Area bounded by parabola and line.
$begingroup$
The problem is stated as "Find m such that the area of the region bounded by y = mx
and y = x^2 - 1 is equal to 36."
I tried solving it by systems of equations:
mx = x^2 - 1 (1)
and the second equation being the integral of [mx - (x^2 - 1)]dx = 36 which gave me:
(mx^2)/2 - (x^3)/3 + x = 36 (2)
then trying to simplify (1) and (2).
Is this the correct way of solving the problem or is there an easier way to solve it. Thanks.
definite-integrals area
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
$endgroup$
add a comment |
$begingroup$
The problem is stated as "Find m such that the area of the region bounded by y = mx
and y = x^2 - 1 is equal to 36."
I tried solving it by systems of equations:
mx = x^2 - 1 (1)
and the second equation being the integral of [mx - (x^2 - 1)]dx = 36 which gave me:
(mx^2)/2 - (x^3)/3 + x = 36 (2)
then trying to simplify (1) and (2).
Is this the correct way of solving the problem or is there an easier way to solve it. Thanks.
definite-integrals area
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
$endgroup$
add a comment |
$begingroup$
The problem is stated as "Find m such that the area of the region bounded by y = mx
and y = x^2 - 1 is equal to 36."
I tried solving it by systems of equations:
mx = x^2 - 1 (1)
and the second equation being the integral of [mx - (x^2 - 1)]dx = 36 which gave me:
(mx^2)/2 - (x^3)/3 + x = 36 (2)
then trying to simplify (1) and (2).
Is this the correct way of solving the problem or is there an easier way to solve it. Thanks.
definite-integrals area
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
$endgroup$
The problem is stated as "Find m such that the area of the region bounded by y = mx
and y = x^2 - 1 is equal to 36."
I tried solving it by systems of equations:
mx = x^2 - 1 (1)
and the second equation being the integral of [mx - (x^2 - 1)]dx = 36 which gave me:
(mx^2)/2 - (x^3)/3 + x = 36 (2)
then trying to simplify (1) and (2).
Is this the correct way of solving the problem or is there an easier way to solve it. Thanks.
definite-integrals area
definite-integrals area
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
asked Nov 26 '18 at 16:55
Dgtal12Dgtal12
61
61
add a comment |
add a comment |
1 Answer
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These are not two simultaneous equations. The two solutions to (1) (if they exist) will give you the bounds of the integral. Then you integrate with those bounds (which ought to remove $x$ completely), and set that equal to $36$.
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
$endgroup$
$begingroup$
Ok. I get that. But how will I get the bounds using an equation with two unknowns? That's the part I'm currently stuck on now.
$endgroup$
– Dgtal12
Nov 26 '18 at 17:08
$begingroup$
@Dgtal12 The bounds are values of $x$ that depend on the value of $m$. So when solving (1), $m$ is some (unknown) constant, and you solve for $x$.
$endgroup$
– Arthur
Nov 26 '18 at 17:23
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
These are not two simultaneous equations. The two solutions to (1) (if they exist) will give you the bounds of the integral. Then you integrate with those bounds (which ought to remove $x$ completely), and set that equal to $36$.
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
$endgroup$
$begingroup$
Ok. I get that. But how will I get the bounds using an equation with two unknowns? That's the part I'm currently stuck on now.
$endgroup$
– Dgtal12
Nov 26 '18 at 17:08
$begingroup$
@Dgtal12 The bounds are values of $x$ that depend on the value of $m$. So when solving (1), $m$ is some (unknown) constant, and you solve for $x$.
$endgroup$
– Arthur
Nov 26 '18 at 17:23
add a comment |
$begingroup$
These are not two simultaneous equations. The two solutions to (1) (if they exist) will give you the bounds of the integral. Then you integrate with those bounds (which ought to remove $x$ completely), and set that equal to $36$.
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
$endgroup$
$begingroup$
Ok. I get that. But how will I get the bounds using an equation with two unknowns? That's the part I'm currently stuck on now.
$endgroup$
– Dgtal12
Nov 26 '18 at 17:08
$begingroup$
@Dgtal12 The bounds are values of $x$ that depend on the value of $m$. So when solving (1), $m$ is some (unknown) constant, and you solve for $x$.
$endgroup$
– Arthur
Nov 26 '18 at 17:23
add a comment |
$begingroup$
These are not two simultaneous equations. The two solutions to (1) (if they exist) will give you the bounds of the integral. Then you integrate with those bounds (which ought to remove $x$ completely), and set that equal to $36$.
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
$endgroup$
These are not two simultaneous equations. The two solutions to (1) (if they exist) will give you the bounds of the integral. Then you integrate with those bounds (which ought to remove $x$ completely), and set that equal to $36$.
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
answered Nov 26 '18 at 16:59
ArthurArthur
113k7109193
113k7109193
$begingroup$
Ok. I get that. But how will I get the bounds using an equation with two unknowns? That's the part I'm currently stuck on now.
$endgroup$
– Dgtal12
Nov 26 '18 at 17:08
$begingroup$
@Dgtal12 The bounds are values of $x$ that depend on the value of $m$. So when solving (1), $m$ is some (unknown) constant, and you solve for $x$.
$endgroup$
– Arthur
Nov 26 '18 at 17:23
add a comment |
$begingroup$
Ok. I get that. But how will I get the bounds using an equation with two unknowns? That's the part I'm currently stuck on now.
$endgroup$
– Dgtal12
Nov 26 '18 at 17:08
$begingroup$
@Dgtal12 The bounds are values of $x$ that depend on the value of $m$. So when solving (1), $m$ is some (unknown) constant, and you solve for $x$.
$endgroup$
– Arthur
Nov 26 '18 at 17:23
$begingroup$
Ok. I get that. But how will I get the bounds using an equation with two unknowns? That's the part I'm currently stuck on now.
$endgroup$
– Dgtal12
Nov 26 '18 at 17:08
$begingroup$
Ok. I get that. But how will I get the bounds using an equation with two unknowns? That's the part I'm currently stuck on now.
$endgroup$
– Dgtal12
Nov 26 '18 at 17:08
$begingroup$
@Dgtal12 The bounds are values of $x$ that depend on the value of $m$. So when solving (1), $m$ is some (unknown) constant, and you solve for $x$.
$endgroup$
– Arthur
Nov 26 '18 at 17:23
$begingroup$
@Dgtal12 The bounds are values of $x$ that depend on the value of $m$. So when solving (1), $m$ is some (unknown) constant, and you solve for $x$.
$endgroup$
– Arthur
Nov 26 '18 at 17:23
add a comment |
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