths.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Substitution: the Chain Rule Backwards · Definite Integrals by Substitution · Area Between Two Curves · Net Change: Displacement and Distance. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Substitution is the chain rule read backwards. You name the inside, convert the differential honestly, and the integral collapses.
| Problem | First move | Final check |
|---|---|---|
| Composite integrand | name the inside u | differentiate your answer |
| Definite integral by substitution | convert both limits | compare with the back-substituted route |
| Area between curves | solve for the intersections | the answer is positive |
| Total distance | solve for where the rate is zero | distance is at least the size of the displacement |
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