The Mean Value Theorem for Integrals, the area function and Part 1 proving its derivative is the integrand, Part 2 evaluating any definite integral from an antiderivative, the chain rule for variable limits, and the net change reading.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 5 — Integration
The Fundamental Theorem of Calculus
Objectives
Five outcomes. The middle two are the theorem itself, and they change what the rest of the course can do.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 472-487 — the section these objectives are drawn from
Warm-up
Chapter 4 answered: which function has this derivative? Sections 5.1 and 5.2 answered: what is the area under this curve? Nothing so far connects them.
Discussion prompt
The area under x squared from 0 to 2 took a page of summation and came to 8 over 3. What is the antiderivative of x squared, evaluated at 2?
Hint: Compute the antiderivative and substitute.
Answer:
The antiderivative is x cubed over 3, and at 2 that is 8 over 3 — the same number the summation produced, and at 0 it is zero.
\[ \int_{0}^{2}x^{2}dx = \tfrac83 \quad \text{and} \quad \frac{x^{3}}{3}\bigg|_{x=2} = \tfrac83 \]
That could be a coincidence. It is not: it happens for every function and every interval, and this section proves it. The consequence is that the area computation of Section 5.1 — a page of summation formulas — collapses to finding an antiderivative and substituting twice.
Concept
The area accumulated under a curve, as a function of where you stop, has the curve itself as its derivative. Consequently every definite integral is computed by evaluating an antiderivative at the two endpoints and subtracting.
the Fundamental Theorem of Calculus — Part 1: differentiating an area function returns the integrand. Part 2: the definite integral of a function equals any antiderivative evaluated at the upper limit minus at the lower.
\[ \frac{d}{dx}\int_{a}^{x}f(t)\,dt = f(x), \qquad \int_{a}^{b}f(x)\,dx = F(b)-F(a) \]
The theorem earns its name by connecting the two halves of the subject. Differentiation and integration were developed to answer unrelated questions — tangents and areas — and this proves they are inverse operations.
Figure (svg): The theorem as a bridge between two apparently unrelated questions
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 488-500
Section
Section 1
Concept
For a continuous function on a closed interval there is a point where the function's value, times the interval's width, equals the integral exactly.
Mean Value Theorem for Integrals — For a continuous function on a closed interval, some interior point has function value equal to the average value — so a rectangle of that height and the interval's width has exactly the integral's area.
\[ \exists c: \; \int_{a}^{b}f(x)\,dx = f(c)(b-a) \]
Section 5.2 introduced this as the statement that a continuous function attains its average value. Here it does real work: it is the step that makes the proof of Part 1 exact rather than approximate.
Figure (svg): The Mean Value Theorem for Integrals: a rectangle of equal area touching the curve
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 488-494 — the Mean Value Theorem for Integrals
Picture it
A curve, its integral, and the height that matches.
Figure (svg): The Mean Value Theorem for Integrals: a rectangle of equal area touching the curve
The dashed height is the average value, and the marked point is where the curve attains it. Both the existence of the height and the fact that the curve reaches it need continuity.
Worked example
Example 5.20. Find the c the theorem promises.
\[ \text{For } f(x)=x^{2} \text{ on } [0,2], \text{ find } c \text{ with } f(c)(b-a) = \int_{0}^{2}f. \]
Recall the integral
Why: From Section 5.1.
\[ \frac{8}{3} \]
Write the equation
Why: Value times width.
\[ c ^{2} \times 2 = \frac{8}{3} \]
Solve for the square
Why: Divide by 2.
\[ c ^{2} = \frac{4}{3} \]
Take the positive root
Why: The interval is non-negative.
\[ c = 2 / \sqrt{3} \]
Evaluate
Why: Approximately.
\[ \text{about } 1.155 \]
Figure (svg): The Mean Value Theorem for Integrals: a rectangle of equal area touching the curve
\[ c = \frac{2}{\sqrt3} \approx 1.155 \]
Verify: confirm the point lies inside and check the area
Why: The value 1.155 is between 0 and 2, as the theorem promises. Checking the area: a rectangle of height four thirds and width 2 has area 8 over 3, matching the integral exactly. Note that the theorem asserts existence without locating the point — here it was findable because the function is simple, and in general it is not, which does not weaken the theorem's use in proofs.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 490-491
Fill the middle
The theorem's equation, with the integral known.
