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What is the integral of the function f(x)=1/e^x+e^x, where x=0 to x=1.
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f(x) = 1/e^x + e^x = e^-x + e^x
Int(e^x) = e^x + C
Using the chain rule, Int(e^-x) = Int[-(e^-x)/(-1)] = e^-x/(-1) + C
so, Int(f(x)) = -e^-x + e^x = e^x - e^-x + C
and from x = 0 to 1 Int(f(x) = (e^1 - e^-1) - (e^0 - e^-0)
= (e - 1/e) - (1 - 1)
= e - 1/e
Posted by kjcdb8er on June 8, 2009 at 5:38 PM (Answer #1)
High School Teacher
We know that Int[ e^(kx)];x=0 tox=1 = (1/k)(e^k -1).
Using the this:
Integral f(x) dx =
[integral e^-x dx], x= 0 to x=1+[integral e^x dx],x=0 to x=1
[(-e^-x)], x=0 to x=1 + [ e^x ],x=0 tox=1
Posted by neela on June 8, 2009 at 7:44 PM (Answer #2)
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