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How do you determine if the improper integral converges or diverges $e^(-2t) dt$ from negative infinity to -1?

1 Answer

Jan 4, 2018

The integral is divergent

Explanation:

The improper ntegral is

$I=int_x^-1e^(-2t)dt$

$=[-1/2e^-(2t)]_x^-1$

$=[1/2e^(-2t)]_-1^x$

$=1/2e^(-2x)-1/2e^2$

Therefore,

$lim_(x->-oo)I=lim_(x->-oo)(1/2e^(-2x)-1/2e^2)=+oo$

The integral is divergent

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