Code Scholars Tutoring · AP Calculus BC

FTC for integral-defined functions

Unit 6 · Topics 6.4, 6.5 · Detailed notes, a worked example, and a self-check with reasoning.

Browse Unit 6 lessons

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.

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

Worked example

If $\displaystyle G(x)=\int_1^{x^2}\cos(t)\,dt$ , find $\displaystyle G'(x)$ .

The integrand is continuous. The FTC evaluates it at

\displaystyle x^2

, and the chain rule multiplies by

\displaystyle 2x

. Thus

\displaystyle G'(x)=2x\cos(x^2)

. There is no need to evaluate the original integral.

Try it yourself

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

Show solution and reasoning

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