Q. 41 - Calculus

Question

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

      $\frac{d}{dx}{\int }_1^8 \ln tdt$

Step-by-Step Solution

Answer

Ans:    $\frac{d}{dx}{\int }_1^8 \ln tdt =0$

Step 1. Given information.

given,

       $\frac{d}{dx}{\int }_1^8 \ln tdt$

Step 2. The objective is to calculate the derivative.

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

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

So,

    $\begin{array}{r}f\left(u\right(t\left)\right)=\ln t\\ f\left(u\right(8\left)\right)=\ln 8\\ u^{\mathrm{\prime }}\left(x\right)=0\\ f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)=\ln 8\left(0\right)=0\end{array}$

Step 3. The derivative expression can be written as,

   $\begin{array}{r}\frac{d}{dx}{\int }_1^8 \ln tdt \\ =0 \left[f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)=0\right]\end{array}$

Therefore, the value is $0$