Question

Let $f:\left[ -1,2 \right]\to \left[ 0,\infty \right)$ be a continuous function such that $f\left( x \right)=f\left( 1-x \right)$ for all $x\in \left[ -1,2 \right]$. Let ${{R}_{1}}=\int\limits_{-1}^{2}{xf\left( x \right)dx}$ and ${{R}_{2}}$ be the area of the region bounded by $y=f\left( x \right)$, $x=-1$, $x=2$ and the X-axis. Then
A. ${{R}_{1}}=2{{R}_{2}}$
B. ${{R}_{1}}=3{{R}_{2}}$
C. $2{{R}_{1}}={{R}_{2}}$
D. $3{{R}_{1}}={{R}_{2}}$

Answer 1

Hint: We try to form the integration for the area of ${{R}_{2}}$. Then we change the variable for the function ${{R}_{1}}=\int\limits_{-1}^{2}{xf\left( x \right)dx}$. We use different theorems of definite integral like $\int\limits_{a}^{b}{f\left( x \right)dx}=-\int\limits_{b}^{a}{f\left( x \right)dx}$ and $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( z \right)dz}$. We change the function using $f\left( x \right)=f\left( 1-x \right)$ and find the relation between the ${{R}_{1}}$ and ${{R}_{2}}$.

Complete step by step solution:
It is given that ${{R}_{2}}$ be the area of the region bounded by $y=f\left( x \right)$, $x=-1$, $x=2$ and the X-axis. We can express it in the form of integration of area under the curve.
So, ${{R}_{2}}=\int\limits_{-1}^{2}{f\left( x \right)dx}$.
It’s also given that ${{R}_{1}}=\int\limits_{-1}^{2}{xf\left( x \right)dx}$.
We need to find the relation between the ${{R}_{1}}$ and ${{R}_{2}}$.
We now change the variable of the integration ${{R}_{1}}=\int\limits_{-1}^{2}{xf\left( x \right)dx}$ with the relation where $x=1-z$.
We take partial differentiation of the relation $x=1-z$. So, $dx=-dz$.
The upper and lower limit changes with the relation

x	-1	2
$z=1-x$	2	-1

Now we replace the values to get ${{R}_{1}}=\int\limits_{-1}^{2}{xf\left( x \right)dx}=\int\limits_{2}^{-1}{\left( 1-z \right)f\left( 1-z \right)\left( -dz \right)}$.
We now use the theorems of definite integral where $\int\limits_{a}^{b}{f\left( x \right)dx}=-\int\limits_{b}^{a}{f\left( x \right)dx}$ and $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( z \right)dz}$.
Therefore, ${{R}_{1}}=\int\limits_{2}^{-1}{\left( 1-z \right)f\left( 1-z \right)\left( -dz \right)}=\int\limits_{-1}^{2}{\left( 1-z \right)f\left( 1-z \right)dz}$.
Now we break the functions as ${{R}_{1}}=\int\limits_{-1}^{2}{\left( 1-z \right)f\left( 1-z \right)dz}=\int\limits_{-1}^{2}{f\left( 1-z \right)dz}-\int\limits_{-1}^{2}{zf\left( 1-z \right)dz}$.
We use the theorem $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( z \right)dz}$.
So, ${{R}_{1}}=\int\limits_{-1}^{2}{f\left( 1-z \right)dz}-\int\limits_{-1}^{2}{zf\left( 1-z \right)dz}=\int\limits_{-1}^{2}{f\left( 1-x \right)dx}-\int\limits_{-1}^{2}{xf\left( 1-x \right)dx}$.
It’s also given that $f\left( x \right)=f\left( 1-x \right)$ which gives
${{R}_{1}}=\int\limits_{-1}^{2}{f\left( 1-x \right)dx}-\int\limits_{-1}^{2}{xf\left( 1-x \right)dx}=\int\limits_{-1}^{2}{f\left( x \right)dx}-\int\limits_{-1}^{2}{xf\left( x \right)dx}$
We replace the values with ${{R}_{1}}=\int\limits_{-1}^{2}{xf\left( x \right)dx}$ and ${{R}_{2}}=\int\limits_{-1}^{2}{f\left( x \right)dx}$.
${{R}_{1}}=\int\limits_{-1}^{2}{f\left( x \right)dx}-\int\limits_{-1}^{2}{xf\left( x \right)dx}={{R}_{2}}-{{R}_{1}}$.
Simplifying we get $2{{R}_{1}}={{R}_{2}}$. The correct option is C.

Note: We need to remember that the transformation from ${{R}_{1}}$ to ${{R}_{2}}$ can also be done inversely. The relation between the variables will remain the same. If a function is strictly positive, the area between it and the X-axis is simply the definite integral.