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
${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then
- A$I = {I_1}$
- B$I = {I_2}$
- C$I = {I_3}$
- D$I = {I_4}$
Explore More
Similar Questions
The value of the integral $\int_{0}^{\pi / 2} (\sin^{100} x - \cos^{100} x) dx$ is
$\int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x=$
$\int_0^{400 \pi} \sqrt{1-\cos 2 x} \, dx =$ (in $\sqrt{2}$)
$\int_0^{2 \pi} \sin ^6 x \cos ^5 x \, dx$ is equal to
The value of $\int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{\left(1+\cos ^{2} x\right)\left(e^{\cos x}+e^{-\cos x}\right)} d x$ is equal to
DifficultJEE MAIN 2022
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo