Question

The area bounded by the curves $y=x\ln x$ and $y=2x-2x^2$ is

• A. $\frac{7}{12}\text{sq. units}$
• B. $\frac{7}{4}\text{sq. units}$
• C. $\frac{5}{4}\text{sq. units}$
• D. $\frac{7}{16}\text{sq. units}$

Answer 1

The correct option is A $\frac{7}{12}\text{sq. units}$

The bounded region is shown in the figure :


So, the required area is :
$=\underset{0}{\overset{1}{\int }}(2x-2x^2-x\ln x)dx$
$={[x^2-\frac{2x^3}{3}]}_0^1-{\left[\frac{x^2}{2}\ln x\right]}_0^1+\underset{0}{\overset{1}{\int }}\frac{1}{x}\cdot \frac{x^2}{2}dx$
$=1-\frac{2}{3}-{[\frac{x^2}{2}\ln x-\frac{x^2}{4}]}_0^1$
$=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\text{sq. units}$