Integration by parts involving square root

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Homework Help Overview

The discussion revolves around the integration of the function x^3√(4-x^2)dx, specifically using the technique of integration by parts. Participants are exploring the steps involved in applying this method and the resulting expressions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to set up the integration by parts formula, with varying choices for u and v. There are discussions about the correctness of their selections and the resulting calculations. Some participants express confusion regarding the correlation between their results and the expected answer.

Discussion Status

There is ongoing dialogue about the steps taken in the integration process, with some participants questioning their own calculations and others suggesting potential errors in signs or algebraic manipulation. A few participants have offered alternative substitution methods that could simplify the problem.

Contextual Notes

Some participants note difficulties in formatting their work clearly, which may hinder understanding. There is also mention of a specific expected answer that participants are trying to reconcile with their results.

learnonthefly
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Homework Statement


|x3sqrt(4-x2)dx


Homework Equations


uv - | vdu


The Attempt at a Solution


u = x2 v = -1/3(4-x2)3/2
du = -2xdx dv = x(4-x2)1/2

uv - | vdu

x2(1/3)(4-x2)3/2 - | 1/3(4-x2)3/2(2xdx)
x2(1/3)(4-x2)3/2 +(1/3)|(4-x2)3/2(2xdx)

u = 4 - x2
du = -2xdx

x2(1/3)(4-x2)3/2 - (1/3)| u3/2du
x2(1/3)(4-x2)3/2-(1/3)2/5u5/2
x2(1/3)(4-x2)3/2-(1/3)(2/5)(4-x2)5/2
x2-(1/3)(4-x2)3/2-(2/15)(4-x2)5/2

Is this all I need? The answer I was supposed to get is -1/15(4-x2)3/2(8+3x2) I guess I don't see how that correlates
 
Last edited:
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Hello learnonthefly. Welcome to PF !

learnonthefly said:

Homework Statement


∫x^3sqrt(4-x^2)dx

Homework Equations


uv - ∫ vdu

The Attempt at a Solution


u = x^2 v = (-)1/3(4-x^2)^3/2
du = 2xdx dv = x(4-x^2)^1/2 dx

uv - ∫ vdu
I made a correction & did some editing above.

x^2(1/3)(4-x^2)^3/2 - | 1/3(4-x)^3/2(2xdx)
x^2(1/3)(4-x^2)^3/2 -(1/3)|(4-x^2)^3/2(2xdx)

u = 4 - x^2
du = 2xdx

x^2(1/3)(4-x^2)^3/2 - (1/3)| u^3/2du
x^2(1/3)(4-x^2)^3/2-(1/3)2/5u^5/2
x^2(1/3)(4-x^2)^3/2-(1/3)(2/5)(4-x^2)^5/2
That is kind of hard to read.

What is the question that you have for us ?
 
learnonthefly said:

Homework Statement


|x^3sqrt(4-x^2)dx

Homework Equations


uv - | vdu

The Attempt at a Solution


u = x^2 v = 1/3(4-x^2)^3/2
du = 2xdx dv = x(4-x^2)^1/2

uv - | vdu

x^2(1/3)(4-x^2)^3/2 - | 1/3(4-x)^3/2(2xdx)
x^2(1/3)(4-x^2)^3/2 -(1/3)|(4-x^2)^3/2(2xdx)

u = 4 - x^2
du = 2xdx

x^2(1/3)(4-x^2)^3/2 - (1/3)| u^3/2du
x^2(1/3)(4-x^2)^3/2-(1/3)2/5u^5/2
x^2(1/3)(4-x^2)^3/2-(1/3)(2/5)(4-x^2)^5/2

Homework Statement


Homework Equations


The Attempt at a Solution


You didn't actually ask a question. And it's tough to read what you've written - please format in Latex, it makes things a lot easier. :smile:

At any rate, your final answer is wrong, but it's probably a simple sign error.

For example, in the first integration by parts, ##v = -\frac{1}{3}{(4-x^2)}^{\frac{3}{2}}##. This is because of the ##-x^2## term within the parentheses.

You should go through your working carefully looking at all the sign errors and correct them.

EDIT: beaten by SammyS :biggrin:
 
Sorry guys, updated and cleaned
 
learnonthefly said:
...

Is this all I need? The answer I was supposed to get is -1/15(4-x2)3/2(8+3x2) I guess I don't see how that correlates
Your result is equivalent to:

[itex]\displaystyle (1/15) (4-x^2)^{3/2} (3 x^2+8)[/itex]

You have a sign error in you integration by substitution part.
 
Yea I see where I messed up I fixed that above. Could the algebraically inclined show me the manipulation to get the answer?
 
learnonthefly said:
Yea I see where I messed up I fixed that above. Could the algebraically inclined show me the manipulation to get the answer?
Factor [itex]\displaystyle \ (4-x^2)^{3/2}\[/itex] from your result & collect terms.
 
I think an easier way to do this would be make u=x^2 right from the beginning. Then after that make another substitution y=4-u. Then it's pretty easy from there.
 

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