Find the value of constant a, given area between parabola and x-axis.












1












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Given $fleft(xright)=-x^2+2x+a$ 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 $int _1^{sqrt{a+1}+1}left(-x^2+2x+aright):dx=8$ 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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  • 1




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
















1












$begingroup$


Given $fleft(xright)=-x^2+2x+a$ 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 $int _1^{sqrt{a+1}+1}left(-x^2+2x+aright):dx=8$ 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.










share|cite|improve this question











$endgroup$








  • 1




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














1












1








1





$begingroup$


Given $fleft(xright)=-x^2+2x+a$ 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 $int _1^{sqrt{a+1}+1}left(-x^2+2x+aright):dx=8$ 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.










share|cite|improve this question











$endgroup$




Given $fleft(xright)=-x^2+2x+a$ 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 $int _1^{sqrt{a+1}+1}left(-x^2+2x+aright):dx=8$ 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.







integration definite-integrals






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edited Dec 18 '18 at 20:39









David G. Stork

11k41432




11k41432










asked Dec 18 '18 at 20:37









Kb1998Kb1998

82




82








  • 1




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














  • 1




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








1




1




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




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










1 Answer
1






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0












$begingroup$

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



$$intlimits_{x = 1 - sqrt{a +1}}^{1 + sqrt{a+1}} -x^2 + 2 x + a dx = 16 $$



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



Plot






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$endgroup$













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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

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



$$intlimits_{x = 1 - sqrt{a +1}}^{1 + sqrt{a+1}} -x^2 + 2 x + a dx = 16 $$



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



Plot






share|cite|improve this answer









$endgroup$













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
















0












$begingroup$

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



$$intlimits_{x = 1 - sqrt{a +1}}^{1 + sqrt{a+1}} -x^2 + 2 x + a dx = 16 $$



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



Plot






share|cite|improve this answer









$endgroup$













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














0












0








0





$begingroup$

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



$$intlimits_{x = 1 - sqrt{a +1}}^{1 + sqrt{a+1}} -x^2 + 2 x + a dx = 16 $$



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



Plot






share|cite|improve this answer









$endgroup$



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



$$intlimits_{x = 1 - sqrt{a +1}}^{1 + sqrt{a+1}} -x^2 + 2 x + a dx = 16 $$



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



Plot







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 21:18









David G. StorkDavid G. Stork

11k41432




11k41432












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


















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
















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




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


















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