Q. 68 - Calculus

Question

Prove Theorem 4.35 in your own words: If $f$ is continuous on $[a,b]$ and $u\left(x\right)$ is a differentiable function, then for all $x\in \left[a,b\right]$, ${\int }_a^{u\left(x\right)}f\left(t\right)dt=f\left(u\left(x\right)\right)u\text{'}\left(x\right)$. Be especially clear about how you use the chain rule.

Step-by-Step Solution

Answer

If $f$ is continuous on $\left[a,b\right]$ and $u\left(x\right)$ is a differentiable function, then for all $x\in \left[a,b\right]$, ${\int }_a^{u\left(x\right)}f\left(t\right)dt=f\left(u\left(x\right)\right)u\text{'}\left(x\right)$.

Step 1. Given information

We have to prove ${\int }_a^{u\left(x\right)}f\left(t\right)dt=f\left(u\left(x\right)\right)u\text{'}\left(x\right)$.

Step 2. Proof of the given question.

Let $F$ be an antiderivative of $f$.

Now,

$$\begin{gathered} \frac{d}{dx}{\int }_a^{u\left(x\right)}f\left(t\right)dt \\ =\frac{d}{dx}\left(F\left(u\left(x\right)\right)\right) \\ =\frac{d}{dx}\left(F\left(u\left(x\right)\right)\right)\times \frac{d}{dx}\left(u\left(x\right)\right) \\ =F\text{'}\left(u\left(x\right)\right)u\text{'}\left(x\right) \\ =f\left(u\left(x\right)\right)u\text{'}\left(x\right) \end{gathered}$$

Therefore, it is proved.