- #1

CellCoree

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does that mean the integral would like this: [tex] \int 4cos(x) - 3sin(x) [/tex] (a=0, b = 0.9*pi) is that correct?

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- Thread starter CellCoree
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- #1

CellCoree

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does that mean the integral would like this: [tex] \int 4cos(x) - 3sin(x) [/tex] (a=0, b = 0.9*pi) is that correct?

- #2

Galileo

Science Advisor

Homework Helper

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Since area is always positive you should split up your integral, and find at which point they intersect.

A picture may help.

- #3

CellCoree

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Galileo said:

Since area is always positive you should split up your integral, and find at which point they intersect.

A picture may help.

ah, 0.9pi is equal to 162 degrees right? i graphed it using my calculator. don't know why, but i thought of 0.9pi as 2.872433.

so would my equation look like this...

[tex] \int 4cos(x) -3sin(x) (a=0,b=.295\pi) [/tex] + [tex] \int 3sin(x) - 4cos(x) (a=.295\pi,b=9\pi)[/tex] is that correct?

- #4

Galileo

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[tex]\int_0^{0.295\pi}(4\cos x-3\sin x)dx+\int_{0.295\pi}^{0.9\pi}(3\sin x-4\cos x)dx[/tex]

- #5

CellCoree

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- 0

lol yeah, i didnt know how to insert a and b into the integral sign. thanks for the help

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