Evaluate the definite integral: $\int_{0}^{1} \left(1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots \infty\right) e^{2x} \, dx$.
- A$e^{2}$
- B$e - 1$
- C$e + 1$
- D$e$
Explore More
Similar Questions
If $\int_0^3 (3x^2 - 4x + 2) dx = k$,then a root of $3x^2 - 4x + 2 = \frac{3k}{5}$ that lies in the interval $[0, 3]$ is
$\int_0^{\pi / 2} \frac{d x}{4+5 \sin x}$
DifficultTS EAMCET 2020
View Solution$\int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx =$
$\int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2}+4 \sin \frac{x}{2}} \, dx$ is equal to
The value of the integral $\int_{-1/2}^{1/2} \left[ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 - 2 \right]^{1/2} dx$ is
Difficult
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