Q. 38 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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 $e^{-x^2}$.