Q. 43 - Calculus

Question

Combining derivatives and integrals: Simplify each of the following as much as possible:

$\frac{d}{dx}{\int }_0^3{e}^{-t^2}dt$

Step-by-Step Solution

Answer

The solution is $\frac{d}{dx}{\int }_0^3{e}^{-t^2}dt=0$

Step 1. Given information

The given integral is-

$\frac{d}{dx}{\int }_0^3{e}^{-t^2}dt$

Step 2. Calculation

The expression is-

$\frac{d}{dx}{\int }_0^3{e}^{-t^2}dt$

The objective is to simplify the expression,

Now, if $f$ is continuous on $[a,b]$ then for all $x\in [a,b],$

$$\begin{gathered} \frac{d}{dx}{\int }_0^{u\left(x\right)}f\left(t\right)dt=f\left(u\right(x\left)\right)u\left(x\right) \\ So, \\ f\left(u\right(t\left)\right)=e^{-t^2} \\ f\left(u\right(3\left)\right)=e^{-9} \\ u\left(x\right)=0 \\ f\left(u\right(x\left)\right)u\left(x\right)=e^{-9}\left(0\right)=0 \end{gathered}$$

The derivative expression can be written as,

$\frac{d}{dx}{\int }_0^3{e}^{-t^2}dt=0$