the area enclosed between the curve y=e^x ,the x axis and the lines x=0 and x=2 is revolved about the x axis . find the volume swept out using intergration
2
i know that , PI ( ex^2 dx
)
0
the promblem is with the e^x to the power 2 if i intergrate it do i have to treat it as 2 functions and if so how do i do that cause never done that in intergration only in differentation. ty for any help
this is the area
this is the volume
try to lean latex it is easy
thats what I write in latex but you should delete the 1 in the [1/math]
then you will have the same as what I write
[tex]\int_{0}^{2} (e^x)dx[1/math]
[tex]\pi\int_{0}^{2} (e^x)^2 dx[1/math]
[tex]\pi\int_{0}^{2} (e^{2x}) dx[1/math]