The Fundamental Theorem of Calculus

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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.

If $f$ 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 Six Pillars of Calculus

Trigonometry Review

Exponential Functions

Logarithms and Inverse functions

Intro to Limits

Limit Laws and Computations

Continuity and the Intermediate Value Theorem

Limits at Infinity

Rates of Change

The Derivative Function

Basic Differentiation Rules

Product and Quotient Rules

Derivatives of Trig Functions

The Chain Rule

Implicit Differentiation

Derivatives of Logs

Derivatives in Science

Related Rates

Linear Approximation and Differentials

Differentiation Review

Absolute and Local Extrema

The Mean Value and other Theorems

$f$ vs. $f'$

Indeterminate Forms and L'Hospital's Rule

Optimization

Newton's Method

Anti-derivatives

The Area under a curve

Definite Integrals

The Fundamental Theorem of Calculus

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