Integral of {sqrt(4-x^2)} from 0 to 2?

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[SOLVED] Integral of {sqrt(4-x^2)} from 0 to 2?

Can someone please explain step by step for this?
I know the answer is pi, I just don't understand the steps towards solving it

First, I'm supposed to set x = 2sin(u)
so then u = arcsin(x/2)
then somehow, {sqrt(4-x^2)} = 2cos(u)
How'd that happen? @__@
I'm really really confused!

Thanks for tolerating XD
 

Answers and Replies

  • #2
Hootenanny
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Can someone please explain step by step for this?
I know the answer is pi, I just don't understand the steps towards solving it

First, I'm supposed to set x = 2sin(u)
so then u = arcsin(x/2)
then somehow, {sqrt(4-x^2)} = 2cos(u)
How'd that happen? @__@
I'm really really confused!

Thanks for tolerating XD
Okay so we have,

[tex]\int^{2}_{0} \sqrt{4-x^2}\;\;dx[/tex]

And if we let [itex]x = 2\sin u[/itex], then we obtain,

[tex]\int^{2}_{0} \sqrt{4-\left(2\sin u\right)^2}\;\;dx[/tex]

[tex] = \int^{2}_{0} \sqrt{4-4\sin^2 u}\;\;dx[/tex]

[tex] = \int^{2}_{0} \sqrt{4\left(1-\sin^2 u\right)}\;\;dx[/tex]

[tex] = \int^{2}_{0} 2\sqrt{1-\sin^2 u}\;\;dx[/tex]

However, we know that [itex]\sin^2\theta+\cos^2\theta = 1 \Rightarrow 1-\sin^2\theta = \cos^2\theta[/itex]. Hence, the integral becomes,

[tex]\int^{2}_{0} 2\cos u \;\;dx[/tex]

Is that clear? Don't forget that before you can integrate, you need to make a change of variable from dx to du.

P.S. We have Homework & Coursework forums for textbook questions and assistance with homework and we have Mathematics forums for general mathematics discussions.
 
  • #3
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Ah! Yes! It helps very much!
Thank you!!
And sorry for the misplaced topic >__<;;
 
  • #4
Hootenanny
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Ah! Yes! It helps very much!
Thank you!!
And sorry for the misplaced topic >__<;;
A pleasure. And don't worry about misplacing your topic, we'll let you off... this time :devil:
 

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