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Given $f\left(x\right)=-x^2+2x+a$ find the exact value of constant $a$ when the area between the curve and x-axis is 16.

I've tried several methods and all of them seem to be too complicated (at least for me) so possibly there is something that I'm missing. What I have tried is finding the x-interception points, which turned out to be $\sqrt{a+1}+1$ and $-\sqrt{a+1}+1$.

Given that $x=1$ divides the parabola into two equal sized areas we would get $\int _1^{\sqrt{a+1}+1}\left(-x^2+2x+a\right)\:dx=8$ which would be too hard for me to solve without straight up using a calculator. I also tried forming a system of linear equations, but it also turned out to be complicated as well.

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    $\begingroup$ Try the substitution $u = x -1$ $\endgroup$
    – Doug M
    Commented Dec 18, 2018 at 20:47

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The limits are indeed at $x = 1 \pm \sqrt{a+1}$. So solve:

$$\int\limits_{x = 1 - \sqrt{a +1}}^{1 + \sqrt{a+1}} -x^2 + 2 x + a\ dx = 16 $$

and find $a = 2 \sqrt[3]{2} \cdot 3^{2/3}-1$.

Plot

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  • $\begingroup$ Thanks, guys. I think I figured out how I can make it a little bit simplier. $\endgroup$
    – Kb1998
    Commented Dec 18, 2018 at 21:21

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