7.6. Improper Integrals http://www.ck12.org
From the figure above, the area of the region to be revolved is given byA=πy^2 =πx^2 e−^2 x. Thus the volume of the
solid is
V=π∫∞
0 x(^2) e− 2 xdx=πlim
l→∞
∫l
0 x
(^2) e− 2 xdx.
As you can see, we need to integrate by parts twice:
∫
x^2 e−^2 xdx=−x
2
2 e
− 2 x+
∫
xe−^2 xdx
=−x
2
2 e
− 2 x−x
2 e
− 2 x−^1
4 e
− 2 x+C.
Thus
V=πllim→∞
[
−x2
2 e− 2 x−x
2 e− 2 x−^1
4 e− 2 x]l
0
=πllim→∞[ 2 x (^2) + 2 x+ 1
− 4 e^2 x
]l
0
=πllim→∞
[ 2 l (^2) + 2 l+ 1
− 4 e^2 l −
1
− 4 e^0]
=πllim→∞[ 2 l (^2) + 2 l+ 1
4 e^2 l +
1
4
]
.
At this stage, we take the limit aslapproaches infinity. Notice that the when you substitute infinity into the function,
the denominator of the expression^2 l^2 −+ 42 el 2 +l^1 ,being an exponential function, will approach infinity at a much faster
rate than will the numerator. Thus this expression will approach zero at infinity. Hence
V=π[
0 +^14
]
=π 4 ,