Q. 47 - Calculus

Question

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

        $\frac{d}{dx}{\int }_x^{x+2} \sin \left(t^2\right)dt$

Step-by-Step Solution

Answer

Ans:     $\frac{d}{dx}{\int }_x^{x+2} \sin \left(t^2\right)dt =\sin \left((x+2{)}^2\right)-\sin \left(x^2\right)$

Step 1. Given information.

given expression,

       $\frac{d}{dx}{\int }_x^{x+2} \sin \left(t^2\right)dt$

Step 2. The objective is to calculate the derivative.

The derivative can be written as,

   $\begin{array}{r}\frac{d}{dx}\left({\int }_x^0 \sin \left(t^2\right)dt+{\int }_0^{x+2} \sin \left(t^2\right)\right)\\ =\frac{d}{dx}\left(-{\int }_0^x \sin \left(t^2\right)dt+{\int }_0^{x+2} \sin \left(t^2\right)\right)\end{array}$

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

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

Step 3. The derivate expression can be written as,

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

Therefore, the answer is $\sin \left((x+2{)}^2\right)-\sin \left(x^2\right)$