Question

Find the volume of revolution when the region enclosed by the functions f(x)= e^x , g(x)=e^-x ,and the two lines is revolved about the x-axis
Solution
V=intergrate e^2x-e^-2x dx
From -1to1
1/2e^+e^-2x when you sub and cal. =0 why please help me

Answer 1

you don't say which two lines, but I infer that they are x = -1 and x=1.

The region is symmetric about the y-axis, so we can just double the area from 0 to 1.

using discs,
v = 2∫[0,1] π(R^2-r^2) dx
where R=e^x and r=e^-x
v = 2∫[0,1] π(e^2x - e^-2x) dx
This is just what you had.

Now, you know that
∫ e^u du = e^u
if we let
u = 2x
du = 2dx
but we only have dx, which is du/2. So,

∫e^(2x) dx = 1/2 ∫ e^u du

That's where the 1/2 comes from, and why the - changes to +, because you are multiplying by -1/2 in the 2nd term.

Not sure what the =0 problem is. e^0 = 1, though, if that helps.