Question

Let $f : \left[-2, 3\right] \to [0, \infty)$ be a continuous function such that $f \left(1-x\right)=f \left(x\right) for all x \in \left[-2, 3\right].$
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 $R_{2} =\int\limits_{-2}^{3} x \, f \left(x\right) dx,$ 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
$R_{2}=\int\limits_{-2}^{3} x f \left(x\right) dx =\int\limits_{-2}^{3} \left(1-x\right) f \left(1-x\right) dx$
$\left[Using\, \int\limits_{a}^{b} f \left(x\right) dx=\int\limits_{a}^{b} f \left(a+b-x\right) dx\right]$
$\Rightarrow R_{2} =\int\limits_{-2}^{3} \left(1-x\right) f \left(x\right) dx$
$\left(\because f\left(x\right)=f\left(1-x\right)on \left[-2, 3\right]\right)$
$\therefore R_{2}+R_{2} =\int\limits_{-2}^{3} x f \left(x\right) dx+\int\limits_{-2}^{3} \left(1-x\right) f\left(x\right) dx$$=\int\limits_{-2}^{3} f \left(x\right) dx =R_{1}$
$\Rightarrow 2R_{2} =R_{1}$