http://www.ck12.org Chapter 7. Integration Techniques
So the volume of the solid isπ/ 4.
Example 6:
Evaluate∫−+∞∞ex+dxe−x.
Solution:
This can be a tough integral! To simplify, rewrite the integrand as
1
ex+e−x=1
e−x(e^2 x+ 1 )=ex
e^2 x+ 1 =ex
1 +(ex)^2.Substitute into the integral:
∫ dx
ex+e−x=∫ ex
1 +(ex)^2 dx.Usingu−substitution, letu=ex,du=exdx.
∫ dx
ex+e−x=∫ du
1 +u^2
=tan−^1 u+C
=tan−^1 ex+C.Returning to our integral with infinite limits, we split it into two regions. Choose as the split point the convenient
x= 0.
∫+∞
−∞dx
ex+e−x=∫ 0
−∞dx
ex+e−x+∫+∞
0dx
ex+e−x.Taking each integral separately,
∫ 0
−∞dx
ex+e−x=l→−lim∞∫ 0
ldx
ex+e−x
=l→−lim∞[tan−^1 ex]^0 l
=l→−lim∞[
tan−^1 e^0 −tan−^1 el]
=π 4 − 0
=π 4.Similarly,