James889
- 190
- 1
Hi,
The area
e^x-1
Is rotated about the y axis, bounded by y=1, x=0 and x=ln2 find the volume of the solid.
And i am clearly making something wrong, so if anyone could verify my work.
~ 2\pi\int_0^{ln2}x\cdot(1-(e^x-1)
-2\pi\int_0^{ln2}xe^x-2x
Integration:
u=x, du=1
dv=e^x, v=e^x
\int xe^x -2x= xe^x -e^x -x^2
e^x(x-1)-x^2\bigg|_0^{ln2} = 2\pi \cdot ln2-1 -(ln2)^2
The answer is supposed to be 2\pi(ln2-1)^2
Thanks!
The area
e^x-1
Is rotated about the y axis, bounded by y=1, x=0 and x=ln2 find the volume of the solid.
And i am clearly making something wrong, so if anyone could verify my work.
~ 2\pi\int_0^{ln2}x\cdot(1-(e^x-1)
-2\pi\int_0^{ln2}xe^x-2x
Integration:
u=x, du=1
dv=e^x, v=e^x
\int xe^x -2x= xe^x -e^x -x^2
e^x(x-1)-x^2\bigg|_0^{ln2} = 2\pi \cdot ln2-1 -(ln2)^2
The answer is supposed to be 2\pi(ln2-1)^2
Thanks!