# Fundamental theorem of calculus review

Review your knowledge of the fundamental theorem of calculus and use it to solve problems.

## What is the fundamental theorem of calculus?

The theorem has two versions.

### a) $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)$

We start with a continuous function $f$ and we define a new function for the area under the curve $y=f(t)$:

What this version of the theorem says is that the derivative of $F$ is $f$. In other words, $F$ is an antiderivative of $f$. Thus, the theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

### b) $\displaystyle\int_a^b\!\! f(x)dx=F(b)\!-\!\!F(a)$

This version gives more direct instructions to finding the area under the curve $y=f(x)$ between $x=a$ and $x=b$. Simply find an antiderivative $F$ and take $F(b)-F(a)$.

*Want to learn more about the fundamental theorem of calculus? Check out this video.*

## Practice set 2: Applying the theorem with chain rule

We can use the theorem in more hairy situations. Let's find, for example, the expression for $\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt$. Note that the interval is between $0$ and $x^3$, not $x$.

To help us, we define $\displaystyle F(x) = \int_{0}^{x}\sin(t) \, dt$. According to the fundamental theorem of calculus, $F'(x)=\sin(x)$.

It follows from our definition that $\displaystyle\int_{0}^{x^3}\sin(t) \, dt$ is $F(x^3)$, which means that $\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt$ is $\dfrac{d}{dx}F(x^3)$. Now we can use the chain rule:

*Want to try more problems like this? Check out this exercise.*