Integrate the following equations using U-Substitution

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Homework Help Overview

The discussion revolves around integrating the function √(x²(x+3)) from -3 to 0 using u-substitution. Participants are exploring the implications of the absolute value in the context of the square root of x squared, particularly for negative values of x.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to apply u-substitution by letting u = x + 3 and changing the bounds accordingly. They express confusion regarding the negative result obtained from the integration, questioning the handling of the square root of x squared for negative x values.

Discussion Status

Some participants have acknowledged a mistake in their reasoning related to the absolute value of x when integrating over a negative range. There is an ongoing exploration of how this affects the final answer, with some guidance provided regarding the interpretation of √(x²) for negative x.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The discussion reflects a need to clarify assumptions about the behavior of the square root function in the context of integration limits.

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Homework Statement


Integrate √(x2(x+3)) from -3 to 0.

0
∫√(x2(x+3)) dx
-32. The attempt at a solution
Here is what I did:
√(x2(x+3))
= x√(x+3)

Let u=x+3, du=dx, x=u-3
Insert the bounds and change the bounds to: 0 to 3.

x√(x+3)=
(u-3)√(u)
= u3/2-3u1/2

Thus we have:
3
∫(u3/2-3u1/2) du
0

Finally:

(2u5/2)/5-(2u3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.
I know the answer should be positive but what I don't understand is why I'm getting a negative answer. Please help.
Thanks in advance.
 
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Xetman said:

Homework Statement


Integrate √(x2(x+3)) from -3 to 0.

0
∫√(x2(x+3)) dx
-3

2. The attempt at a solution
Here is what I did:
√(x2(x+3))
= x√(x+3)

Let u=x+3, du=dx, x=u-3
Insert the bounds and change the bounds to: 0 to 3.

x√(x+3)=
(u-3)√(u)
= u3/2-3u1/2

Thus we have:
3
∫(u3/2-3u1/2) du
0

Finally:

(2u5/2)/5-(2u3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.
I know the answer should be positive but what I don't understand is why I'm getting a negative answer. Please help.
Thanks in advance.
Be careful.

[itex]\displaystyle \sqrt{x^2} = |x|[/itex]

In fact, for x<0, [itex]\displaystyle\ \sqrt{x^2} = -x\ .[/itex]
 
Xetman said:

Homework Statement


Integrate √(x2(x+3)) from -3 to 0.

0
∫√(x2(x+3)) dx
-3


2. The attempt at a solution
Here is what I did:
√(x2(x+3))
= x√(x+3)

Let u=x+3, du=dx, x=u-3
Insert the bounds and change the bounds to: 0 to 3.

x√(x+3)=
(u-3)√(u)
= u3/2-3u1/2

Thus we have:
3
∫(u3/2-3u1/2) du
0

Finally:

(2u5/2)/5-(2u3/2) from 0 to 3.
Thus I get: -4.15.

However the correct answer is positive 4.15.
I know the answer should be positive but what I don't understand is why I'm getting a negative answer. Please help.
Thanks in advance.

It's positive because if x is negative then sqrt(x^2)=(-x). Right?
 
Oh yeah. Silly mistake XD thanks.
 

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