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eek
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#1
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#1
could anyone help on this question please:

integrate x(x-1)^4

thank you!
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Jonny W
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#2
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#2
(int) x.(x - 1)^4 dx
= x.(1/5)(x - 1)^5 - (int) 1.(1/5)(x - 1)^5 dx
= (1/5)x(x - 1)^5 - (1/30)(x - 1)^6 + c
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Gaz031
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#3
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(Original post by eek)
could anyone help on this question please:

integrate x(x-1)^4

thank you!
Take:
v=x, dv/dx=1, du/dx=(x-1)^4, u=(1/5)(x-1)^5, (standard form. You can use a substitution to prove this to yourself for any values n, a in (x-a)^n)

Hence:
INT x(x-1)^4 = (1/5)(x)(x-1)^5 - INT[(1/5)(x-1)^5]
= (x/5)(x-1)^5 - (1/5)(1/6)(x-1)^6
= (1/5)(x-1)^5[ x - (1/6)(x-1) ] + C

Edit:
The last line is just the factoring of (1/5)(x-1)^5 out as it is a common factor. You could similarly change it to:
(1/30)(x-1)^5 [6x - (x-1) = (1/30)(5x+1)(x-1)^5, which is a neater form of the answer and is thus the one given in the book.
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eek
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#4
Report Thread starter 14 years ago
#4
gaz - i'm not sure how you got the last line and also the answer in the book is

(5x+1)(x-1)^5/30

????
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Gaz031
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#5
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Edited my post above to show how it is equal to the answer given in the book. Note i've simply factored out the fractions and expressions.
Also, if you divide the outside of the brackets by 6 you'll multiply the inside by 6 so the expression is exactly the same.
It's usually preferable for there to be as few fractions as possible.
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Nima
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#6
Report 14 years ago
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(Original post by eek)
gaz - i'm not sure how you got the last line and also the answer in the book is

(5x+1)(x-1)^5/30

????
JonnyW: (1/5)x(x - 1)^5 - (1/30)(x - 1)^6 + c
= [6x(x-1)^5]/30 - [(x-1)^6]/30 + c
= [6x(x-1)^5 - (x-1)^6]/30 + c
= {[(x-1)^5][6x - (x-1)]}/30 + c
= [(5x+1)(x-1)^5]/30 + c
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eek
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#7
Report Thread starter 14 years ago
#7
ah ok!! thank you!
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