Fill in the blanks
f(c)\cdot 2 = \frac83 \;\Longrightarrow\; f(c) = 4/3
Why: The required height is the integral over the width, which is the average value. Finding where the function takes that value is the second step, and it is what the theorem guarantees is possible.
Worked example
Checkpoint 5.20. A function that misses its average.
\[ \text{Show the theorem can fail for a discontinuous function.} \]
Take a step function
Why: Zero then two.
\[ 0\text{ on } [0, 1], 2\text{ on } [1, 2] \]
Compute its integral
Why: One rectangle of height 2 and width 1.
\[ 2 \]
Compute the required height
Why: Integral over width.
\[ 1 \]
Look for a point with that value
Why: The function takes only 0 and 2.
Conclude
Why: The theorem fails.
Figure (svg): The solution to Worked example why continuity is needed shown as a ladder of expressions, one row per legal move
\[ f_{\text{avg}}=1 \text{ but } f \text{ takes only } 0 \text{ and } 2 \]
Verify: identify where the proof breaks
Why: The proof bounds the average between the function's minimum and maximum, then invokes the Intermediate Value Theorem to say every value between is attained. That last step needs continuity, and a step function jumps straight past the intermediate values. So the failure is precise and traceable rather than mysterious — and it is the same dependency Section 4.4's Mean Value Theorem has.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 491-493
Trap
\[ c = \frac{a+b}{2} = 1 \]
Take the interval's midpoint
Why: The student assumes the average is attained halfway.
\[ f(1) = 1 \ne \tfrac43 \]
The midpoint's value is 1, not four thirds. The theorem locates a point by its VALUE, not by its position.
\[ f(c) = \tfrac43 \;\Longrightarrow\; c = \tfrac{2}{\sqrt3} \approx 1.155 \]
Solve for the value, then find where the function takes it
Why: Which input gives it depends entirely on the function.
The midpoint happens to work for a linear function, which is probably where the intuition comes from. For anything curved it does not, and the parabola here shows the point sitting well to the right of centre.
Ranking
The Mean Value Theorem for Integrals.
Put in order
Why: Step d is the one that needs continuity, which is why a step function is a counterexample. The bounding in step b comes straight from Section 5.2's properties, doing genuine work rather than sitting decoratively in a list.
Two truths and a lie
All three are about the theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The midpoint works for a linear function and generally not otherwise — for x squared on the interval from 0 to 2 the point is about 1.155, well right of the midpoint at 1. The theorem specifies a value, not a position.
Prediction
Commit before reasoning.
Predict first
Section 5.2 already stated this. Why is it restated at the start of Section 5.3?
Correct: Because it makes Part 1's proof exact.
\[ \int_{x}^{x+h}f = f(c)\,h \text{ for some } c\in[x,x+h] \]
Why: Part 1's argument says that widening the interval by a small amount h adds a sliver of area that is almost a rectangle of height f of x. This theorem replaces almost with exactly: the sliver's area IS the function's value at some point inside times h. Continuity then forces that point toward x as h shrinks, and the limit is f of x. Without it the argument would be a picture rather than a proof.
Section
Section 2
Concept
Fix the left end and let the right end move. The area accumulated is a function of the right end, and its derivative is the original function evaluated there.
the area function — The integral from a fixed left endpoint to a variable upper limit, regarded as a function of that limit. Part 1 states its derivative is the integrand evaluated at the upper limit.
\[ F(x)=\int_{a}^{x}f(t)\,dt \;\Longrightarrow\; F'(x)=f(x) \]
The letter inside must differ from the upper limit. The variable of integration is a placeholder swept over the interval while x names where the sweep stops, and using x for both makes the statement unreadable.
Figure (svg): The area function: area accumulated from a fixed left end to a moving right end
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 494-502 — Fundamental Theorem, Part 1
Picture it
Area accumulated as a function of where you stop.
Figure (svg): The area function: area accumulated from a fixed left end to a moving right end
As the right edge moves right, the shaded area grows — and the rate at which it grows is exactly the curve's height at the edge, which is what Part 1 asserts.
