How do you integrate #(e^(x-2))^2/e^(2x)#?

2 Answers
Oct 1, 2016

Answer:

Use properties of exponents to rewrite.

Explanation:

#(e^(x-2))^2 = e^(2(x-2)) = e^(2x-4) = e^(2x)e^(-4)#

So #(e^(x-2))^2/e^(2x) = (e^(2x)e^-4)/e^(2x) = e^-4#

Thus #int (e^(x-2))^2/e^(2x) dx = int e^-4 dx = x e^-4 +C#

Oct 1, 2016

Answer:

#x/e^4 + c#

Explanation:

#int(e^(x-2))^2/e^(2x) dx = int(e^(2x-4))/e^(2x) dx#

#= inte^(2x)/(e^(2x)*e^4) dx#

#= int cancel(e^(2x))/(cancel(e^(2x))e^4) dx#

Thus the integral reduces to:

#int 1/e^4 dx#

Since #1/e^4# is a constant:

#= 1/e^4 int dx = 1/e^4*x +c#

#=x/e^4 + c#