Integral of $\sqrt[3]{x}$ from 8 to 27

[Dell]

given the integral


$$\int$$e^{$$\sqrt[3]{x}$$}dx from 8 to 27

i called $$\sqrt[3]{x}$$=t
and integrate now from 2-3

x=t³
dx=3t²dt

$$\int$$e^t3t²dt
=3$$\int$$e^tt²dt

u=t²
du=2tdt

dv=e^tdt
v=e^t

$$\int$$udv=uv-$$\int$$vdu

3$$\int$$e^tt²dt=3(t²e^t-$$\int$$e^t2tdt)

is this correct so far?? do i now need to, again integrate in parts, now

u=t
du=dt
dv=e^tdt
v=e^t
?

 

[Dick]

Yes, integrate by parts again. It looks like you are doing just fine so far.