Chapter 7.3, Problem 41E

To determine

To find the value of $f(x)$

Answer to Problem 41E

Option (C)

Explanation of Solution

Given information: $\frac{dy}{dx}=4x\ln (x)$ and $x=0;[y=2\text{and}y=5]$

Formula used: ${\displaystyle \int uvdx=u{\displaystyle \int vdx-{\displaystyle \int \left(u^{\prime }{\displaystyle \int vdx}\right)}}}dx$

Calculation :

The given expression is

  $dy=4x\ln (x)dx$

Integrating both the expression becomes as:

  ${\displaystyle \int dy}=4{\displaystyle \int x\ln (x)dx}$

  $y=4\left[\ln (x)\left(\frac{x^2}{2}\right)-{\displaystyle \int \left(\frac{1}{x}\right)\left(\frac{x^2}{2}\right)}\right]+C$

  $=4\left[\ln (x)\left(\frac{x^2}{2}\right)-{\displaystyle \int \left(\frac{x}{2}\right)}\right]+C$

  $y=4\left[\ln (x)\left(\frac{x^2}{2}\right)-\left(\frac{x^2}{4}\right)\right]+C$

When $x=0;[y=2\text{and}y=5]$ and $C=3$

Hence, $f(x)=4\left[\ln (x)\left(\frac{x^2}{2}\right)-\left(\frac{x^2}{4}\right)\right]+3$