Question 1145735
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We are only interested in the domain interval between the two points of intersection of the two curves. See Figure.


 *[illustration BetweenTwoQuadraticsCropped.jpg]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 3x\ +\ 2\ =\ -x^2\ +\ 3x\ +\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 6x\ =\ 0]


So the two curves intersect at *[tex \LARGE (0,\,2)] and *[tex \LARGE (3,\,2)]


The area under *[tex \LARGE y\ =\ -x^2\ +\ 3x\ +\ 2] is given by


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int_0^3\,-x^2\ +\ 3x\ +\ 2\,dx]


from this area we need to subtract the area under *[tex \LARGE y\ =\ x^2\ -\ 3x\ +\ 2] or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int_0^3\,x^2\ -\ 3x\ +\ 2\,dx]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int_0^3\,-x^2\ +\ 3x\ +\ 2\,dx\ -\ \int_0^3\,x^2\ -\ 3x\ +\ 2\,dx]


But applying linearity gives us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int_0^3\,-2x^2\ +\ 6x\ \,dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \[-\frac{2x^3}{3}\ +\ 3x^2\]_0^3]


You can do the arithmetic.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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