# Thread: Area under a curve

1. Okay, well I still don't really know how to integrate functions like that, and not all of them are semicircles. For instance, integrating (2-x)^6....I'm not sure how you would do that other than the backwards chain rule, and I'm not sure how you do that either...

2. Originally Posted by Ackbeet
Then 6.28 might not be the more accurate answer, but it is the better answer. In engineering, at least, you better give an answer that your boss is happy with. If he wants decimals, give him decimals. Otherwise, it won't be as useful to him.
Haha yeah, I know what you mean. It's kind of irritating because technically it isn't as correct...but still. You're right.

3. Here is the integration you wanted

$\displaystyle \int^{2}_{-2}\sqrt{4-x^2}dx$

let $\displaystyle x = 2\sin{\u}$ and $\displaystyle dx = 2\cos{u}du$

and sub

$\displaystyle \int(\sqrt{4-4\sin^2{u}})(2\cos{u})du$

$\displaystyle \int(4\cos^2{u})du$

The other method was a lot easier lol.

$\displaystyle \int(2\cos{2u}+2)du$

$\displaystyle sin2u +2u$

$\displaystyle 2\sin{u}\cos{u} +2u$

$\displaystyle u =arcsin(\frac{x}{2})$

$\displaystyle \cos{u}=\sqrt{1-\frac{x^2}{4}}$

now plug all back in

$\displaystyle 2(\frac{x}{2})(\sqrt{1-\frac{x^2}{4}})+2arcsin(\frac{x}{2})$

now just use your limits from -2 to 2

or any method?

5. It was supposed to be to practice integration, but it's all the same for getting the answer.

6. Originally Posted by bobsanchez
It was supposed to be to practice integration, but it's all the same for getting the answer.
Have you moved on to trigonometric substitutions?

pick any point on the circumference not on an axis,
join the point to the origin (circle centre) and drop a perpendicular to the x-axis
then you have a right-angled triangle.

By Pythagoras' theorem

$\displaystyle \left(\sqrt{4-x^2}\right)^2+x^2=4=2^2$

Label the acute angle at the origin $\displaystyle \theta$

$\displaystyle \displaystyle\huge\ sin\theta=\frac{\sqrt{4-x^2}}{2}\ \Rightarrow\ \sqrt{4-x^2}=2sin\theta$

$\displaystyle \displaystyle\huge\ cos\theta=\frac{x}{2}\ \Rightarrow\ x=2cos\theta\ \Rightarrow\ dx=-2sin\theta d\theta$

$\displaystyle x=0,\ \theta=\frac{{\pi}}{2}$

$\displaystyle x=2,\ \theta=0$

This allows you to integrate the trigonometric integral

$\displaystyle \displaystyle\huge\int_{\frac{{\pi}}{2}}^{0}2sin\t heta(-2sin\theta)d\theta=4\int_{0}^{\frac{{\pi}}{2}}sin^ 2\theta d\theta$

$\displaystyle =\displaystyle\huge\ 2\int_{0}^{\frac{{\pi}}{2}}(1-cos2\theta)d\theta$

7. Haven't done trigonometric subs yet...let me dissect your post and see what I get.

8. ## Another idea...

I did this using trig substitution.

9. Originally Posted by bobsanchez
Haven't done trigonometric subs yet...let me dissect your post and see what I get.
If you haven't then your teacher probably expected you to recognize that this was a semi-circle and use $\displaystyle (1/2)\pi r^2$.

10. Originally Posted by bobsanchez
Okay, well I still don't really know how to integrate functions like that, and not all of them are semicircles. For instance, integrating (2-x)^6....I'm not sure how you would do that other than the backwards chain rule, and I'm not sure how you do that either...
In that case, you can integrate with respect to 2-x,
or rename it "u".

You would have a "power" integral if it was $\displaystyle \int{x^6}dx$

$\displaystyle u=2-x$

$\displaystyle \frac{du}{dx}=-1$

$\displaystyle du=-dx$

$\displaystyle dx=-du$

Now, you have a "power" integral.

$\displaystyle \int{(2-x)^6}dx=-\int{u^6}du$

It's about getting it into a form you can integrate.

The advantage of integrating the "semi-circle" function is..

since you would know the area of a semicircle,
then you will be "dead sure" your integration is incorrect if you do not get an answer of

$\displaystyle \frac{{\pi}r^2}{2}$

11. Originally Posted by bobsanchez
Oh, sorry. I had the question mark there because I didn't know where you were going with it, not whether or not it's a semicircle. Um...I don't really know. It's doesn't quite touch the x-axis according to my graphing calculator, and as a result it is sort of messing with all of the estimations I've attempted, and I don't know how to integrate a function like that.
If you're studying calculus it is expected that:

1. You can sketch graphs (like $\displaystyle y = \sqrt{1 - x^2}$) by hand, without having to use a graphing calculator. If you cannot do this you are strongly advised to review pre-calculus.

2. You understand the limitations (due to resolution etc.) of graphs drawn using a graphics calculator.

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