An integration that doesn't quite make sense.

**Showcase_22** · September 11th 2008, 04:52 AM

The question is copied directly from my book but the working is my own.

The answer isn't 0 and I have a feeling that my limits are wrong. I'm not really sure how else to write the limits in order for me to get an answer than isn't 0.

help would be greatly appreciated.

**Laurent** · September 11th 2008, 07:12 AM

To find your error, I looked for the line where you start integrating functions taking negative values. This is the line with $\int_0^\pi \cos t\sin t dt$ . To go from the previous line into this one, you wrote $(\cos^2 t)^{1/2}=\cos t$ , whereas it should have been $(\cos^2 t)^{1/2}=|\cos t|$ .

Laurent.

**Showcase_22** · September 11th 2008, 07:17 AM

Okay then. I have absolutely no idea how to integrate moduli.

The last one I did was easy and it was just finding the area of two triangles and add them together. I don't think I can do the same for this.

I suppose what i'm asking is "how do you integrate moduli?"

**Moo** · September 11th 2008, 08:40 AM

Originally Posted by Showcase_22

Okay then. I have absolutely no idea how to integrate moduli.

The last one I did was easy and it was just finding the area of two triangles and add them together. I don't think I can do the same for this.

I suppose what i'm asking is "how do you integrate moduli?"

**Showcase_22** · September 11th 2008, 09:48 AM

My big worry is that i'm not sure if my limits are right.

I'd be integrating a cos x along the way that will give me a sign x.

Sin 0=0 and sin pi=0 so it would give me an area of 0.

**Showcase_22** · September 11th 2008, 09:58 AM

Is this what I need to integrate:

?

I just noticed something else. In the other post you write "mod cos x < (or equal to) 0" (I haven't figured out how to quote properly). Isn't mod cos x always greater than 0? If any of the values were negative then they wouldn't be needed in the question anyway since it wants the area above the x axis.

?

**Laurent** · September 11th 2008, 11:57 AM

Originally Posted by Showcase_22

In the other post you write "mod cos x < (or equal to) 0"

This was of course a typo for: $\cos x\leq 0\Leftrightarrow |\cos x|=-\cos x$ .

To explicit Moo's hint a little, here is an example of computation of an integral with an absolute value: $\int_{-1}^2|x|dx=\int_{-1}^0 |x|dx +\int_0^2 |x|dx=\int_{-1}^0 (-x) dx+\int_0^2 x dx=\cdots$ . Try to adapt this to your situation.

**Showcase_22** · September 12th 2008, 01:17 AM

I decided to draw a graph of what I needed:

...and it suddenly all made sense! I was supposed to use the symmetry of the graph (integrate the function from 0 to pi/2) and then double it)!!

\m/ MAD SKILLZ! \m/

I got it right! thanks a bundle!

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