Q. 39 - Calculus

Question

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

  

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

Step-by-Step Solution

Answer

Ans:    $\frac{d}{dx}{\int }_1^{\sqrt{x}} x\ln tdt = {\int }_1^{\sqrt{x}} \ln t+\frac{1}{2}\sqrt{x}\ln \sqrt{x}$

Step 1. Given information.

given, 

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

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, 

    $\begin{array}{r}f\left(u\right(t\left)\right)=\ln \sqrt{t}\\ f\left(u\right(x\left)\right)=\ln \sqrt{x}\\ u^{\mathrm{\prime }}\left(x\right)=\frac{1}{2}\sqrt{x}\\ f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)=\ln \sqrt{x}\frac{1}{2}\sqrt{x}=\frac{1}{2}\sqrt{x}\ln \sqrt{x}\end{array}$

Step 3. The derivative expression can be written as,

$\begin{array}{r}\frac{d}{dx}{\int }_1^{\sqrt{x}} x\ln tdt \\ ={\int }_1^{\sqrt{x}} \ln t+\frac{1}{2}\sqrt{x}\ln \sqrt{x} \left[f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)=\frac{1}{2}\sqrt{x}\ln \sqrt{x}\right]\end{array}$

Therefore, the answer is ${\int }_1^{\sqrt{x}} \ln t+\frac{1}{2}\sqrt{x}\ln \sqrt{x} .$