Question

If $y=f\left(x\right)$ makes positive intercept $2$ and $0$ unit $x$ and $y$ encloses an area of $3/4$ square unit with the axes then ${\int }_0^{2}x{f}^{\prime }\left(x\right)dx$ is:

• 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 ${\int }_0^{2}f\left(x\right)dx=3/4$
$\therefore$ ${\int }_0^{2}x{f}^{\prime }\left(x\right)dx=x{\int }_0^2{f}^{\prime }\left(x\right)dx-{\int }_0^{2}f\left(x\right)dx={\left(xf\left(x\right)\right)}_0^2-3/4$
$=2f\left(2\right)-\frac{3}{4}=0-\frac{3}{4}$
$\therefore f\left(2\right)=0$, curve having intercept $=-3/4$