Let $f:[-2,3] \rightarrow[0, \infty)$ be a continuous function such that $f(1-x)=f(x)$ for all $x \in[-2,3]$…
Let $f:[-2,3] \rightarrow[0, \infty)$ be a continuous function such that $f(1-x)=f(x)$ for all $x \in[-2,3]$.
If $\mathrm{R}_1$ is the numerical value of the area of the region bounded by $y=f(x), x=-2, x=3$ and the axis of $x$ and $R_2=\int_{-2}^3 x f(x) d x$, then :
$3 R_1=2 R_2$
$2 \mathrm{R}_1=3 \mathrm{R}_2$
$\mathrm{R}_1=\mathrm{R}_2$
$\mathrm{R}_1=2 \mathrm{R}_2$
Solution
We have
$
\begin{aligned}
\mathrm{R}_2 &=\int_{-2}^3 x f(x) d x=\int_{-2}^3(1-x) f(1-x) d x \\
& {\left[\mathrm{U} \operatorname{sing} \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right] } \\
\Rightarrow \mathrm{R}_2 &=\int_{-2}^3(1-x) f(x) d x \\
\therefore \mathrm{R}_2+\mathrm{R}_2 &=\int_{-2}^3 x f(x) d x+\int_{-2}^3(1-x) f(x) d x \\
\Rightarrow 2 \mathrm{R}_2=\mathrm{R}_1
\end{aligned}
$