Question

Find the volume of the region bounded by the graph y=e^(-x), y=0, x=0, x=1 revolved around the x-axis?

Answer 1

When I found the integral of pif(x)^2, it was negative: any idea why I'm getting a negative volume?

Answer 2

@ganeshie8

Answer 3

cant really be negative if you are squaring the function, it has to be positive

Answer 4

hey satellite can you help me with a wuestion after this?

Answer 5

as long as it isn't geometry sure

Answer 6

yes!!!

Answer 7

The integral of \[\int\limits_{0}^{1}\pi e^{-2x} dx=(- \pi e^{-2x})/2\]

Answer 8

Right?

Answer 9

\[\pi \int _0^1e^{-2x}dx\] i guess

Answer 10

But that gives me negative volume when I plug in one, which of course doesn't make sense.

Answer 11

did you forget to plug in zero and subtract?

Answer 12

It's from 0 to 1, not -1 to 0

Answer 13

\[F(1)-F(0)\] is what you need to compute

Answer 14

it is positive for sure

Answer 15

Yes, and I'm saying that pi*1/(-2e^(2)) is a negative number.

Answer 16

it doesn't matter that \(F(1)\) is negative
\[F(1)-F(0)\] is positive

Answer 17

\[\int_0^1\pi e^{-2x}~dx=\pi\left[-\frac{1}{2}e^{-2x}\right]_0^1=-\pi\left(e^{-2}-e^{0}\right)=\pi\left(1-\frac{1}{e^2}\right)\]
and \(\dfrac{1}{e^2}<1\), so \(1-\dfrac{1}{e^2}>0\).

Answer 18

... divided by 2...

Answer 19

anti derivative is
\[-\frac{1}{2e^{2x}}\]

Answer 20

That's all I was wondering. I forgot that e^0 =/= 0. Thanks.

Answer 21

at 1 you get
\[-\frac{1}{e^2}\] at 0 you get
\[-\frac{1}{2}\] then
\[-\frac{1}{e^2}-(-\frac{1}{2})\] is positive