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So part a I've done, and it's $(3a, 8a^2)$ and $(-a,0)$

I've tried integrating it to find the area: $\int_{-a}^{3a} 3a^2+2ax-x^2 \ dx = [3a^2x+ax^2-\frac{1}{3}x^3]_{-a}^{3a}$

$(3a^2\cdot(3a)+a\cdot(3a)^2-\frac{1}{3}(3a)^3)-(3a^2\cdot(-a)+a\cdot(-a)^2-\frac{1}{3}(-a)^3)$

$=(9a^3+9a^3-9a)-(-3a^3+a^3+\frac{1}{3}a^3)$

$=9a^3-(\dfrac{-5}{3}a^3)=\frac{32}{3}$

The answer in the back is $\frac{64}{3}$ so I'm out by a factor of 2. It may be that the answer is wrong, and I can't see what I've done wrong.

Also - can I have a hint for part c?

Jim
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1 Answers1

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If you subtract the lower curve from the upper curve in order to find the area bounded by the curves, you get $$(5a^2 +4ax - x^2) - (x^2 - a^2) = 6a^2 +4ax - 2x^2 = 2(3a^2 + 2ax - x^2)$$

So your original integrand is off by a factor of $2$. Correct for that, and you'll have the desired area.

For part $(c)$, first try experimenting with a few simple values for $a$: say $a = -1, 1, 2$, to see that the portion of the area that is above the line $y = 0$ remains unchanged, regardless of the value of $a$.

To formalize this, find where the curves intersect the $x$-axis (when they intersect the line $y = 0$), and calculate the area (integrate) of the region that's above the line $y = 0$, using appropriate bounds for $x$. Divide this result by the correct area computed in part $(b)$.

amWhy
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