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The Fundamental Theorem of Calculus

Slopes and areas are two sides of one coin.

Derivatives measure how fast something changes. Integrals add up tiny pieces into a total. They look like separate problems, but they undo each other. The rate at which area under a curve grows is the height of the curve, and that one fact turns hard limits of sums into quick subtraction.

On this page: accumulation functions with a graph you can drag, both parts of the theorem and why they are true, evaluating integrals, derivatives of integrals with variable limits, net change and average value, six worked examples, nine practice problems and an eight-question quiz.

Drag the area Jump to practice

As x slides right, the shaded area A(x) under sin t grows while the curve is above the axis and shrinks once it dips below. The height f(x) is the rate of change of A.

Section IAccumulation functions

Fix a starting point a and let the right edge x move. The signed area under f from a to x is a new function of x, the accumulation function:

A(x) = ∫_a^x f(t) dt

The letter t is just a placeholder inside the integral, so that x can name the moving edge. Pieces above the axis add to A; pieces below subtract.

This explorer needs JavaScript turned on. Here is what it shows. For f(t) = 2 − t from 0, A(x) = 2x − x²/2: A rises to its peak A(2) = 2 where f crosses zero, then falls back to A(4) = 0. For f(t) = sin t from 0, A(x) = 1 − cos x: A(π) = 2 and A(2π) = 0. For f(t) = t² − 1 from 0, A(x) = x³/3 − x: A falls to its lowest value, −2/3, at x = 1, and A(2.5) ≈ 2.708333. In every case the slope of A at x equals the height f(x).

Things to notice: A increases wherever f is positive and decreases wherever f is negative. Where f crosses zero, A has a peak or a valley. The steeper the bottom graph, the taller the top one.

Section IIThe Fundamental Theorem, Part 1

Part 1

If f is continuous on a ≤ x ≤ b, then A(x) = ∫_a^x f(t) dt is differentiable there, and

A′(x) = f(x)

Why it is true. Nudge the right edge from x to x + h. The area grows by a thin strip. Its width is h and its height is almost exactly f(x), so

A(x + h) − A(x) ≈ f(x) · h

Divide by h: the difference quotient of A is about f(x). As h → 0 the strip becomes a perfect rectangle, and the approximation becomes exact. In words: the rate at which area accumulates is the height of the curve.

Moving the edge from x to x + h adds a strip of width h and height about f(x). So A(x + h) − A(x) ≈ f(x)·h, and dividing by h gives A′(x) = f(x).

So every continuous function has an antiderivative: its own accumulation function. Integration and differentiation undo each other.

Section IIIThe Fundamental Theorem, Part 2

Part 2

If f is continuous on a ≤ x ≤ b and F is any antiderivative of f, then

∫_a^b f(x) dx = F(b) − F(a)

Why it is true. By Part 1, A(x) = ∫_a^x f(t) dt is an antiderivative of f. Two antiderivatives of the same function differ by a constant, so F(x) = A(x) + C. Then

F(b) − F(a) = (A(b) + C) − (A(a) + C) = A(b) − A(a) = A(b)

because A(a) = ∫_a^a f(t) dt = 0. And A(b) is exactly the integral from a to b. The constant C always cancels, so any antiderivative works.

We write the subtraction with brackets: [F(x)]_a^b = F(b) − F(a).

Section IVEvaluating definite integrals

Part 2 turns a limit of Riemann sums into three steps:

Find an antiderivative F of the integrand (no + C needed).
Evaluate it at the upper limit and at the lower limit.
Subtract: upper minus lower.

For example, ∫₀² x² dx = [x³/3]₀² = 8/3 − 0 = 8/3. That is the exact area the Riemann sums in the Integrals unit were closing in on. The answer is still a signed area: parts below the axis subtract, just as before.

Section VDerivatives of integrals with variable limits

Part 1 handles an upper limit of plain x. When the upper limit is a function g(x), combine Part 1 with the chain rule:

ddx ∫_a^g(x) f(t) dt = f(g(x)) · g′(x)

When x is in the lower limit, flip the limits first, which changes the sign. With variables in both limits, split at a constant c:

ddx ∫_g(x)^h(x) f(t) dt = f(h(x)) h′(x) − f(g(x)) g′(x)

The constant starting point never shows up in the answer, because shifting where the area starts only adds a constant to A.

