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Integral Calculus> Find the integral, limit ( 0 to infinity)...

question mark

Find the integral, limit ( 0 to infinity)
∫[2e^-x] dx
[] isintegral part function

Kawal , 11 Years ago

Grade 11

1 Answers

Jitender Singh

Ans:Hello student, please find answer to your question
$I = \int_{0}^{\infty }[2e^{-x}]dx$
Since
$0 < [2e^{-x}] \leq 2, \forall x\in [0, \infty )$
$[2e^{-x}] = 1 , 0 < x < ln2$
$[2e^{-x}] = 0 , \forall x > ln2$
Hence,
$I =\int_{0}^{ln2}1.dx + \int_{ln2}^{\infty }0.dx$
$I = [x]_{0}^{ln2}$
$I = {ln2}$

Approved

Last Activity: 11 Years ago

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