Understanding Integration with a Constant in the Limits

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Homework Help Overview

The discussion revolves around the integration of the function \(\int_0^h \frac{1}{2\sqrt{hx}}dx\), where \(h\) is treated as a constant. Participants are exploring the manipulation of constants within the integral and the implications of pulling out terms during integration.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning how to correctly factor constants out of the integral, specifically the term \(\frac{1}{2\sqrt{hx}}\) and its relationship to \(\frac{1}{2\sqrt{h}}\). There is a discussion about whether only \(\frac{1}{\sqrt{h}}\) can be factored out, and the reasoning behind treating \(2\sqrt{h}\) as a constant.

Discussion Status

Some participants have provided insights into the algebraic manipulation of the integral, suggesting that constants can be factored out. However, there is still some uncertainty regarding the treatment of the constants involved, and multiple interpretations of the problem are being explored.

Contextual Notes

Participants are working under the assumption that \(h\) is a constant during the integration process, and there may be confusion regarding the rules of factoring constants in integrals.

vorcil
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I need to figure out,

[tex]\int_0^h \frac{1}{2\sqrt{hx}}dx[/tex]

If h is a constant,

how do i do this?

my book shows that I can pull out,

[tex]\frac{1}{2\sqrt{h}} \int \frac{1}{\sqrt{x}}dx[/tex]

How does the 2 from [tex]\frac{1}{2\sqrt{hx}}[/tex] come out with the [tex]\sqrt{h}[/tex]?

I thought I would've only been able to pull out 1/root h,

like this,

[tex]\frac{1}{\sqrt{h}} \int \frac{1}{2\sqrt{x}}dx[/tex]

-

why does 2 root h get assigned constant? instead of only h
 
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1/2 is a constant, 1/√h is a constant

it must follow that

1/2√h is constant as well.
 
vorcil said:
I need to figure out,

[tex]\int_0^h \frac{1}{2\sqrt{hx}}dx[/tex]

If h is a constant,

how do i do this?

my book shows that I can pull out,

[tex]\frac{1}{2\sqrt{h}} \int \frac{1}{\sqrt{x}}dx[/tex]

How does the 2 from [tex]\frac{1}{2\sqrt{hx}}[/tex] come out with the [tex]\sqrt{h}[/tex]?

I thought I would've only been able to pull out 1/root h,

like this,

[tex]\frac{1}{\sqrt{h}} \int \frac{1}{2\sqrt{x}}dx[/tex]

-

why does 2 root h get assigned constant? instead of only h
The basic idea is that [itex]\int k*f(x) dx = k*\int f(x) dx[/itex].

The rest in your problem is just algebra.
[tex]\frac{1}{2\sqrt{hx}} = \frac{1}{2*\sqrt{h}\sqrt{x}} = \frac{1}{2\sqrt{h}} \frac{1}{\sqrt{x}}[/tex]

Integration is being done with respect to x (i.e., with x as the variable), so h is just another constant in this process.
 
cheers
 

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