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Quick integration question

by vorcil
Tags: integration
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vorcil
#1
Apr13-10, 08:05 PM
P: 394
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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rock.freak667
#2
Apr13-10, 08:09 PM
HW Helper
P: 6,202
1/2 is a constant, 1/√h is a constant

it must follow that

1/2√h is constant as well.
Mark44
#3
Apr13-10, 08:14 PM
Mentor
P: 21,311
Quote Quote by vorcil View Post
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.

vorcil
#4
Apr13-10, 08:20 PM
P: 394
Quick integration question

cheers


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