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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
If ${I_1} = \int\limits_0^1 {{e^{ - x}}} {\cos ^2}\,x\,dx\,;\,{I_2} = \int\limits_0^1 {{e^{ - {x^2}}}} {\cos ^2}\,x\,dx$ and $\,{I_3} = \int\limits_0^1 {{e^{ - {x^3}}}} dx$ ; then
  • A
    $I_2>I_3>I_1$
  • B
    $I_3>I_1>I_2$
  • C
    $I_2>I_1>I_3$
  • ✓
    $I_3>I_2>I_1$
✓

Answer

Correct option: D.
$I_3>I_2>I_1$
d
Given:

$I_{1}=\int_{0}^{1} e^{-x} \cos ^{2} x d x$

$I_{2}=\int_{0}^{1} e^{-x^{2}} \cos ^{2} x d x$ and

$I_{3}=\int_{0}^{1} e^{-x^{3}} d x$

For $x \in(0,1)$

$\Rightarrow x>x^{2}$ or $-x<-x^{2}$

and $x^{2}>x^{3}$ or $-x^{2}<-x^{3}$

$\therefore {e^{ - {x^2}}} < {e^{ - {x^3}}}$ and ${e^{ - x}} < {e^{ - {x^2}}}$

$ \Rightarrow {e^{ - x}} < {e^{ - {x^2}}} < {e^{ - x}}$

$\Rightarrow e^{-x^{3}}>e^{-x^{2}}>e^{-x}$

$\Rightarrow I_{3}>I_{2}>I_{1}$

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