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anonymous
 5 years ago
What did I do wrong? Calc. Substitution
Problem: [21< (this is from x=2 to x=1) x(x1)^1/2 dX
My Answer and work:
Substitute U for x1
du/dx = 1
du=dx
[21 x (u)^1/2
(x^2)/2 * (u^3/2)/(2/3) ]21
(1/2(2)^2 * (3/2)1^(3/2))  (1/2(1)^2*(3/2)(0)^(3/2)
This comes out to 1.5 but the answers in the back of the book say it is 16/15. Can someone show me where I went wrong and get me to the right answer?
anonymous
 5 years ago
What did I do wrong? Calc. Substitution Problem: [21< (this is from x=2 to x=1) x(x1)^1/2 dX My Answer and work: Substitute U for x1 du/dx = 1 du=dx [21 x (u)^1/2 (x^2)/2 * (u^3/2)/(2/3) ]21 (1/2(2)^2 * (3/2)1^(3/2))  (1/2(1)^2*(3/2)(0)^(3/2) This comes out to 1.5 but the answers in the back of the book say it is 16/15. Can someone show me where I went wrong and get me to the right answer?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You left an x when you substituted. Even integrated it. You did u=x1. So x=u+1 If you do this, you end up with (u+1)*sqrt(u) which you can distribute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ohh thanks, I didnt realize that you had to pay attention to that X. Will you finish the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure. (u+1)*sqrt(u) Distribute u^(3/2)+u^(1/2) Integrate and you get: (2/5)u^(5/2)+(2/3)u^(3/2) We also change the limits of integration using u=x1 So we are integrating from u=1 to u=0 Evaluating this we get 16/15 If integrating from x=1 to x=2, this same procedure gives 16/15
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