Let f:[-2,3] arrow[0, ∞) be a continuous function such that f(1-x)=f(x) for all x ∈[-2,3] . If 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=∫_ -2 ^3 x f(x)

Correct answer: D. R_1=2 R_2 We have aligned R_ 2 &=∫_ -2 ^3 x f(x) d x=∫_ -2 ^3(1-x) f(1-x) d x & [U sing ∫_a^b f(x) d x=∫_a^b f(a+b-x) d x] R_ 2 &=∫_ -2 ^3(1-x) f(x) d x R_ 2 + R_ 2 &=∫_ -2 ^3 x f(x) d x+∫_ -2 ^3(1-x) f(x) d x 2 R_ 2 = R_ 1 aligned

JEE Main 201325 AprilOnline

Exam: JEE Main
Year: 2013
Date: 25 April
Mode: Online
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2013 · 25 April · Online · Mathematics · Area Under Curves

Let f:[-2,3] arrow[0, ∞) be a continuous function such that...

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 :

Options

A $3 R_1=2 R_2$
B $2 \mathrm{R}_1=3 \mathrm{R}_2$
C $\mathrm{R}_1=\mathrm{R}_2$
D $\mathrm{R}_1=2 \mathrm{R}_2$

Correct answer

D. $\mathrm{R}_1=2 \mathrm{R}_2$

Step-by-step 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}

Practice Area Under Curves on Quantrex Academy →

More from Area Under Curves