Guides / Integrals

The fundamental theorem of calculus: derivatives of integrals, and the chain rule on the limit

The fundamental theorem says that integrating and differentiating undo each other. It has two parts: one tells you the derivative of an integral whose upper limit is a variable, and the other lets you evaluate a definite integral with any antiderivative. Both are short, and the mistakes are about the limits.

Why it works

Think of F(x) = ∫_a^x f(t) dt as the area that has piled up under f from a to x. Move x a little to the right, by h, and you add a thin strip about f(x) tall and h wide. So the area grows at the rate f(x), and that is what F′(x) = f(x) says. The starting point a is fixed, so it doesn't affect how fast the area is growing.

If the upper limit is x² instead of x, the edge of the area moves at the rate 2x, not 1, so the area grows 2x times as fast. That's the chain rule. Part 2 follows from Part 1: any two antiderivatives differ by a constant, and the constant cancels in F(b) − F(a).

Example 1: the upper limit is x

Find F′(x) and F′(2) for F(x) = ∫₁^x √(1 + t³) dt.

1. F′(x) = √(1 + x³)Part 1: replace t with the upper limit. There's no antiderivative to find, and this integrand doesn't have a simple one.
2. F′(2) = √(1 + 8) = 3The lower limit, 1, plays no part.

Example 2: a function as the upper limit

Find ddx ∫₀^x2 cos t dt.

1. cos(x²) · ddx(x²)Put the upper limit into the integrand, then multiply by the limit's derivative.
2. = 2x cos(x²)The chain rule factor is 2x.
3. Check: ∫₀^x2 cos t dt = sin(x²), and (sin(x²))′ = 2x cos(x²)This integrand has an antiderivative, so you can confirm the answer the long way.

Example 3: both limits move

Find ddx ∫_x^2x 1t dt for x > 0.

1. 12x · 2 − 1x · 1Upper limit: f(2x) times (2x)′. Lower limit: minus f(x) times (x)′.
2. = 1x − 1xSimplify each term.
3. = 0The integral is ln(2x) − ln x = ln 2 for every x, a constant, so its derivative really is 0.

Example 4: evaluating a definite integral

Evaluate ∫₁⁴ (3√x − 1x²) dx.

1. F(x) = 2x^3/2 + 1xAn antiderivative: 3x^1/2 → 3 · x^3/2/(3/2), and −x⁻² → x⁻¹. No + C needed; it would cancel.
2. F(4) = 2 · 8 + 14 = 654, F(1) = 2 + 1 = 34^3/2 = 8. The integrand is continuous on [1, 4], so Part 2 applies.
3. 654 − 3 = 534 = 13.25Top limit minus bottom limit.

Common mistakes, and the exact line they happen on

1. Leaving out the chain rule

The integral from 0 to x² in Example 2.

1. ddx ∫₀^x2 cos t dt = cos(x²)This is the mistake. "Put the limit in" is only the whole rule when the limit is plain x. Here the limit moves at the rate 2x, so the answer is 2x cos(x²). At x = 1 that's 2 cos 1 ≈ 1.081, not cos 1 ≈ 0.540. Differentiating sin(x²) shows the factor.

2. Subtracting the integrand at the lower limit

The integral from 1 to x in Example 1.

1. F′(x) = √(1 + x³) − √(1 + 1³) = √(1 + x³) − √2This is the mistake. This copies "F(b) − F(a)" from Part 2 into Part 1. But √2 is a constant, so subtracting it can't belong in a derivative: the lower limit 1 doesn't move, and it contributes nothing. F′(2) is 3, not 3 − √2 ≈ 1.586.

3. Using Part 2 across a break in the integrand

Evaluate ∫₋₁¹ 1x² dx.

1. An antiderivative of 1/x² is −1/xRight so far.
2. [−1x]₋₁¹ = −1 − 1 = −2This is the mistake. 1/x² is positive everywhere, so its integral can't be negative. Part 2 needs f to be continuous on the whole interval, and 1/x² blows up at x = 0. The integral is improper, and here it diverges: ∫ from 0 to 1 of 1/x² is already infinite. Before using F(b) − F(a), check for a zero in a denominator between the limits.

Practice

1. Find ddx ∫₂^x (t³ + 1)⁵ dt.

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   (x³ + 1)⁵. No need to expand or integrate anything.

2. Find ddx ∫₀^{sin x} √(1 + t²) dt.

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   √(1 + sin²x) · cos x. The chain rule factor is the derivative of sin x.

3. Find ddx ∫x21 e−t dt.

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   The variable is in the lower limit, so there's a minus sign: −e^−x2 · 2x = −2xe^−x2.

4. Evaluate ∫₀^π sin x dx.

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   [−cos x]₀^π = −cos π + cos 0 = 1 + 1 = 2.

5. Evaluate ∫₀² (3x² − 2x) dx.

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   [x³ − x²]₀² = (8 − 4) − 0 = 4.