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

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Jan 4, 2018

Answer:

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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