Worked example
Example 5.22. Part 1 applied directly.
\[ \text{Find } F'(x) \text{ for } F(x)=\int_{1}^{x}\sqrt{t^{3}+1}\,dt. \]
Identify the integrand
Why: The function inside.
\[ \sqrt{t ^{3} + 1} \]
Check the form
Why: Constant lower limit, x as the upper.
\[ \text{Part } 1\text{ applies directly} \]
Apply Part 1
Why: Substitute x for the placeholder.
\[ \sqrt{x ^{3} + 1} \]
Note what was not needed
Why: No antiderivative was found.
State
Why: The derivative.
\[ \sqrt{x ^{3} + 1} \]
Figure (svg): The area function: area accumulated from a fixed left end to a moving right end
\[ F'(x)=\sqrt{x^{3}+1} \]
Verify: notice what makes this remarkable
Why: The integrand has no elementary antiderivative — there is no formula in the standard functions whose derivative is the square root of x cubed plus one. So F cannot be written down in closed form at all, and yet its derivative is known exactly and immediately. Part 1 gives information about a function that cannot be expressed, which is why it is used to DEFINE functions such as the error function and the logarithmic integral.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 496-497
Fill the middle
An area function with a constant lower limit.
Fill in the blanks
F(x)=\int_\sqrt{x^3+1}^___\sqrt___+1}\,dt \;\Longrightarrow\; F'(x)=___
Why: Part 1 says the derivative is the integrand at the upper limit. No antiderivative was needed, and for this integrand none exists in elementary terms.
Worked example
Checkpoint 5.22. The proof in five steps.
\[ \text{Prove } \frac{d}{dx}\int_{a}^{x}f(t)\,dt = f(x) \text{ for continuous } f. \]
Write the difference quotient
Why: The definition of the derivative.
\[ \frac{[F(x + h) - F(x)]}{h} \]
Simplify by additivity
Why: The two areas differ by a sliver.
\[ (\frac{1}{h}) \int\text{ from } x\text{ to } x + h\text{ of } f \]
Apply the Mean Value Theorem for Integrals
Why: The sliver is exactly a rectangle.
\[ (\frac{1}{h}) f(c) h = f(c) \]
Note where c lies
Why: Between x and x + h.
Take the limit using continuity
Why: f(c) tends to f(x).
\[ F'(x) = f(x) \]
Figure (svg): Why the area function's derivative is the integrand: a thin sliver
\[ F'(x)=\lim_{h\to0}f(c)=f(x) \]
Verify: identify where each hypothesis was used
Why: Additivity came from Section 5.2's properties; the Mean Value Theorem for Integrals turned the sliver into an exact rectangle; and continuity was used twice — once inside that theorem, and once at the end to conclude that f of c approaches f of x as c is squeezed toward it. Every hypothesis does work, and dropping continuity breaks the argument in two separate places.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 497-499
Error analysis
A student writes an area function.
Annotate
On: \( F(x)=\int_{a}^{x}f(x)\,dx \)
The clash is not merely stylistic: with one letter doing both jobs the statement of Part 1 becomes unreadable, since it is impossible to say what is being substituted where.
Ranking
From the difference quotient to the result.
Put in order
Why: Step c is where the previous idea pays off, turning an approximate picture into an exact equation. Continuity is used in step e and also inside step c, which is why the theorem needs it so firmly.
Sorting
It needs a constant lower limit and a bare x above.
Sort into buckets
Sort each expression.
The last case needs everything: split at a constant, reverse the lower piece, and apply the chain rule to both. It is the next idea's business, and all of it reduces to Part 1 plus rules already known.
Prediction
Commit before reasoning.
Predict first
The worked example differentiated an integral whose antiderivative cannot be written down. Why does that matter?
Correct: Because it gives exact information about functions with no closed form.
\[ \text{erf}(x)=\frac{2}{\sqrt\pi}\int_{0}^{x}e^{-t^{2}}dt \;\Longrightarrow\; \text{erf}'(x)=\frac{2}{\sqrt\pi}e^{-x^{2}} \]
Why: Many important functions are defined precisely as integrals — the error function of statistics, the logarithmic integral of number theory — because no elementary formula exists for them. Part 1 hands their derivatives over immediately, so they can be analysed with all of Chapter 4's machinery despite never being written explicitly. That is a far deeper consequence than convenience.
