Integral from 0 to 1: x^2 (4x^3+2)^3 dx = x^3/3(x^4)^3 = F(1) - F(0)

F(1) = 1/3

F(0) = 0

= 1/3

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- Apr 14th 2008, 10:43 AMshepherdm1270check evaluation of indefinate integral
Integral from 0 to 1: x^2 (4x^3+2)^3 dx = x^3/3(x^4)^3 = F(1) - F(0)

F(1) = 1/3

F(0) = 0

= 1/3 - Apr 14th 2008, 10:58 AMearboth
- Apr 14th 2008, 11:03 AMSean12345
Or you could think of it as

Hence

- Apr 14th 2008, 11:21 AMshepherdm1270
I'm not understanding where you got the 1/48

- Apr 14th 2008, 11:26 AMshepherdm1270
- Apr 14th 2008, 11:41 AMMoo
Hello,

;) - Apr 14th 2008, 11:56 AMSean12345
- Apr 14th 2008, 10:52 PMearboth
- Apr 14th 2008, 11:08 PMcigster
Introduce a new variable:

u = 4x^3 + 2

They you can get dx=du/12x^2. You insert both of these two into you initial equation and get:

x^2*u^3*du/12x^2

Due to new variable you must change limits into 2 to 6, getting final equation

u^4/48Inserting the limits you obtain the result 26.67 or 80/3.