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

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In summary, the integral of {sqrt(4-x^2)} from 0 to 2 can be solved by first setting x = 2sin(u) and then making a change of variable from dx to du. This results in the integral becoming 2cos(u) from 0 to 2, which simplifies to just pi.
  • #1
momogiri
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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
 
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  • #2
momogiri said:
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
Ah! Yes! It helps very much!
Thank you!
And sorry for the misplaced topic >__<;;
 
  • #4
momogiri said:
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:
 

1. What is the formula for the integral of sqrt(4-x^2) from 0 to 2?

The formula for the integral of sqrt(4-x^2) from 0 to 2 is ∫sqrt(4-x^2)dx = (1/2)arcsin(x/2) + (1/4)x*sqrt(4-x^2) + C.

2. How do you solve the integral of sqrt(4-x^2) from 0 to 2?

To solve the integral of sqrt(4-x^2) from 0 to 2, you can use the formula ∫sqrt(4-x^2)dx = (1/2)arcsin(x/2) + (1/4)x*sqrt(4-x^2) + C and plug in the upper and lower limits of integration, which in this case are 0 and 2. Then, you can simplify the equation and solve for the constant C.

3. Why do we use the formula (1/2)arcsin(x/2) + (1/4)x*sqrt(4-x^2) + C for the integral of sqrt(4-x^2) from 0 to 2?

The formula (1/2)arcsin(x/2) + (1/4)x*sqrt(4-x^2) + C comes from the substitution method, where we substitute u = 4-x^2 in the original equation. This substitution allows us to simplify the integral and solve for the constant C, making it easier to find the solution.

4. Can you use a different method to solve the integral of sqrt(4-x^2) from 0 to 2?

Yes, there are other methods that can be used to solve the integral of sqrt(4-x^2) from 0 to 2, such as the trigonometric substitution method or the integration by parts method. However, the formula (1/2)arcsin(x/2) + (1/4)x*sqrt(4-x^2) + C is the most commonly used and efficient method for this specific integral.

5. What is the significance of the integral of sqrt(4-x^2) from 0 to 2?

The integral of sqrt(4-x^2) from 0 to 2 represents the area under the curve of the function sqrt(4-x^2) between the limits of 0 and 2. This area has many real-world applications, such as in calculating the volume of a sphere or finding the arc length of a circle. It also helps in understanding the properties and behavior of the function in this specific range.

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