Section
Section 3
Concept
When the upper limit is a function of x rather than x itself, Part 1 combines with the chain rule: the integrand at that function, times the function's derivative. A variable lower limit contributes a minus sign.
variable limits of integration — When a limit of integration is a function of x, differentiating the integral requires the chain rule; when the variable is in the lower limit, the reversal convention supplies a minus sign.
\[ \frac{d}{dx}\int_{a}^{u(x)}f = f(u(x))\,u'(x) \]
Nothing new is being introduced. Part 1 handles the bare case, the chain rule of Section 3.6 handles the composition, and Section 5.2's reversal convention handles the lower limit.
Figure (svg): The chain rule applied when the upper limit is itself a function
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 498-505 — variable limits of integration
Picture it
Each built from Part 1 plus something already known.
Figure (svg): The chain rule applied when the upper limit is itself a function
The fourth case is the general one, and it is handled by splitting at any constant and treating each piece by the rules above. No new theorem is needed at any point.
Worked example
Example 5.24. Part 1 with the chain rule.
\[ \text{Find } \frac{d}{dx}\int_{1}^{x^{2}}\cos t\,dt. \]
Name the outer function
Why: The area function of the upper limit.
\[ F(u)\text{ with } u = x ^{2} \]
Apply Part 1 to F
Why: Its derivative is the integrand.
\[ F'(u) = \cos u \]
Apply the chain rule
Why: Times the inner derivative.
\[ \cos(u) \times u' \]
Substitute the inner function
Why: And its derivative.
\[ \cos(x ^{2}) \times 2 x \]
State
Why: The derivative.
\[ 2 x \cos(x ^{2}) \]
Figure (svg): The chain rule applied when the upper limit is itself a function
\[ \frac{d}{dx}\int_{1}^{x^{2}}\cos t\,dt = 2x\cos(x^{2}) \]
Verify: check by computing the integral explicitly
Why: Since cosine does have an elementary antiderivative, this can be checked directly: the integral is sine of x squared minus sine of 1, and differentiating by the chain rule gives cosine of x squared times 2x — matching. The check is available here only because the integrand is simple, but it confirms the method for the many cases where no such check exists.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 500-501
Fill the middle
An integral whose upper limit is a square.
Fill in the blanks
\frac2x___\int____^___}\cos t\,dt = \cos(x^___)\cdot___
Why: The upper limit is a function, so its derivative multiplies. Omitting the factor is the section's commonest error, and the same one recurs throughout Section 5.5.
Worked example
Checkpoint 5.24. Split at a constant.
\[ \text{Find } \frac{d}{dx}\int_{x}^{x^{3}}e^{t^{2}}dt. \]
Split at any constant
Why: Say zero.
\[ \int\text{ from } x\text{ to } 0\text{ plus } \int\text{ from } 0\text{ to } x ^{3} \]
Reverse the first
Why: Swapping limits flips the sign.
\[ -\int\text{ from } 0\text{ to } x\text{ plus } \int\text{ from } 0\text{ to } x ^{3} \]
Differentiate the first
Why: Part 1.
\[ -e ^{x ^{2}} \]
Differentiate the second
Why: Part 1 and the chain rule.
\[ e ^{x ^{6}} \times 3 x ^{2} \]
Combine
Why: Add.
\[ 3 x ^{2} e ^{x ^{6}} - e ^{x ^{2}} \]
Figure (svg): The solution to Worked example both limits varying shown as a ladder of expressions, one row per legal move
\[ 3x^{2}e^{x^{6}}-e^{x^{2}} \]
Verify: check the structure of the answer
Why: The pattern is the integrand at the upper limit times that limit's derivative, minus the same at the lower limit — which is the general rule, and it looks exactly like the evaluation formula of Part 2 with chain-rule factors attached. That resemblance is not accidental: both come from the same theorem. Note that the choice of splitting constant was arbitrary and cancels out, since a different choice changes both pieces by equal and opposite amounts.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 501-503
Trap
\[ \frac{d}{dx}\int_{1}^{x^{2}}\cos t\,dt = \cos(x^{2}) \]
Substitute the upper limit and stop
Why: The student applies Part 1 as though the limit were bare x.
The upper limit is a function, so the chain rule contributes its derivative as a factor. The answer is short by a factor of 2x.
\[ = \cos(x^{2})\cdot 2x \]
Substitute the upper limit AND multiply by its derivative
Why: Part 1 states the bare case; a composition needs the chain rule too.
