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Method for double integrals?
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How do I go about solving double integrals of this type :
$$I = int_0^À dx int_y^À x^4 sin(yx^2)~ mathrm dy$$
I just need some direction or a full answer I can follow?
calculus integration
asked Jul 24 at 15:00
K.M.
17912
add a comment |Â
up vote
-2
down vote
favorite
How do I go about solving double integrals of this type :
$$I = int_0^À dx int_y^À x^4 sin(yx^2)~ mathrm dy$$
I just need some direction or a full answer I can follow?
calculus integration
asked Jul 24 at 15:00
K.M.
17912
2
Is your real question how to integrate things of the form $x^4sin (ax^2)$ with respect to $x$? Because that's the hard part.
– The Count
Jul 24 at 15:03
The limits of integration can't involve the differential. $x$ is a limit of your $dx$ integral.
– B. Goddard
Jul 24 at 17:18
@TheCount yes but instead of a constant $a$ I have a variable $y$
– K.M.
Jul 24 at 17:52
From the point of view of the first integral as you had originally written it, $y$ is a constant. That's a good conceptual point to grasp. It is now a totally different question....
– The Count
Jul 25 at 1:35
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
How do I go about solving double integrals of this type :
$$I = int_0^À dx int_y^À x^4 sin(yx^2)~ mathrm dy$$
I just need some direction or a full answer I can follow?
calculus integration
asked Jul 24 at 15:00
K.M.
17912
How do I go about solving double integrals of this type :
$$I = int_0^À dx int_y^À x^4 sin(yx^2)~ mathrm dy$$
I just need some direction or a full answer I can follow?
calculus integration
asked Jul 24 at 15:00
K.M.
17912
edited Jul 25 at 14:23
asked Jul 24 at 15:00
K.M.
17912
asked Jul 24 at 15:00
K.M.
17912
asked Jul 24 at 15:00
K.M.
17912
17912
2
Is your real question how to integrate things of the form $x^4sin (ax^2)$ with respect to $x$? Because that's the hard part.
– The Count
Jul 24 at 15:03
The limits of integration can't involve the differential. $x$ is a limit of your $dx$ integral.
– B. Goddard
Jul 24 at 17:18
@TheCount yes but instead of a constant $a$ I have a variable $y$
– K.M.
Jul 24 at 17:52
From the point of view of the first integral as you had originally written it, $y$ is a constant. That's a good conceptual point to grasp. It is now a totally different question....
– The Count
Jul 25 at 1:35
add a comment |Â
2
Is your real question how to integrate things of the form $x^4sin (ax^2)$ with respect to $x$? Because that's the hard part.
– The Count
Jul 24 at 15:03
The limits of integration can't involve the differential. $x$ is a limit of your $dx$ integral.
– B. Goddard
Jul 24 at 17:18
@TheCount yes but instead of a constant $a$ I have a variable $y$
– K.M.
Jul 24 at 17:52
From the point of view of the first integral as you had originally written it, $y$ is a constant. That's a good conceptual point to grasp. It is now a totally different question....
– The Count
Jul 25 at 1:35
2
2
Is your real question how to integrate things of the form $x^4sin (ax^2)$ with respect to $x$? Because that's the hard part.
– The Count
Jul 24 at 15:03
Is your real question how to integrate things of the form $x^4sin (ax^2)$ with respect to $x$? Because that's the hard part.
– The Count
Jul 24 at 15:03
The limits of integration can't involve the differential. $x$ is a limit of your $dx$ integral.
– B. Goddard
Jul 24 at 17:18
The limits of integration can't involve the differential. $x$ is a limit of your $dx$ integral.
– B. Goddard
Jul 24 at 17:18
@TheCount yes but instead of a constant $a$ I have a variable $y$
– K.M.
Jul 24 at 17:52
@TheCount yes but instead of a constant $a$ I have a variable $y$
– K.M.
Jul 24 at 17:52
From the point of view of the first integral as you had originally written it, $y$ is a constant. That's a good conceptual point to grasp. It is now a totally different question....
