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STD 12 - 8. Application and integration — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 8. Application and integration1 Mark
MCQ
Let $f:\left[ { - 2,3} \right] \to \left[ {0,\infty } \right)$ be a continuous function such that $f(1-x) = f(x)$ for all $x \in \left[ { - 2,3} \right]$ . 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 ${R_2} = \int\limits_{ - 2}^3 {x\,f\left( x \right)} dx$ , then
  • A
    $3R _1= 2R_2$
  • B
    $2R _1= 3R_2$
  • C
    $R _1= R_2$
  • ✓
    $R _1= 2R_2$
✓

Answer

Correct option: D.
$R _1= 2R_2$
d
We have

${{\rm{R}}_2} = \int\limits_{ - 2}^3 {xf(x)dx} $

$ = \int\limits_{ - 2}^3 {(1 - x)f(1 - x)dx} $

$\left[ {{\rm{Using}}\int\limits_a^b {f\left( x \right)dx}  = \int\limits_a^b {f\left( {a + b - x} \right)dx} } \right]$

$ \Rightarrow {{\rm{R}}_2} = \int\limits_{ - 2}^3 {(1 - x)f\left( x \right)dx} $

$(\because f(x)=f(1-x) \text { on }[-2,3])$

$\therefore {{\rm{R}}_2} + {R^2} = \int\limits_{ - 2}^3 {xf\left( x \right)dx + } \int\limits_{ - 2}^3 {(1 - x)f\left( x \right)dx} $

$ = \int\limits_{ - 2}^3 {f\left( x \right)dx = {R_1}} $

$\Rightarrow 2 \mathrm{R}_{2}=\mathrm{R}_{1}$

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