Section VINet change and average value

Part 2 has a direct reading for any quantity Q with rate of change Q′:

∫_a^b Q′(t) dt = Q(b) − Q(a)

The integral of a rate is the net change. Integrate velocity and you get displacement; integrate a flow rate and you get the volume added. For total distance travelled, integrate speed |v(t)| instead, so backwards motion counts too.

The average value of f on a ≤ x ≤ b is the height of the rectangle with the same width and the same signed area:

f_avg = 1b − a ∫_a^b f(x) dx

If f is continuous, it actually reaches its average at some point c in the interval (the Mean Value Theorem for integrals).

Section VIIWorked examples

Example 1: evaluating an integral

Evaluate ∫₁³ (3x² − 2x) dx.

F(x) = x³ − x²Power rule, term by termF(3) = 27 − 9 = 18; F(1) = 1 − 1 = 0Upper and lower limits18 − 0 = 18Answer

✓ Numerical check: a midpoint sum with n = 1000 gives 17.999998, within 0.000002 of 18.

Example 2: Part 1

Find d over dx ∫₂^x √(1 + t³) dt.

The integrand is continuous for t ≥ −1Part 1 applies near any x > −1√(1 + x³)Answer

✓ This integral has no simple formula, yet its derivative is easy. At x = 2, a difference quotient with h = 0.0001 gives 3.000, matching √9 = 3.

Example 3: chain rule with a variable limit

Find d over dx ∫₀^x² cos t dt.

Upper limit g(x) = x², g′(x) = 2xIdentify the inside functioncos(x²) · 2xAnswer

✓ Check: the integral is sin(x²), and (sin(x²))′ = 2x cos(x²).

Example 4: variables in both limits

Find d over dx ∫_x^x² 1 over t dt for x > 0.

1x² · 2x − 1x · 1Upper term minus lower term2x − 1x = 1xAnswer

✓ Check: the integral is ln(x²) − ln x = ln x, whose derivative is 1/x.

Example 5: displacement and distance

A particle moves with velocity v(t) = t² − 4t + 3 m/s for 0 ≤ t ≤ 3. Find its displacement and the total distance travelled.

Antiderivative: t³/3 − 2t² + 3tPosition, up to a constant∫₀³ v dt = 9 − 18 + 9 = 0 mDisplacement: back where it startedv = (t − 1)(t − 3): positive on 0 to 1, negative on 1 to 3Sign of the velocity∫₀¹ v dt = 4/3; ∫₁³ v dt = −4/3Each legDistance = 4/3 + 4/3 = 8/3 ≈ 2.667 mAnswer

✓ A midpoint sum of |v(t)| with n = 1000 gives 2.666667.

Example 6: average value

Find the average value of f(x) = x² on 0 ≤ x ≤ 3, and where it is reached.

13 ∫₀³ x² dx = 13 · 9 = 3Integral ÷ widthx² = 3 at x = √3 ≈ 1.732The point c

✓ The rectangle of height 3 and width 3 has area 9, the same as the area under the curve.

Section VIIICommon mistakes

Subtracting in the wrong order. It is F(upper) − F(lower), even when the upper limit is smaller.
Forgetting the chain rule. d/dx of an integral up to x² is f(x²) · 2x, not just f(x²).
Dropping the sign for a lower limit. x in the lower limit gives −f(x).
Using Part 2 across a break. The integrand must be continuous on the whole interval: 1/x² from −1 to 1 has no finite area.
Confusing displacement with distance. Integrate |v| for distance.

Section IXPractice problems

Try each one on paper first, then open the answer to check your steps.

