Functions defined by integrals.

Theorem 8.8.1 (The Fundamental Theorem of Calculus)   Let $ f$ be a function, continuous on the interval $ I$ . For $ x_0 \in I$ , we define the function $ F$ as follows:

$ \forall x \in I, F(x)= \int_{x_0}^{x} f(t) \; dt$ .

Then $ F$ is differentiable on $ I$ and $ F'=f$ .

With another notation:
\fbox{
$\frac {d}{dx} \left( \int_{x_0}^x f(t) \; dt \right) = f(x)$
}

Example 8.8.2       

Let $ F(x)=\int_x^{2x} \sin t \; dt$ . We prove that $ F$ is differentiable on $ \mathbb{R}$ and we will compute the first derivative of $ f$ .

As the sine function is continuous on $ \mathbb{R}$ , the function $ F$ is well-defined on $ \mathbb{R}$ . For any real $ a$ , we have:

$\displaystyle F(x)=\int_x^{2x} \sin t \; dt = \int_x^a \sin t \; dt + \int_a^{2...
...nt_a^x \sin t \; dt}_{F_1(x)} + \underbrace{\int_a^{2x} \sin t \; dt}_{F_2(x)}.$    

The function $ F_1$ is differentiable on $ \mathbb{R}$ , by Theorem 5.9 and Theorem 8.1. We have:

$\displaystyle \forall x \in \mathbb{R}, \; F'_1(x)=\underbrace{2}_{\text{inner derivative}} \underbrace{\sin 2x}_{\text{outer derivative}}.$    

The function $ F_2$ is differentiable on $ \mathbb{R}$ , by Theorem 8.1; we have:

$\displaystyle \forall x \in \mathbb{R}, \; F'_2(x)=\sin x.$    

Thus $ F$ is differentiable on $ \mathbb{R}$ and

$\displaystyle \forall x \in \mathbb{R}, \; F'(x)=2 \sin 2x - \sin x.$    

Actually, in Example 8.2, we could have computed an explicit formula for $ F(x)$ . In the Example 8.3, this should be impossible.

Example 8.8.3       

Let $ F(x)=\int_x^{2x} \sqrt{e^t - \sin t} \; dt$ . The function $ f$ defined by $ f(t)=\sqrt{e^t - \sin t}$ is continuous on $ [0,+\infty )$ , therefore by Theorem 8.1, the function $ F$ is differentiable on $ [0,+\infty )$ and we have:

$\displaystyle \forall x \in [0,+ \infty ), F'(x)= 2 \sqrt{e^{2x} - \sin 2x } - \sqrt{e^t - \sin t}.$    

Noah Dana-Picard 2007-12-28