Find the value of constant a, given area between parabola and x-axis.
$begingroup$
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.
integration definite-integrals
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
asked Dec 18 '18 at 20:37
Kb1998Kb1998
82
$endgroup$
add a comment |
$begingroup$
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.
integration definite-integrals
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
asked Dec 18 '18 at 20:37
Kb1998Kb1998
82
$endgroup$
1
$begingroup$
Try the substitution $u = x -1$
$endgroup$
– Doug M
Dec 18 '18 at 20:47
add a comment |
$begingroup$
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.
integration definite-integrals
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
asked Dec 18 '18 at 20:37
Kb1998Kb1998
82
$endgroup$
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
integration definite-integrals
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
asked Dec 18 '18 at 20:37
Kb1998Kb1998
82
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
asked Dec 18 '18 at 20:37
Kb1998Kb1998
82
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
edited Dec 18 '18 at 20:39
David G. Stork
11k41432
11k41432
asked Dec 18 '18 at 20:37
Kb1998Kb1998
82
asked Dec 18 '18 at 20:37
Kb1998Kb1998
82
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $a = 2 sqrt[3]{2} cdot 3^{2/3}-1$.
answered Dec 18 '18 at 21:18
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Thanks, guys. I think I figured out how I can make it a little bit simplier.
$endgroup$
– Kb1998
Dec 18 '18 at 21:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $a = 2 sqrt[3]{2} cdot 3^{2/3}-1$.
answered Dec 18 '18 at 21:18
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Thanks, guys. I think I figured out how I can make it a little bit simplier.
$endgroup$
– Kb1998
Dec 18 '18 at 21:21
add a comment |
$begingroup$
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 $a = 2 sqrt[3]{2} cdot 3^{2/3}-1$.
answered Dec 18 '18 at 21:18
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Thanks, guys. I think I figured out how I can make it a little bit simplier.
$endgroup$
– Kb1998
Dec 18 '18 at 21:21
add a comment |
$begingroup$
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 $a = 2 sqrt[3]{2} cdot 3^{2/3}-1$.
answered Dec 18 '18 at 21:18
David G. StorkDavid G. Stork
11k41432
$endgroup$
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 $a = 2 sqrt[3]{2} cdot 3^{2/3}-1$.
answered Dec 18 '18 at 21:18
David G. StorkDavid G. Stork
11k41432
answered Dec 18 '18 at 21:18
David G. StorkDavid G. Stork
11k41432
answered Dec 18 '18 at 21:18
David G. StorkDavid G. Stork
11k41432
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
add a comment |
$begingroup$
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
add a comment |
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1
$begingroup$
Try the substitution $u = x -1$
$endgroup$
– Doug M
Dec 18 '18 at 20:47