Area under a curve

**bobsanchez** · August 6th 2010, 01:04 PM

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...

**bobsanchez** · August 6th 2010, 01:05 PM

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.

**11rdc11** · August 6th 2010, 01:11 PM

Here is the integration you wanted

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

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

and sub

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

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

The other method was a lot easier lol.

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

$sin2u +2u$

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

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

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

now plug all back in

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

now just use your limits from -2 to 2

**Archie Meade** · August 6th 2010, 01:32 PM

So, was this a question to be answered solely with integration techniques,
or any method?

**bobsanchez** · August 7th 2010, 06:20 AM

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

**Archie Meade** · August 7th 2010, 06:55 AM

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?

If you draw a quarter-circle in the first quadrant, radius=2,
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

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

Label the acute angle at the origin $\theta$

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

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

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

$x=2,\ \theta=0$

This allows you to integrate the trigonometric integral

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

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

Double your final answer.

**bobsanchez** · August 7th 2010, 08:50 AM

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

**Suomik1988** · August 7th 2010, 09:22 AM

I did this using trig substitution.

**HallsofIvy** · August 7th 2010, 11:22 AM

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 $(1/2)\pi r^2$ .

**Archie Meade** · August 7th 2010, 03:57 PM

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 $\int{x^6}dx$

$u=2-x$

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

$du=-dx$

$dx=-du$

Now, you have a "power" integral.

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

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

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

**mr fantastic** · August 7th 2010, 04:48 PM

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 $y = \sqrt{1 - x^2}$ ) 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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