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Find the volume of the solid of revolution formed by rotating the bounded region about the x-axis.

f(x)=e-x, y=0, x=-2, x=1.

1 Answer

3 votes

Answer:

27.11514 pi

Explanation:

Given is a function exponential as


f(x) = e^(-x)

The region bounded by the above curve, y =0 , x=-2 x =1 is rotated about x axis.

The limits for x are -2 and 1

The volume when rotated through x axis is found by


\pi\int\limits^b_a {f(x)^2} \, dx

Here a = -2 and b =1

volume =
\pi\int\limits^1_(-2) {(e^-x)^2} \, dx

=
\pi ((e^(-2x) )/(-2) )\\= (\pi)/(2) (-e^1+e^4)\\= 27.11514 \pi

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