The Fundamental Theorem of Calculus

We can solve harder problems involving derivatives of integral functions. For example, what do we do when our upper limit of integration is not simply our variable $x$ , as it must be to use FTC 1, but is rather a function of $x$ ? We will use FTC 2 to solve this FTC 1 problem.

Example: Compute ${\displaystyle\frac{d}{dx} \int_1^{x^2} \tan^{-1}(s)\, ds.}$

Solution: Let $F(x)$ be the antiderivative of $\tan^{-1}(x)$ . Finding a formula for $F(x)$ is hard, but we don't actually need the antiderivative , since we will not integrate. Recall that by FTC 2,

$\int_1^{x^2} \tan^{-1}(s) \,ds = F\left(x^2\right) - F(1),$ so

$\frac{d}{dx} \int_1^{x^2} \tan^{-1}(s)\,ds = \frac{d}{dx}\left(F\left(x^2\right) - F(1)\right).$
By the chain rule, we now get

$F'(x^2)\,2x-0= 2x\, f(x^2) = 2x \tan^{-1}\left(x^2\right).$
since

$F'=f$ by our assumption.

This method generalizes, but please do not try to memorize this; you do not need to, because it is simply applying FTC 2 and the chain rule, as you see in the box below and in the following video.

$f$ is a continuous function with antiderivative

$F$ , and

$g$ and

$h$ are differentiable functions, then

$\frac{d}{dx} \int_{g(x)}^{h(x)} f(s)\, ds = \frac{d}{dx} \Big[F\left(h(x)\right) - F\left(g(x)\right)\Big]$

$\hspace{3cm}\quad\quad\quad= F'\left(h(x)\right) h'(x) - F'\left(g(x)\right) g'(x)$

$\hspace{3cm}\quad\quad = f\left(h(x)\right) h'(x) - f\left(g(x)\right) g'(x).$

There is an an alternate way to solve these problems, using FTC 1 and the chain rule. We will illustrate using the previous example.

Example: Compute ${\displaystyle\frac{d}{dx} \int_1^{x^2} \tan^{-1}(s)\, ds.}$

Solution: We let $u=x^2$ and let $g(u)=\int_1^u\tan^{-1}(s)\, ds$ , and use the fact that $\frac{d}{dx}g(u)=g'(u)\frac{du}{dx}$ to get

$\frac{d}{dx}\int_1^{x^2} \tan^{-1}(s)\, ds=\frac{d}{dx} \left(\int_1^{u} \tan^{-1}(s)\, ds\right)=\frac{d}{dx}\left(g(u)\right)=g'(u)\frac{du}{dx},$
By FTC 1,

$g'(u)=\tan^{-1}(u)$ , and

$\frac{du}{dx}=\frac{d}{dx}(x^2)=2x$ , giving us

$\tan^{-1}(u)\frac{du}{dx}=\tan^{-1}(x^2)\cdot 2x.$

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

More FTC 1