Question

Let $f:\left[-2,3\right]\to [0,\infty )$ be a continuous function such that $f\left(1-x\right)=f\left(x\right)forallx\in \left[-2,3\right].$
If $R_1$ is the numerical value of the area of the region bounded by $y=f\left(x\right),x=-2,x=3$ and the axis of x and $R_2=\underset{-2}{\overset{3}{\int }}xf\left(x\right)dx,$ then:

• A. $3R_1=2R_2$
• B. $2R_1=3R_2$
• C. $R_1=R_2$
• D. $R_1=2R_2$

Answer 1

Answer: (D)

We have
$R_2=\underset{-2}{\overset{3}{\int }}xf\left(x\right)dx=\underset{-2}{\overset{3}{\int }}\left(1-x\right)f\left(1-x\right)dx$
$\left[Using\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx\right]$
$\Rightarrow R_2=\underset{-2}{\overset{3}{\int }}\left(1-x\right)f\left(x\right)dx$
$\left(\because f\left(x\right)=f\left(1-x\right)on\left[-2,3\right]\right)$
$\therefore R_2+R_2=\underset{-2}{\overset{3}{\int }}xf\left(x\right)dx+\underset{-2}{\overset{3}{\int }}\left(1-x\right)f\left(x\right)dx$$=\underset{-2}{\overset{3}{\int }}f\left(x\right)dx=R_1$
$\Rightarrow 2R_2=R_1$