Download MARKS App - Trusted by 15,00,000+ IIT JEE & NEET aspirants! Download Now
Search any question & find its solution
Question: Answered & Verified by Expert
$\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\cdots\right.$ upto $\left.\infty\right) e^{2 x} d x=$
MathematicsDefinite IntegrationMHT CETMHT CET 2020 (19 Oct Shift 2)
Options:
  • A $e^{2}$
  • B $e-1$
  • C $e+1$
  • D $e$
Solution:
1786 Upvotes Verified Answer
The correct answer is: $e-1$
(B)
$\int_{0}^{1}\left(1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots \infty\right) e^{2 x} d x$
$=\int_{0}^{1} e^{-x} e^{2 x} d x=\int_{0}^{1} e^{x} d x$
$=\left[e^{x}\right]_{0}^{1}=e^{1}-e^{0}=e-1$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.