If $I$ is the greatest of the definite integrals
${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then

  • A
    $I = {I_1}$
  • B
    $I = {I_2}$
  • C
    $I = {I_3}$
  • D
    $I = {I_4}$

Explore More

Similar Questions

$\int_0^\pi x \sin^3 x \cos^2 x \, dx =$

$\int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\tan x}} = $

If $f(x) = f(\Pi + e - x)$ and $\int_{e}^{\Pi} f(x) dx = \frac{2}{e + \Pi}$,then $\int_{e}^{\Pi} x f(x) dx$ is equal to

The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is

$\int_{1/2}^{2} \frac{1}{x} \sin \left( x - \frac{1}{x} \right) dx = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo