Mixing
How do I find the area bounded by a $y=frac{4}{3}x^2+frac{12}{3}x-3$ and $y=sqrt{x}$
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I am having difficulties with this problem:$y=frac{4}{3}x^2+frac{12}{3}x-3$ and $y=sqrt{x}$.
Graphng two of the functions I get the following:
Graphing both functions shows that they intersect at $x=0.771$.
Now, the thing thats bugging me about this problem is that both graph's don't seem to have a bounded area. For this particular situation, $y=sqrt{x}$ is only valid for values that are positive only.
I took a rough guess and did the integral like this:
$$int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)$$
$$int_{0}^{0.771}(sqrt{x})-frac{4}{3}x^2-frac{12}{3}x+3)$$
$$intsqrt{x}-frac{4}{3}int(x^2)-frac{12}{3}int(x)+int(3)$$
$$frac{x^{frac{3}{2}}}{3/2}-frac{4x^3}{9}-frac{12x^2}{6}+3x$$
Evaluating the integral:
$$[frac{(0.771)^{frac{3}{2}}}{3/2})-frac{4(0.771)^3}{9}-frac{12(0.771)^2}{6}+3(0.771)]^{0.771}-[0]^{0}$$
$$therefore int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)=0.9204 text{square units}$$
I don't know if my approach is correct. I need help.
calculus integration area curves
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
$endgroup$
add a comment |
$begingroup$
I am having difficulties with this problem:$y=frac{4}{3}x^2+frac{12}{3}x-3$ and $y=sqrt{x}$.
Graphng two of the functions I get the following:
Graphing both functions shows that they intersect at $x=0.771$.
Now, the thing thats bugging me about this problem is that both graph's don't seem to have a bounded area. For this particular situation, $y=sqrt{x}$ is only valid for values that are positive only.
I took a rough guess and did the integral like this:
$$int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)$$
$$int_{0}^{0.771}(sqrt{x})-frac{4}{3}x^2-frac{12}{3}x+3)$$
$$intsqrt{x}-frac{4}{3}int(x^2)-frac{12}{3}int(x)+int(3)$$
$$frac{x^{frac{3}{2}}}{3/2}-frac{4x^3}{9}-frac{12x^2}{6}+3x$$
Evaluating the integral:
$$[frac{(0.771)^{frac{3}{2}}}{3/2})-frac{4(0.771)^3}{9}-frac{12(0.771)^2}{6}+3(0.771)]^{0.771}-[0]^{0}$$
$$therefore int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)=0.9204 text{square units}$$
I don't know if my approach is correct. I need help.
calculus integration area curves
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
$endgroup$
$begingroup$
@joseabp91 Oh righto. So the bounded area is that tiny little portion of the graph from x=0 to x=0.771 right?
$endgroup$
– AugieJavax98
Jan 17 at 15:11
$begingroup$
Does the problem say $y = sqrt{x}$ or $x = y^2$, because it makes a difference
$endgroup$
– caverac
Jan 17 at 15:18
$begingroup$
@caverac problem says $y=sqrt{x}$. Am I allowed to rearrange it to that?
$endgroup$
– AugieJavax98
Jan 17 at 15:20
$begingroup$
The thing here is that $y = color{red}{+}sqrt{x}$ is only one of the solutions of $y^2 = x$ (your blue line above), but there's another one, namely $y = color{red}{-}sqrt{x}$ which will completely close the region. I would have interpreted the problem like that, because as it is right now it is a bit ambiguous (as you rightly noticed). But that's just my opinion
$endgroup$
– caverac
Jan 17 at 15:24
$begingroup$
Yes because the function which you want to integrate is just positive between $0$ and $0.771$. The final solution must be $1.37$.
$endgroup$
– joseabp91
Jan 17 at 15:32
add a comment |
$begingroup$
I am having difficulties with this problem:$y=frac{4}{3}x^2+frac{12}{3}x-3$ and $y=sqrt{x}$.
Graphng two of the functions I get the following:
Graphing both functions shows that they intersect at $x=0.771$.
