Integrating x^3 (x^2+20)^1/2: Steps & Answer

In summary: Check it out and give it a try. Your posts will look much better, and people will be more likely to read and respond to them.
  • #1
NSB3
2
0

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
 
Last edited:
Physics news on Phys.org
  • #2
NSB3 said:
made u the whole radical
Please post your working for that attempt.
 
  • #3
haruspex said:
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
NSB3 said:

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
NSB3 said:
u=x^2 +20
I think u2=x2+20 is a little simpler.
 
  • #6
Mark44 said:
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
NSB3 said:
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
NSB3 said:
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/
 

1. What is the purpose of integrating x^3 (x^2+20)^1/2?

The purpose of integrating x^3 (x^2+20)^1/2 is to find the area under the curve represented by the given function. This process is known as integration and is an important tool in mathematics and science.

2. What are the steps to integrate x^3 (x^2+20)^1/2?

The steps to integrate x^3 (x^2+20)^1/2 are as follows:

  1. Use the power rule to rewrite x^3 as x^2 * x
  2. Use the chain rule to rewrite (x^2+20)^1/2 as (x^2+20)^1/2 * 2x
  3. Distribute the x^2 term to get x^3 * (x^2+20)^1/2 * 2x
  4. Use the power rule again to rewrite (x^2+20)^1/2 as (x^2+20)^3/2 * (2x)^2
  5. Simplify the expression to get 2x^3 * (x^2+20)^3/2
  6. Add a constant term to the end of the expression to represent the indefinite integral

3. Can you provide a visual representation of integrating x^3 (x^2+20)^1/2?

Yes, here is a graph showing the function x^3 (x^2+20)^1/2 and its integral:

4. What is the final answer to integrating x^3 (x^2+20)^1/2?

The final answer to integrating x^3 (x^2+20)^1/2 is 2/5 * (x^2+20)^5/2 + C, where C is the constant term representing the indefinite integral.

5. How can integrating x^3 (x^2+20)^1/2 be applied in real-world situations?

Integrating x^3 (x^2+20)^1/2 can be used in various fields such as physics, engineering, and economics to calculate the area under a curve and solve real-world problems. For example, it can be used to calculate work done by a variable force, determine the volume of an irregular shape, or find the cost of producing a certain quantity of goods.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
525
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
705
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
160
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
735
  • Calculus and Beyond Homework Help
Replies
10
Views
203
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
Back
Top