Area Question 21
Question 21
- Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(x)=f(1-x), \forall x \in[-1,2]$. If $R_{1}=\int_{-1}^{2} x f(x) d x$ and $R_{2}$ are the area of the region bounded by $y=f(x), x=-1, x=2$ and the $X$-axis. Then,
(2011) (a) $R_{1}=2 R_{2}$ (c) $2 R_{1}=R_{2}$ (b) $R_{1}=3 R_{2}$ (d) $3 R_{1}=R_{2}$
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Solution:
- $R_{1}=\int_{-1}^{2} x f(x) d x$
Using $\quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
$$ \begin{aligned} & R_{1}=\int_{-1}^{2}(1-x) f(1-x) d x \ \therefore \quad & R_{1}=\int_{-1}^{2}(1-x) f(x) d x \end{aligned} $$
$$ [f(x)=f(1-x), \text { given }] $$
Given, $R_{2}$ is area bounded by $f(x), x=-1$ and $x=2$.
$$ \therefore \quad R_{2}=\int_{-1}^{2} f(x) d x $$
On adding Eqs. (i) and (ii), we get
$$ 2 R_{1}=\int_{-1}^{2} f(x) d x $$
From Eqs. (iii) and (iv), we get
$$ 2 R_{1}=R_{2} $$