Evaluate ∫₀² (x³ + 1) dx. FTC Part 2
Show answer and steps
F(x) = x⁴/4 + xAn antiderivativeF(2) − F(0) = (4 + 2) − 0 = 6Answer
✓ A midpoint sum with n = 1000 gives 5.999998, within 0.000002 of the exact answer.
Evaluate ∫₀^π/2 cos x dx. FTC Part 2
Show answer and steps
F(x) = sin xSince (sin x)′ = cos xsin(π/2) − sin 0 = 1 − 0 = 1Answer
✓ The region is one quarter of a cosine arch, all above the axis, so a positive answer makes sense.
Evaluate ∫₁⁴ √x dx. FTC Part 2
Show answer and steps
F(x) = ⅔ x^3/2Power rule with n = ½⅔(8 − 1) = 14/3 ≈ 4.666667Answer
✓ 4^3/2 = 8 and 1^3/2 = 1. A midpoint sum with n = 1000 gives 4.666667.
Evaluate ∫₁^e 1 over x dx. FTC Part 2
Show answer and steps
F(x) = ln xFor x > 0ln e − ln 1 = 1 − 0 = 1Answer
✓ The area under 1/x from 1 to e is exactly 1, which is one way to define e.
Find d over dx ∫₁^x (t² + 1)⁵ dt. FTC Part 1
Show answer and steps
The integrand is continuous and the upper limit is xPart 1 applies directly(x² + 1)⁵Answer
✓ No need to expand or integrate: replace t with x.
Find d over dx ∫₀^x³ sin t dt. Chain rule
Show answer and steps
Upper limit u = x³, du/dx = 3x²The limit is a function of xsin(x³) · 3x² = 3x² sin(x³)Answer
✓ Check: the integral equals 1 − cos(x³), whose derivative is sin(x³) · 3x².
Find d over dx ∫_x⁵ e^t² dt. Lower limit
Show answer and steps
∫_x⁵ = −∫₅^xFlip the limits to put x on top−e^x²Answer
✓ Moving x to the right makes the region narrower, so the integral shrinks: the minus sign is expected.
Water flows into a tank at r(t) = 20 − 2t litres per minute for 0 ≤ t ≤ 10. The tank starts with 50 litres. How much is in it at t = 10? Net change
Show answer and steps
∫₀¹⁰ (20 − 2t) dt = [20t − t²] = 200 − 100 = 100Litres added50 + 100 = 150 litresAnswer
✓ The rate falls from 20 to 0 L/min, so the inflow is a triangle with area ½ · 10 · 20 = 100.
Find the average value of sin x on 0 ≤ x ≤ π. Average value
Show answer and steps
1 over π − 0 ∫₀^π sin x dx = 2 over πAverage = integral ÷ width2/π ≈ 0.63662Answer
✓ The average lies between the minimum 0 and maximum 1, a little below ⅔.

Section XSelf-guided quiz: eight quick questions

Choose one answer for each question, then press “Check my answers.” Your best score is saved only in this browser.

1. If A(x) = ∫₀^x f(t) dt and f is continuous, what is A′(x)?

f′(x)f(x)F(x) − F(0)0

2. What is ∫₀¹ x² dx?

12/31/21/3

3. Find ddx ∫₁^x cos t dt.

cos xsin xcos x − cos 1−sin x

4. Find ddx ∫₀^x² e^t dt.

e^x²e^2x2x e^x²x² e^x

5. A particle has velocity v(t) in m/s, and ∫₀⁵ v(t) dt = 12. What does 12 mean?

Its speed at t = 5 is 12 m/sIt travelled 12 m in total, alwaysIts displacement from t = 0 to t = 5 is 12 mIts acceleration is 12 m/s²

6. What is the average value of f(x) = 2x on 0 ≤ x ≤ 4?

48162

7. Use Part 2 to find ∫₀^π sin x dx.

01π2

8. If F′ = f, F(1) = 3 and F(4) = 10, what is ∫₁⁴ f(x) dx?

1373/10−7

Section XIBarrow, Newton and Leibniz

By the 1660s, mathematicians could find many tangents and many areas, but by separate tricks. Isaac Barrow, the first Lucasian Professor of Mathematics at Cambridge, developed a method of tangents that came close to calculus; his geometry lectures were published as Lectiones Geometricae in 1670. He is credited with being the first to recognise that integration and differentiation are inverse operations, although he never stated the Fundamental Theorem explicitly.

Isaac Newton wrote a tract on fluxions in October 1666. In it he calculated areas by reversing differentiation, and historians regard it as containing the first clear statement of the Fundamental Theorem. For Newton, integrating meant finding the quantity (the “fluent”) whose rate of change was given.

Gottfried Wilhelm Leibniz reached the same connection independently, thinking of an integral as a sum of infinitely thin pieces. By 1675 he had settled on writing ∫ y dy = ½ y², exactly as we do today. His notation, d and ∫, made the inverse relationship easy to see and to use, which is why it is the notation on this page.

NotesSources and credits

J. J. O’Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews: A history of the calculus, on Barrow’s awareness of the inverse relationship without an explicit statement, Newton’s October 1666 tract and its first clear statement of the theorem, and Leibniz’s notation by 1675.
MacTutor: Isaac Barrow (biography), on the Lucasian chair, Lectiones Geometricae (1670) and integration and differentiation as inverse operations.
MacTutor: Isaac Newton (biography), on the method of fluxions and integration as the inverse of differentiation.
MacTutor: Gottfried Wilhelm Leibniz (biography).

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