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2 curves in an integral 
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#1
Feb204, 09:12 PM

P: 244

When you are trying to find an integral and you are dealing with two curves, and the formula calls for [itex] f(x) [/itex], I know that you do [itex] f(x)  g(x) [/tex] and put in that value for the f(x) in the original formula.
When the formula calls for [itex] f^2 (x) [/itex], do you do [itex] (f^2 (x)  g^2 (x)) [/itex] or [itex] (f(x)g(x))^2 [/itex] ? 


#2
Feb304, 07:18 AM

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P: 39,568

This is what comes of memorizing formulas without understanding.
"When you are trying to find an integral and you are dealing with two curves". You understand, don't you, that an integral does not necessarily have anything to do with "two curves" I think you are talking about finding the area between two curves. In that case you integrate f(x) g(x) because you are using that distance as the height of the "rectangles" in the Riemann sum. "When the formula calls for [itex] f^2 (x) [/itex] , do you do [itex] (f^2 (x)  g^2 (x)) [/itex] or [itex] (f(x)g(x))^2 [/itex]?" Well, I don't know because I have no idea what formula you are talking about or what kind of problem you are doing. I do recall that one method of finding a "volume of revolution" involves integrating πf(x)^{2}dx because you are thinking of f(x) as the radius of a circle so that πf(x)^{2} is the area of the circle and πf(x)^{2}dx is the volume of the flat disk. If the axis of rotation is outside the figure, then you can think of it as one circle inside another (a "washer"). You could find the area of the washer by calculating the area of the outer circle and then subtracting the area of the inner circle: πf(x)^{2} πg(x)^{2}. Then the volume is the integral of π(f(x)^{2} g(x)^{2})dx. But you never just "do" something without understanding why you do it. 


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