The check that catches this is the one from Chapter 3: if the inside is anything but a bare x, a chain-rule factor is owed. The same omission is the commonest error in Section 5.5's substitution method.
Matching
Part 1 with each complication.
Match the pairs
Why: The last is the general rule and it contains the other three as special cases. Its shape — upper contribution minus lower contribution — mirrors Part 2's evaluation formula, which is no coincidence.
Ranking
Both limits functions of x.
Put in order
Why: The splitting constant is genuinely arbitrary: choosing a different one changes both pieces by equal and opposite amounts. Everything here is Part 1 plus the reversal convention plus the chain rule — no new theorem is involved.
Prediction
Commit before reasoning.
Predict first
Differentiating an integral whose LOWER limit is x produces a minus sign. Why?
Correct: Because reversing the limits flips the sign.
\[ \int_{x}^{b}f = -\int_{b}^{x}f \;\Longrightarrow\; \frac{d}{dx} = -f(x) \]
Why: Part 1 is stated with the variable on top, so an integral with x below must first be reversed — and Section 5.2's convention says that negates it. Geometrically it also makes sense: moving the LEFT edge rightward removes area rather than adding it, so the accumulated amount decreases. The algebra and the picture agree.
Section
Section 4
Concept
Any antiderivative of the integrand, evaluated at the upper limit minus at the lower, gives the definite integral. The arbitrary constant cancels, so any antiderivative will do.
the evaluation theorem — The definite integral of a continuous function equals any antiderivative evaluated at the upper limit minus its value at the lower limit.
\[ \int_{a}^{b}f(x)\,dx = F(b)-F(a), \quad F'=f \]
The constant cancels because it appears in both terms with opposite signs. That is why the plus C, essential in Section 4.10, is simply omitted when evaluating a definite integral.
Figure (svg): Part 2 in use: the whole computation of Section 5.1, in two lines
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 502-510 — Fundamental Theorem, Part 2
Picture it
The same area, computed both ways.
Figure (svg): Part 2 in use: the whole computation of Section 5.1, in two lines
The left column works only for functions with a summation formula; the right works for every function with a known antiderivative, which is an incomparably larger class.
Worked example
Example 5.26. Section 5.1's page of work, replaced.
\[ \text{Evaluate } \int_{0}^{2}x^{2}\,dx \text{ using Part 2.} \]
Find an antiderivative
Why: The reversed power rule.
\[ x ^{3} / 3 \]
Evaluate at the upper limit
Why: Substitute 2.
\[ \frac{8}{3} \]
Evaluate at the lower limit
Why: Substitute 0.
\[ 0 \]
Subtract
Why: Upper minus lower.
\[ \frac{8}{3} \]
Note the constant was unnecessary
Why: It would cancel.
Figure (svg): Part 2 in use: the whole computation of Section 5.1, in two lines
\[ \int_{0}^{2}x^{2}dx = \frac{x^{3}}{3}\bigg|_{0}^{2} = \frac83 \]
Verify: confirm the constant genuinely cancels
Why: Using x cubed over 3 plus 7 as the antiderivative gives 8 over 3 plus 7, minus 0 plus 7 — and the sevens cancel, leaving 8 over 3. So any antiderivative gives the same answer, which is why the constant is dropped for definite integrals. Compare with Section 5.1: five steps of summation formulas replaced by two substitutions, and this method works for every integrand with a known antiderivative.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 504-505
Fill the middle
An antiderivative evaluated at both limits.
Fill in the blanks
\frac0}___\bigg|____^___ = \frac83 - ___
Why: Upper limit minus lower limit. The value at 0 is zero here, which is why the answer is simply 8 over 3 — but the subtraction must still be written, since it is not always so convenient.
