Posted by **LILY** on Saturday, April 14, 2012 at 3:07am.

Evaluate the definite integral

∫(0,2) (x-1)^25 dx..

thats how i got stuck

u=x-1, then du=dx

=∫(0,2) u^25du

=(1/26)u^26. i dont know what to do with integral (2,0)..

- COLLEGE CALCULUS. HELP! -
**Steve**, Saturday, April 14, 2012 at 5:03am
just keep working away. You have the correct answer

1/26 (x-1)^26 [0,2]

= 1/26 [(1)^26 - (-1)^26]

= 1/26 [1 - 1]

= 0

- COLLEGE CALCULUS. HELP! -
**Reiny**, Saturday, April 14, 2012 at 8:13am
you have to break up your integral into two parts, since there is an x-intercept in your domain from 0 to 2, namely x = 1

think of it as finding the area from x=0 to x=1 plus the area from x=1 to x=2

we get ∫-(x-1)^25 dx from 0 to 1

= (-1/26)(x-1)^26 | from 0 to 1

= 0 - (-1/26)(-1)^26 = 1/26

and

∫(x-1)^25 dx from 1 to 2

= (1/26)(x-1)^26 | from 1 to 2

= (1/26)(1)^26 - (1/26)(0) = 1/26

so the total integral is 2/26 or 1/13

## Answer this Question

## Related Questions

- Calculus - Evaluate the following definite integral: integral at a = -1, b=2 -...
- Calculus AP - hi again im really need help TextBook: James Stewart:Essential ...
- Calculus AP - integral of [sqrt(u) - 2 u^2]/u du i got stuck ∫u^-1 (sqrt(...
- Calculus - 5) Evaluate the definite integral. On the integral from 1 to e^7 &#...
- Calculus - 5) Evaluate the definite integral. On the integral from 1 to e^7 &#...
- Calculus - Evaluate the triple integral ∫∫∫_E (x+y)dV where E ...
- Calculus - Evaluate the triple integral ∫∫∫_E (x)dV where E is...
- Calculus - Evaluate the triple integral ∫∫∫_E (xy)dV where E ...
- Calculus - 20) Evaluate the definite integral. On the integral from e to e^3 &#...
- Calculus - Evaluate the triple integral ∫∫∫_E (z)dV where E is...

More Related Questions