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

The value of $\int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$ for $x \in (0, \pi/2)$ is:

Let $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ and $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,then:

$\int\limits_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx = $

Evaluate the integral: $\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x$

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of the integral $\int_{-\pi / 2}^{\pi / 2} [[x] - \sin x] \, dx$ is equal to:

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