Worked example
Checkpoint 5.26. Signed area, computed quickly.
\[ \text{Evaluate } \int_{0}^{2\pi}\sin x\,dx \text{ and the total area.} \]
Find an antiderivative
Why: Section 4.10's table.
\[ -\cos x \]
Evaluate at the upper limit
Why: Cosine of 2pi is 1.
\[ -1 \]
Evaluate at the lower limit
Why: Cosine of 0 is 1.
\[ -1 \]
Subtract
Why: Upper minus lower.
\[ 0 \]
For total area, split at pi
Why: Two pieces, magnitudes added.
\[ 2 + 2 = 4 \]
Figure (svg): The solution to Worked example an integral with a sign change shown as a ladder of expressions, one row per legal move
\[ \int_{0}^{2\pi}\sin x\,dx = 0, \quad \text{area} = 4 \]
Verify: confirm both against Section 5.2's geometric argument
Why: Section 5.2 obtained the same two numbers from the picture alone, without any antiderivative. So Part 2 reproduces the geometric reasoning and does it faster — but it does not remove the need to think about signs. The integral still gives signed area, and splitting at the zeros is still required for a total. The theorem changes how integrals are computed, not what they mean.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 506-507
Error analysis
A student evaluates a definite integral.
Annotate
On: \( \int_{1}^{3}2x\,dx = x^{2}\bigg|_{1}^{3} = 1 - 9 = -8 \)
The sign check is the fastest guard. A positive integrand over an interval with the limits in the usual order must give a positive answer, so a negative result signals the subtraction was reversed.
Sorting
Differentiating an integral, or evaluating one?
Sort into buckets
Sort each task.
The two parts point in opposite directions: Part 1 differentiates an integral, Part 2 integrates using a derivative. Knowing which is wanted is usually clear from whether the answer should be a function or a number.
Two truths and a lie
All three are about Part 2.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Part 2 changes how an integral is computed, not what it means: the answer is still net signed area, so a total area still requires splitting at the integrand's zeros. The sine over a full period still evaluates to zero, quickly.
Prediction
Commit before reasoning.
Predict first
Section 4.10 insisted the constant was essential. Why is it dropped here?
Correct: Because it cancels in the subtraction.
\[ [F(b)+C]-[F(a)+C] = F(b)-F(a) \]
Why: Evaluating F plus C at the upper limit and subtracting F plus C at the lower gives F of b plus C minus F of a minus C, and the constants cancel exactly. It was genuinely essential in Section 4.10, where the answer was a family of functions and the constant was the degree of freedom an initial condition consumed. Here the answer is a number and the constant has nowhere to go.
Section
Section 5
Concept
Read the evaluation formula with the integrand as a derivative: integrating a rate of change over an interval gives the net change in the quantity. That reading is what makes the theorem useful far beyond geometry.
the net change interpretation — Since the integrand can always be regarded as some function's derivative, the definite integral of a rate of change over an interval equals the total change in the quantity over that interval.
\[ \int_{a}^{b}F'(x)\,dx = F(b)-F(a) \]
This is why velocity integrates to displacement, power to energy and flow rate to volume. Section 5.4 develops the reading into a method and calls it the Net Change Theorem.
Figure (svg): The theorem read as net change: a rate integrated gives the total change
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 506-514 — interpreting the theorem
Picture it
Four pairs of rate and total.
Figure (svg): The theorem read as net change: a rate integrated gives the total change
Every row is the same theorem with different words. The formula was proved once and applies wherever a quantity and its rate of change both appear.
Worked example
Example 5.28. The rate reading applied.
\[ \text{A particle has velocity } v(t)=t^{2}-4 \text{ m/s. Find its displacement over } [0,3]. \]
Recognise the reading
Why: Velocity is the position's derivative.
Find an antiderivative
Why: The reversed power rule.
\[ t ^{3} / 3 - 4 t \]
Evaluate at 3
Why: Substitute.
\[ 9 - 12 = -3 \]
Evaluate at 0
Why: Substitute.
\[ 0 \]
Subtract
Why: Upper minus lower.
\[ -3\text{ metres} \]
Figure (svg): The theorem read as net change: a rate integrated gives the total change
\[ \int_{0}^{3}(t^{2}-4)\,dt = -3 \]
Verify: interpret the sign and contrast with distance
Why: The negative displacement means the particle ends up 3 metres to the LEFT of where it started. It moved backwards for the first two seconds, when the velocity was negative, and forwards afterwards — but not far enough to make up the ground. The total distance travelled is a different and larger number, requiring a split at t equal to 2 where the velocity changes sign, and it comes to about 7.67 metres.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 508-510
Matching
The net change reading.
Match the pairs
Why: Every pair is the same theorem in different words. Recognising the pattern is what lets one proof serve mechanics, hydraulics, electrical engineering and economics without any adaptation.