Now, the thing thats bugging me about this problem is that both graph's don't seem to have a bounded area. For this particular situation, $y=sqrt{x}$ is only valid for values that are positive only.
I took a rough guess and did the integral like this:
$$int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)$$
$$int_{0}^{0.771}(sqrt{x})-frac{4}{3}x^2-frac{12}{3}x+3)$$
$$intsqrt{x}-frac{4}{3}int(x^2)-frac{12}{3}int(x)+int(3)$$
$$frac{x^{frac{3}{2}}}{3/2}-frac{4x^3}{9}-frac{12x^2}{6}+3x$$
Evaluating the integral:
$$[frac{(0.771)^{frac{3}{2}}}{3/2})-frac{4(0.771)^3}{9}-frac{12(0.771)^2}{6}+3(0.771)]^{0.771}-[0]^{0}$$
$$therefore int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)=0.9204 text{square units}$$
I don't know if my approach is correct. I need help.
calculus integration area curves
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
$endgroup$
I am having difficulties with this problem:$y=frac{4}{3}x^2+frac{12}{3}x-3$ and $y=sqrt{x}$.
Graphng two of the functions I get the following:
Graphing both functions shows that they intersect at $x=0.771$.
Now, the thing thats bugging me about this problem is that both graph's don't seem to have a bounded area. For this particular situation, $y=sqrt{x}$ is only valid for values that are positive only.
I took a rough guess and did the integral like this:
$$int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)$$
$$int_{0}^{0.771}(sqrt{x})-frac{4}{3}x^2-frac{12}{3}x+3)$$
$$intsqrt{x}-frac{4}{3}int(x^2)-frac{12}{3}int(x)+int(3)$$
$$frac{x^{frac{3}{2}}}{3/2}-frac{4x^3}{9}-frac{12x^2}{6}+3x$$
Evaluating the integral:
$$[frac{(0.771)^{frac{3}{2}}}{3/2})-frac{4(0.771)^3}{9}-frac{12(0.771)^2}{6}+3(0.771)]^{0.771}-[0]^{0}$$
$$therefore int_{0}^{0.771}(sqrt{x})-(frac{4}{3}x^2+frac{12}{3}x-3)=0.9204 text{square units}$$
I don't know if my approach is correct. I need help.
calculus integration area curves
calculus integration area curves
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
edited Jan 18 at 13:19
N. F. Taussig
44.4k93357
44.4k93357
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
asked Jan 17 at 14:47
AugieJavax98AugieJavax98
315210
315210
$begingroup$
@joseabp91 Oh righto. So the bounded area is that tiny little portion of the graph from x=0 to x=0.771 right?
$endgroup$
– AugieJavax98
Jan 17 at 15:11
$begingroup$
Does the problem say $y = sqrt{x}$ or $x = y^2$, because it makes a difference
$endgroup$
– caverac
Jan 17 at 15:18
$begingroup$
@caverac problem says $y=sqrt{x}$. Am I allowed to rearrange it to that?
$endgroup$
– AugieJavax98
Jan 17 at 15:20
$begingroup$
The thing here is that $y = color{red}{+}sqrt{x}$ is only one of the solutions of $y^2 = x$ (your blue line above), but there's another one, namely $y = color{red}{-}sqrt{x}$ which will completely close the region. I would have interpreted the problem like that, because as it is right now it is a bit ambiguous (as you rightly noticed). But that's just my opinion
$endgroup$
– caverac
Jan 17 at 15:24
$begingroup$
Yes because the function which you want to integrate is just positive between $0$ and $0.771$. The final solution must be $1.37$.
$endgroup$
– joseabp91
Jan 17 at 15:32
add a comment |
$begingroup$
@joseabp91 Oh righto. So the bounded area is that tiny little portion of the graph from x=0 to x=0.771 right?
$endgroup$
– AugieJavax98
Jan 17 at 15:11
$begingroup$
Does the problem say $y = sqrt{x}$ or $x = y^2$, because it makes a difference
$endgroup$
– caverac
Jan 17 at 15:18
$begingroup$
@caverac problem says $y=sqrt{x}$. Am I allowed to rearrange it to that?
