Question 224067
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I'm not sure if this is what your instructor had in mind when s/he assigned this problem, but it is the quickest and easiest solution that comes to my mind.  If there are any other more direct solutions out there, I would welcome one of the other tutors weighing in.


{{{drawing(
500, 500, -5, 5, -1, 9,
grid(1),
graph(
500, 500, -5, 5, -1, 9,
2x^2,
3x^2+5,
8)
)}}}


{{{drawing(
500, 500, -5, 5, -1, 9,
grid(1),
graph(
500, 500, -5, 5, -1, 9,
-2x^2+8,
-3x^2+3,
8)
)}}}


We are concerned with the area inside of the *[tex \Large 2x^2] parabola, outside the *[tex \Large 3x^2 + 5] parabola, and below <i>y</i> = 8.  A tricky integration in my mind.  But if you turn things upside down it becomes much more straightforward.  The area bounded by *[tex \Large -2x^2 + 8], *[tex \Large -3x^2+ 3], and the <i>x</i>-axis is the same area.  And the area calculation would simply be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \int_{-2}^2\,-2x^2+8\,dx\ -\ \int_{-1}^{1}\,-3x^2+3\,dx]


I presume you can handle these simple integrals.  If you are having difficulty with them, write back and I'll help you.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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