Here is a question:
Find the volume of the solid formed by rotating the region enclosed by the graph of the function and the $x$ axis through $2\pi$ radians about the $x$ axis in the given integral:
y = $e^{-x}$ for $x$ from $[0,1]$
I am getting the following answer:
Volume = $\int_0^1\pi e^{-2x} dx$
The antiderivative is $-\frac{\pi}{2} e^{-2x}$
I Evaluate this in the interval $0$ to $1$ and I get an answer of $-\frac{\pi}{2} (e^{-2}+1)$ However the given solution is $\frac{\pi}{2}(1-e^{-2}),$ so I am messing up my signs somewhere. Please could some body help me out! Much appreciated :).
As of the antiderivative, everything you had was correct. Carrying on from there, you would have:
$$-\frac{\pi}{2}e^{-2x}\Bigg|_0^1 = -\frac{\pi}{2}\left(e^{-2} - e^0\right) = -\frac{\pi}{2}\left(e^{-2} - 1\right)$$
which is equivalent to the given answer.