Integration Problem

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Homework Statement


the integral of x^3 (x^2 + 20)^1/2

Homework Equations


use u substitution

The Attempt at a Solution


I think I have finally figured the problem out, can you confirm if this is the correct answer please?

u=x^2 +20 x= sqrt(u-20)
du= 2x dx
integral of x^3 * sqrt( u) du/2x
cancel the x's and move the 1/2 in front of the integral
plug in the sqrt(u-20) for x
1/2 integral of (sqrt(u-20))^2 * sqrt(u) du
1/2 integral of u-20 * sqrt(u) du
now I distribute the sqrt(u) to the (u-20) and get
1/2 integral of u^3/2 - 2u^1/2
then I integrated getting
1/2[2/5u^5/2 - 4/3u^3/2]
finally getting 1/2[2/5(x^2+20)^5/2 - 4/3 (x^2+20)^3/2] + C
 
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Answers and Replies

  • #2
haruspex
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made u the whole radical
Please post your working for that attempt.
 
  • #3
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Please post your working for that attempt.
Moderator's Note: There was a previous version that the OP edited when he finally figured out what to do.
 
  • #4
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Homework Statement


the integral of x^3 (x^2 + 20)^1/2

Homework Equations


use u substitution

The Attempt at a Solution


I think I have finally figured the problem out, can you confirm if this is the correct answer please?

u=x^2 +20 x= sqrt(u-20)
du= 2x dx
integral of x^3 * sqrt( u) du/2x
cancel the x's and move the 1/2 in front of the integral
plug in the sqrt(u-20) for x
1/2 integral of (sqrt(u-20))^2 * sqrt(u) du
1/2 integral of u-20 * sqrt(u) du
now I distribute the sqrt(u) to the (u-20) and get
1/2 integral of u^3/2 - 2u^1/2
then I integrated getting
1/2[2/5u^5/2 - 4/3u^3/2]
finally getting 1/2[2/5(x^2+20)^5/2 - 4/3 (x^2+20)^3/2] + C
The first term is correct, but should be simplified. The second term's coefficient is off.

You can check to see if your answer is correct by differentiating your answer. If the derivative equals the original integrand, then all is good.
 
  • #5
haruspex
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u=x^2 +20
I think u2=x2+20 is a little simpler.
 
  • #6
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The first term is correct, but should be simplified. The second term's coefficient is off.

You can check to see if your answer is correct by differentiating your answer. If the derivative equals the original integrand, then all is good.
I realized that I dropped the 0 on 20 and put 2 on the 4th line from the bottom. So then I got 40/3 as coefficient instead of 4/3 so my new final answer after distributing the 1/2 is
((x^2+20)^5/2)/5 - (20(x^2+20)^3/2)/3 +C is it right to distribute the 1/2 in?
 
  • #7
Ray Vickson
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I realized that I dropped the 0 on 20 and put 2 on the 4th line from the bottom. So then I got 40/3 as coefficient instead of 4/3 so my new final answer after distributing the 1/2 is
((x^2+20)^5/2)/5 - (20(x^2+20)^3/2)/3 +C is it right to distribute the 1/2 in?
What did you get when you differentiated your "answer"? Did you get your original integrand, or did you get something else? If you got your integrand, then your answer is correct (if you differentiated correctly); otherwise, it is incorrect (or else you differentiated incorrectly). Please report what you got.
 
  • #8
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I realized that I dropped the 0 on 20 and put 2 on the 4th line from the bottom. So then I got 40/3 as coefficient instead of 4/3 so my new final answer after distributing the 1/2 is
((x^2+20)^5/2)/5 - (20(x^2+20)^3/2)/3 +C is it right to distribute the 1/2 in?
This is correct, now, but as Ray and I said, you should differentiate your answer - you don't need us to confirm your answer (provided that you can differentiate correctly). Also, you don't distribute the 1/2 - you distribute the 1/3.

I would simplify the answer to make it clearer and cleaner by putting the constants at the front of the two terms, like so:
(1/5)(x2 + 20)^(5/2) - (20/3)(x2 + 20)^(3/2) + C
Better:
(1/5)(x2 + 20)5/2 - (20/3)(x2 + 20)3/2 + C
Best (using LaTeX):
##\frac{1}{5}(x^2 + 20)^{5/2} - \frac{20}{3}(x^2 + 20)^{3/2} + C##

We have a page on how to get started with LaTeX: https://www.physicsforums.com/help/latexhelp/
 

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