Worked example
Checkpoint 5.28. What the theorem unified.
\[ \text{Why does this theorem carry that name?} \]
Recall the two origins
Why: Tangent problems and area problems.
Note they had no visible link
Why: One is local, one is global.
State what the theorem says
Why: They are inverse operations.
Note the computational consequence
Why: Areas from antiderivatives.
Note the conceptual consequence
Why: One subject, not two.
Figure (svg): The solution to Worked example why it is called fundamental shown as a ladder of expressions, one row per legal move
\[ \text{differentiation and integration are inverse} \]
Verify: appreciate the historical scale of the surprise
Why: Archimedes computed areas by exhaustion around 250 BC and Fermat was finding tangents in the 1630s; the two traditions ran for nearly two millennia with no suspicion of a connection. Newton and Leibniz established the link in the 1660s and 1670s, and it is what turned a collection of techniques into a subject. The name is not marketing: without this theorem the two halves of the course would remain unrelated.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 510-512
Trap
\[ \int_{0}^{3}(t^{2}-4)\,dt = -3 \;\Longrightarrow\; \text{it travelled } 3 \text{ m} \]
Read the integral as a distance
Why: The student ignores the sign's meaning.
A distance cannot be negative, and the particle actually covered about 7.67 metres — it went backwards, then forwards.
\[ \text{distance} = \int_{0}^{3}|v(t)|\,dt \approx 7.67 \]
Split where the velocity changes sign and add magnitudes
Why: Displacement is signed; distance is not.
This is Section 5.2's signed-against-total distinction in physical clothing. The theorem computes either, but only after the question is settled — and the two answers here differ by more than a factor of two.
Fill the middle
The evaluation formula, with the integrand read as a derivative.
Fill in the blanks
\int_F(a)^___F'(x)\,dx = F(b)-___
Why: Integrating a rate of change over an interval gives the net change in the quantity. It is Part 2 with the letters read differently, and it is why the theorem reaches so far beyond geometry.
Sorting
The distinction survives the theorem.
Sort into buckets
Sort each quantity by what the plain integral gives.
Part 2 made the computation quick but changed nothing about this distinction. Asking which question is being posed remains the first step, exactly as in Section 5.2.
Prediction
Commit before reasoning.
Predict first
What does the theorem's name refer to?
Correct: That it unifies two independently developed subjects.
\[ \text{tangents (local)} \;\longleftrightarrow\; \text{areas (global)} \]
Why: Area problems date to Archimedes and tangent problems to the seventeenth century, and the two traditions had no known connection. This theorem proves they are inverse operations, which both collapses the computation of areas to antidifferentiation and makes calculus one subject rather than two. Frequency of use is a consequence, not the reason for the name.
Comparison
Fill the blanks. They point in opposite directions.
Comparison matrix
| Part 1 | Part 2 | |
|---|---|---|
| Does what | differentiates an integral | evaluates an integral |
| Input | an integral with a variable limit | an integrand and two numbers |
| Output | a function | a number |
| Needs | no antiderivative at all | an antiderivative |
The last row is the practical difference. Part 1 works even when no antiderivative can be written down, which is why functions like the error function are defined by integrals in the first place.
Pattern
Given an integral to differentiate or evaluate.
Step five survives the theorem intact. Part 2 made the arithmetic fast but the integral still returns net signed area, and asking which question was posed remains the first thing to settle.
Stewart, Calculus: Early Transcendentals 8e, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 392-401
Check
Part 1.
Check your understanding
What is the derivative of the integral from 1 to x of the square root of t cubed plus one?
Answer: A
Why: Part 1 substitutes the upper limit into the integrand.
Check
Variable limits.
Check your understanding
What is the derivative of the integral from 1 to x^2 of cos t?
Answer: A
Why: Part 1 gives the cosine at the upper limit, and the chain rule contributes 2x.
Check
Part 2.
Check your understanding
Why can the arbitrary constant be dropped when evaluating a definite integral?
Answer: A
Why: F(b) + C minus F(a) + C leaves F(b) - F(a).
Real world
A hospital tracks a patient's kidney function by an infusion whose concentration in the blood is modelled continuously. The clinician needs total drug exposure over six hours, and the pharmacokinetic model gives concentration as a function of time.
