Question

Use the Second Fundamental Theorem of Calculus, if needed, to calculate each of the derivatives given below.

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

Short answer

Ans: $\frac{d}{dx}{\int }_0^x e^{-t^2}dt=e^{-x^2}$

Step-by-step solution

Step 1. Given information.

given,

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

Step 2. The objective is to calculate the derivative.

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

$\frac{d}{dx}{\int }_a^x f\left(t\right)dt=f\left(x\right)$

So,

$\begin{array}{r}f\left(t\right)=e^{-t^2}\\ f\left(x\right)=e^{-x^2}\end{array}$

Step 3. The derivative expression can be written as,

$\begin{array}{r}\frac{d}{dx}{\int }_0^x e^{-t^2}dt\\ =e^{-x^2}\left[f\left(x\right)=e^{-x^2}\right]\end{array}$

Therefore, the answer is $e^{-x^2}$.