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Exam 2 Question

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Collin Li:
Sorry, wasn't thinking properly. I edited my post to provide the correct answer.

reg:
VCAA is right.

First, we integrate cos(x) across the interval, and we get 1 as a result.
We then divide by the interval,

1 / (pi / 2) =
2 / pi

cara.mel:
You should be able to do translations by hand. It's a similar thing for all graphs.

Q2 I don't know how to explain it but I can do it. It's the integral of cos x (total area) divided by how wide it is (so then it would be 'how high does it get if you spread the area out evenly if that makes any sense). Like I said I don't know how to explain it, I just know it works

so integral cos(x) => sin(pi/2)-sin(0) = 1
Then divide this by (pi/2 - 0) = 2/pi.

Someone explain that better ty :K

costargh:
Thanks

Collin Li:
Hmm... that's weird.

I would have done that question last year and got it right (80/80 on exam 2). I was thinking of the average slope.

Yes, the integration method is the correct way. The sum of all possible values from 0, pi/2 is represented by the integral from 0 to pi/2, and then divided by the width: (pi/2 - 0) will give you the average value.

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