Figuring out the Volume of Revolution and avoiding $u$-substitution

Question

I'm trying to figure out how to solve this question.

Calculate the volume of revolution when V is revolved around the x-axis. Give an exact answer.

$f(x)=\frac{2e^x}{1+e^x}$

The area V is enclosed by the graph of $f(x)$ and lines $x=2$ and $y=1$

I know we should integrate the function this way to find the volume when the graph is revolved around the x-axis: $$\pi\int_?^? f(x)^2dx$$

But well, first of all, I'm not sure which $x$ intervals are meant here. I only have the one $x=2$.

Secondly, I did not learn $u$ substitution, not really sure how I'm supposed to integrate $\frac{2e^x}{1+e^x}$ (which is what we are supposed to do without $u$-substitution).

Thanks in advance for any help!

Answer 1

HINT

• make a sketch of $f(x)$ which is an increasing funtion which is $<1$ for $x<0$
• then with the given data the set up should be

$$\pi\int_{-\infty}^0 [f(x)]^2\,dx+\pi\int_{0}^2 1^2\,dx$$