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Code Scholars Tutoring · AP Calculus AB

FTC for integral-defined functions

Unit 6 · Topics 6.4, 6.5. Original AP Calculus AB notes, worked example, and self-check.

Understand the idea

When the integrand is continuous, an accumulation function differentiates to its integrand at the upper limit. A moving upper limit adds a chain-rule factor. Moving lower and upper limits produce two terms. Determine increase, extrema, and concavity of an accumulation function by analyzing its derivative, not by trying to find an elementary antiderivative first.

ddx∫ag(x)f(t) dt=f(g(x))g′(x)\frac d{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))g'(x)

Worked example

If G(x)=∫1x2cos⁡(t) dt\displaystyle G(x)=\int_1^{x^2}\cos(t)\,dt, find G′(x)\displaystyle G'(x).

The integrand is continuous. The FTC evaluates it at x2\displaystyle x^2, and the chain rule multiplies by 2x\displaystyle 2x. Thus G′(x)=2xcos⁡(x2)\displaystyle G'(x)=2x\cos(x^2). There is no need to evaluate the original integral.

Try it yourself

For H(x)=∫0x(t2−4) dt\displaystyle H(x)=\int_0^x(t^2-4)\,dt, where does H\displaystyle H have a local minimum?

Show solution and reasoning

H′(x)=x2−4\displaystyle H'(x)=x^2-4 changes from negative to positive at x=2\displaystyle x=2, giving a local minimum there. At −2\displaystyle -2 it changes from positive to negative, giving a local maximum.

Unit 6 MCQNext: Integral properties and the evaluation theorem →