Area Ques 30
Question
- 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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Answer:
Correct Answer: 30.(c)
Solution:
Formula:
- $R _{1}=\int _{-1}^{2} x f(x) d x \quad …….(i)$
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 \quad …….(ii) \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 \quad …….(iii) $
On adding Eqs. (i) and (ii), we get
$ 2 R _{1}=\int _{-1}^{2} f(x) d x \quad …….(iv) $
From Eqs. (iii) and (iv), we get
$ 2 R _{1}=R _{2} $
BETA
