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:

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:

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

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

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

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

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

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

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

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:

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