- #1

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In any case, why?

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- Thread starter kasse
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- #1

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In any case, why?

- #2

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So the total area = Int |f(x)-g(x)|dx

- #3

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"Ordinary" integration will do.

- #4

HallsofIvy

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Here, y= x is above y= x

[tex]\int_0^3 (x- (x^2- 2x))dx= \int_0^3 (3x- x^2)dx[/tex]

Of course, you

[tex]\int_{x=0}^3 \int_{y= x^2- 2x}^x dy dx[/tex]

but after the first integration, it reduces to the single integral form.

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