COLLEGE CALCULUS. HELP!

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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 don't know what to do with integral (2,0)..

  • COLLEGE CALCULUS. HELP! -

    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! -

    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

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