Evaluate the definite integral $\int_{0}^{1} x e^{x^{2}} d x$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Let $I = \int_{0}^{1} x e^{x^{2}} d x$.
Substitute $x^{2} = t$,which implies $2x \, dx = dt$ or $x \, dx = \frac{1}{2} dt$.
Change the limits of integration:
When $x = 0$,$t = 0^{2} = 0$.
When $x = 1$,$t = 1^{2} = 1$.
Substituting these into the integral,we get:
$I = \int_{0}^{1} e^{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int_{0}^{1} e^{t} dt$.
The integral of $e^{t}$ is $e^{t}$.
Applying the limits:
$I = \frac{1}{2} [e^{t}]_{0}^{1} = \frac{1}{2} (e^{1} - e^{0})$.
Since $e^{0} = 1$,we have:
$I = \frac{1}{2} (e - 1)$.

Explore More

Similar Questions

$\int_1^{e^2} \frac{dx}{x(1+\log x)^2} = $

The integral $\int_{\pi /6}^{\pi /3} {\sec ^{2/3} x \, \csc ^{4/3} x \, dx}$ is equal to

$\int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{\frac{1}{3}}} dx$ is equal to

If $\alpha = \int_0^1 \left(e^{9x + 3 \tan^{-1} x}\right) \left(\frac{12 + 9x^2}{1 + x^2}\right) dx$,where $\tan^{-1} x$ takes only principal values,then the value of $\left(\log_e |1 + \alpha| - \frac{3\pi}{4}\right)$ is

Evaluate the following integral: $\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t \,d t$

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