$endgroup$
– AugieJavax98
Jan 17 at 15:20
$begingroup$
The thing here is that $y = color{red}{+}sqrt{x}$ is only one of the solutions of $y^2 = x$ (your blue line above), but there's another one, namely $y = color{red}{-}sqrt{x}$ which will completely close the region. I would have interpreted the problem like that, because as it is right now it is a bit ambiguous (as you rightly noticed). But that's just my opinion
$endgroup$
– caverac
Jan 17 at 15:24
$begingroup$
Yes because the function which you want to integrate is just positive between $0$ and $0.771$. The final solution must be $1.37$.
$endgroup$
– joseabp91
Jan 17 at 15:32
$begingroup$
@joseabp91 Oh righto. So the bounded area is that tiny little portion of the graph from x=0 to x=0.771 right?
$endgroup$
– AugieJavax98
Jan 17 at 15:11
$begingroup$
@joseabp91 Oh righto. So the bounded area is that tiny little portion of the graph from x=0 to x=0.771 right?
$endgroup$
– AugieJavax98
Jan 17 at 15:11
$begingroup$
Does the problem say $y = sqrt{x}$ or $x = y^2$, because it makes a difference
$endgroup$
– caverac
Jan 17 at 15:18
$begingroup$
Does the problem say $y = sqrt{x}$ or $x = y^2$, because it makes a difference
$endgroup$
– caverac
Jan 17 at 15:18
$begingroup$
@caverac problem says $y=sqrt{x}$. Am I allowed to rearrange it to that?
$endgroup$
– AugieJavax98
Jan 17 at 15:20
$begingroup$
@caverac problem says $y=sqrt{x}$. Am I allowed to rearrange it to that?
$endgroup$
– AugieJavax98
Jan 17 at 15:20
$begingroup$
The thing here is that $y = color{red}{+}sqrt{x}$ is only one of the solutions of $y^2 = x$ (your blue line above), but there's another one, namely $y = color{red}{-}sqrt{x}$ which will completely close the region. I would have interpreted the problem like that, because as it is right now it is a bit ambiguous (as you rightly noticed). But that's just my opinion
$endgroup$
– caverac
Jan 17 at 15:24
$begingroup$
The thing here is that $y = color{red}{+}sqrt{x}$ is only one of the solutions of $y^2 = x$ (your blue line above), but there's another one, namely $y = color{red}{-}sqrt{x}$ which will completely close the region. I would have interpreted the problem like that, because as it is right now it is a bit ambiguous (as you rightly noticed). But that's just my opinion
$endgroup$
– caverac
Jan 17 at 15:24
$begingroup$
Yes because the function which you want to integrate is just positive between $0$ and $0.771$. The final solution must be $1.37$.
$endgroup$
– joseabp91
Jan 17 at 15:32
$begingroup$
Yes because the function which you want to integrate is just positive between $0$ and $0.771$. The final solution must be $1.37$.
$endgroup$
– joseabp91
Jan 17 at 15:32
add a comment |
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$begingroup$
@joseabp91 Oh righto. So the bounded area is that tiny little portion of the graph from x=0 to x=0.771 right?
$endgroup$
– AugieJavax98
Jan 17 at 15:11
$begingroup$
Does the problem say $y = sqrt{x}$ or $x = y^2$, because it makes a difference
$endgroup$
– caverac
Jan 17 at 15:18
$begingroup$
@caverac problem says $y=sqrt{x}$. Am I allowed to rearrange it to that?
$endgroup$
– AugieJavax98
Jan 17 at 15:20
$begingroup$
The thing here is that $y = color{red}{+}sqrt{x}$ is only one of the solutions of $y^2 = x$ (your blue line above), but there's another one, namely $y = color{red}{-}sqrt{x}$ which will completely close the region. I would have interpreted the problem like that, because as it is right now it is a bit ambiguous (as you rightly noticed). But that's just my opinion
$endgroup$
– caverac
Jan 17 at 15:24
$begingroup$
Yes because the function which you want to integrate is just positive between $0$ and $0.771$. The final solution must be $1.37$.
$endgroup$
– joseabp91
Jan 17 at 15:32