# What is the integral of the function f(x)=1/e^x+e^x, where x=0 to x=1.

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### 2 Answers

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**

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

[-(1/e)+1]+[e-1]

e-1/e