Discussion prompt
Explain which part of the theorem applies, what quantity the integral gives, and why the derivative reading also matters here.
Hint: Exposure is concentration accumulated over time.
Answer:
Total exposure is the integral of concentration over the six hours — the quantity clinicians call the area under the curve, and the name is literal. Part 2 computes it: find an antiderivative of the concentration model and evaluate at 6 and 0.
\[ \text{AUC} = \int_{0}^{6}C(t)\,dt = F(6)-F(0) \]
Part 1 matters too, for a different reason. Define the accumulated exposure from time zero to time x as a function of x. Part 1 says its derivative is the concentration at time x — so the rate at which exposure accumulates is exactly the current concentration, which is what makes the running total meaningful to monitor.
That reading has a practical consequence: exposure accumulates fastest when concentration peaks, so the peak is where the accumulated total is most sensitive to timing errors. A dose given an hour late changes the running total most during the peak window.
Note that pharmacokinetic models often have no elementary antiderivative, in which case Part 2 is unavailable and the numerical methods of Section 5.1 are used on measured data instead — which is exactly the fallback that section's transfer problem described.
Commit first
Answer, then rate your confidence honestly.
Predict first
What does Part 1 give that Part 2 cannot?
Correct: The derivative of an integral with no elementary antiderivative.
\[ \frac{d}{dx}\int_{0}^{x}e^{-t^{2}}dt = e^{-x^{2}} \quad \text{though } \int e^{-x^{2}}dx \text{ has no elementary form} \]
Why: Part 2 requires an antiderivative to be found, so it stalls when none can be written in elementary terms. Part 1 needs none at all: it hands over the derivative of the area function immediately, whatever the integrand. That is why functions such as the error function and the logarithmic integral are defined as integrals and analysed with Part 1 — they cannot be written any other way, and Part 1 makes them tractable regardless.
Explain it
They cannot see why finding an antiderivative should have anything to do with area.
Discussion prompt
In four sentences or fewer, give them the idea.
Hint: Think about area accumulating as you sweep right.
Answer:
Ask them to imagine sweeping right under a curve and watching the shaded area grow. When the curve is high the area grows quickly, and when it is low the area grows slowly — so the RATE at which area accumulates is the curve's height.
That is exactly the statement that the area function's derivative is the curve. Reading it backwards, an antiderivative tells you the accumulated area, and subtracting its values at the two ends gives the area between them.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the Mean Value Theorem, remember it specifies a value and not a position. For Part 1, keep the inside letter different from the limit. For variable limits, ask whether the limit is a bare x and attach a factor if not. For Part 2, check the sign against the integrand before reporting. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw two boxes labelled with the tangent question and the area question, an arrow between them both ways, and the evaluation formula beneath as the bridge. Below, sketch a curve with area shaded from a fixed left end to a moving right end, mark the right end with a variable, and write the area function with the correct choice of letters. Beside it, draw the sliver picture for the proof of Part 1 and write the five steps in the margin, marking where continuity is used. In the middle of the page, write the four variable-limit cases with their derivatives, circling where each chain-rule factor and minus sign comes from. In the lower half, evaluate the parabola's area both ways side by side — the Section 5.1 summation in five lines and the Part 2 evaluation in two — and write one sentence on what the comparison shows. At the bottom, write the net change reading with four rate-and-total pairs beside it.
If your area-function sketch uses the same letter inside the integral as on the limit, redraw it — that clash is exactly what makes the statement of Part 1 impossible to read, and getting it right on paper fixes it for good.
Recap
Five things, and the middle two changed what the rest of the course can compute.
| If you see | Then |
|---|---|
| An integral to differentiate | Part 1: substitute the upper limit |
| A function as a limit | Attach its derivative as a factor |
| The variable in the lower limit | Reverse, and pick up a minus |
| A definite integral to evaluate | Part 2: antiderivative, upper minus lower |
| A plus C in a definite integral | Drop it: it cancels |
| A rate of change integrated | The answer is the net change |
| A total rather than a net wanted | Split at the integrand's zeros first |
Section 5.4 collects the basic integration formulas and develops the net change reading into a method, applying it to motion, growth and accumulated quantities of every kind.
OpenStax Calculus Volume 1, §5.3 The Fundamental Theorem of Calculus §5.3, pp. 472-487 — everything on these slides traces back here
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