Calculus 2. Is this the right answer?

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The question is: Find the area of the region bounded by #y=xe^(-x^2/2)# and its asymptote. For my final answer, I got an area equal to 1 and I was hoping if someone could tell me if this is correct. Thanks in advance!

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Answer 1 · 2017-03-15T00:20:23

#int_0^(oo)(xe^(-x^2/2))dx=1#

The area is 1.

#y=xe^(-x^2/2)#
graph{xe^(-x^2/2) [-6.244, 6.243, -3.12, 3.123]}

Since the problem is referring to the curve bounded within #"quadrant I"#, we are solving for:
#int_0^(oo)(xe^(-x^2/2))dx#

(Fix notation in advance by writing as a limit of a definite integral bounded by #0# and #Q#)
#=lim_(Q->oo)[int_0^(Q)(xe^(-x^2/2))dx]#

#=lim_(Q->oo)[-e^(-x^2/2)]_0^(Q)#

#=lim_(Q->oo)[(-e^(-Q^2/2))-(-e^(-0^2/2))]#

#=lim_(Q->oo)[-e^(-Q^2/2)+1]#

#=lim_(Q->oo)[frac{-1}{e^((Q^2/2))}+1]#

#=1#