Evaluate the definite integral from 1 to 0?

Question

Evaluate the definite integral from 1 to 0

(e^{8 x} - 8 x)^{4} (e^{8 x} - 1) dx =

Answer 1

use a ti-89 and go to calculus, integrate, then put in the (function, x, 0, 1) and then make it negative because you cant have the integral from 1 to 0. It has to be from 0 to 1 and so you flip it and make it negative.

Answer 2

Call e^8x-8x = z
so 8[e^8x-1) dx = dz and the integral becomes
1/8*Int z^4dz = 1/40z^5 taken between e^8 - 8 and 1
=1/40[1-(e^8-8)^5]

Answer 3

as D((e^{8 x} - 8 x))=8(e^{8 x} - 1) we have
let t=e^{8 x} - 8 x) then dt=8(e^{8 x} - 1)
so th e problem can be written as
Int (t^4)/8 dt
=t^5/(8x5)=t^5/40

giving answer (e^{8 x} - 8 x))^5/40 +c
integrating 0 to 1 you get
(e^1-8)^5/40-(e^0-8x0)^5/40
=((e-8)^5-1)/40

Answer 4

definite, you should use a u substitution to remedy this one. in case you enable u = 5x+8, du = 5 dx, so dx = du/5. replace decrease back into a million/5x+8 dx to get a million/(5u) du the vital of one million/(5u) du is ln(u)/5. you may positioned 5x+8 decrease back into u to have ln(5x+8)/5. evaluate from 0 to at least a million.