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Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(x)=f(1-x)$ for all $x \in[-1,2]$. Let $R_1=\int_{-1}^2 x f(x) d x$ and $R_2$ be the area of the region bounded by $y=f(x), x=-1, x=2$ and the $X$-axis. Then,
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Verified Answer
The correct answer is:
$2 R_1=R_2$
$2 R_1=R_2$
Here, $R_1=\int_{-1}^2 x f(x) d x$
Using, $\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$$
\begin{aligned}
& R_1=\int_{-1}^2(1-x) f(1-x) d x, \\
\therefore \quad & \left.R_1=\int_{-1}^2(1-x) f(x) d x \quad \ldots(1-x)\right]
\end{aligned}
$$
Given, $R_2$ is area bounded by
$$
\begin{aligned}
& f(x), x-1 \text { and } x=2 \\
\therefore & R_2=\int_{-1}^2 f(x) d x
\end{aligned}
$$
Adding Eqs. (i) and (ii), we get
$$
2 R_1=\int_{-1}^2 f(x) d x
$$
$\therefore$ From Eqs. (iii) and (iv), we get
$$
2 R_1=R_2
$$
Using, $\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$$
\begin{aligned}
& R_1=\int_{-1}^2(1-x) f(1-x) d x, \\
\therefore \quad & \left.R_1=\int_{-1}^2(1-x) f(x) d x \quad \ldots(1-x)\right]
\end{aligned}
$$
Given, $R_2$ is area bounded by
$$
\begin{aligned}
& f(x), x-1 \text { and } x=2 \\
\therefore & R_2=\int_{-1}^2 f(x) d x
\end{aligned}
$$
Adding Eqs. (i) and (ii), we get
$$
2 R_1=\int_{-1}^2 f(x) d x
$$
$\therefore$ From Eqs. (iii) and (iv), we get
$$
2 R_1=R_2
$$
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