Question

If $y=f\left(x\right)$ makes $+ve$ intercept of $2$ and $0$ unit on $x$ and $y$ axes respectively and enclosed an area of $\frac{3}{4}$ square unit in the first quadrant, then $\underset{0}{\overset{2}{\int }}x{f}^{\prime }\left(x\right)dx$ is equal to

• A. $\frac{3}{2}$
• B. $1$
• C. $\frac{5}{4}$
• D. $-\frac{3}{4}$

Answer 1

The correct option is D $-\frac{3}{4}$

Given that $\underset{0}{\overset{2}{\int }}f\left(x\right)dx=\frac{3}{4}$
Using byparts, we have:
$\int x{f}^{{}^{\prime }}\left(x\right)dx=xf\left(x\right)-\int f\left(x\right)dx$
Now,
$\underset{0}{\overset{2}{\int }}x{f}^{{}^{\prime }}\left(x\right)dx=\left[xf\right(x){]}_0^2-\underset{0}{\overset{2}{\int }}f\left(x\right)dx$
$=2f\left(2\right)-\frac{3}{4}$
$=0-\frac{3}{4}=-\frac{3}{4}(\because f(2)=0)$