– The Count
Jul 25 at 1:35
From the point of view of the first integral as you had originally written it, $y$ is a constant. That's a good conceptual point to grasp. It is now a totally different question....
– The Count
Jul 25 at 1:35
add a comment |Â
1 Answer
1
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0
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Change order of integration. $I=int_0^x x^4dxint_0^xsin(yx^2)dy=int_0^x x^4frac1-cos(x^3)x^2dx$. Let $u=x^3$. Then $I=frac13int_0^x(1-cos(u))du=frac13(x^3-sin(x^3))$.
For clarity you should use different symbols for variables of integration and integration limits, when they are not the same.
answered Jul 24 at 16:53
herb steinberg
93529
I changed the limits to À and $y$
– K.M.
Jul 25 at 14:26
Changing the limit to $pi$ looks strange. The final answer would then be $frac13(pi^3-sin(pi^3))$.
– herb steinberg
Jul 25 at 15:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Change order of integration. $I=int_0^x x^4dxint_0^xsin(yx^2)dy=int_0^x x^4frac1-cos(x^3)x^2dx$. Let $u=x^3$. Then $I=frac13int_0^x(1-cos(u))du=frac13(x^3-sin(x^3))$.
For clarity you should use different symbols for variables of integration and integration limits, when they are not the same.
answered Jul 24 at 16:53
herb steinberg
93529
I changed the limits to À and $y$
– K.M.
Jul 25 at 14:26
Changing the limit to $pi$ looks strange. The final answer would then be $frac13(pi^3-sin(pi^3))$.
– herb steinberg
Jul 25 at 15:46
add a comment |Â
up vote
0
down vote
Change order of integration. $I=int_0^x x^4dxint_0^xsin(yx^2)dy=int_0^x x^4frac1-cos(x^3)x^2dx$. Let $u=x^3$. Then $I=frac13int_0^x(1-cos(u))du=frac13(x^3-sin(x^3))$.
For clarity you should use different symbols for variables of integration and integration limits, when they are not the same.
answered Jul 24 at 16:53
herb steinberg
93529
I changed the limits to À and $y$
– K.M.
Jul 25 at 14:26
Changing the limit to $pi$ looks strange. The final answer would then be $frac13(pi^3-sin(pi^3))$.
– herb steinberg
Jul 25 at 15:46
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Change order of integration. $I=int_0^x x^4dxint_0^xsin(yx^2)dy=int_0^x x^4frac1-cos(x^3)x^2dx$. Let $u=x^3$. Then $I=frac13int_0^x(1-cos(u))du=frac13(x^3-sin(x^3))$.
For clarity you should use different symbols for variables of integration and integration limits, when they are not the same.
answered Jul 24 at 16:53
herb steinberg
93529
Change order of integration. $I=int_0^x x^4dxint_0^xsin(yx^2)dy=int_0^x x^4frac1-cos(x^3)x^2dx$. Let $u=x^3$. Then $I=frac13int_0^x(1-cos(u))du=frac13(x^3-sin(x^3))$.
For clarity you should use different symbols for variables of integration and integration limits, when they are not the same.
answered Jul 24 at 16:53
herb steinberg
93529
edited Jul 25 at 15:47
answered Jul 24 at 16:53
herb steinberg
93529
answered Jul 24 at 16:53
herb steinberg
93529
answered Jul 24 at 16:53
herb steinberg
93529
93529
I changed the limits to À and $y$
– K.M.
Jul 25 at 14:26
Changing the limit to $pi$ looks strange. The final answer would then be $frac13(pi^3-sin(pi^3))$.
– herb steinberg
Jul 25 at 15:46
add a comment |Â
I changed the limits to À and $y$
– K.M.
Jul 25 at 14:26
Changing the limit to $pi$ looks strange. The final answer would then be $frac13(pi^3-sin(pi^3))$.
– herb steinberg
Jul 25 at 15:46
I changed the limits to À and $y$
– K.M.
Jul 25 at 14:26
I changed the limits to À and $y$
– K.M.
Jul 25 at 14:26
Changing the limit to $pi$ looks strange. The final answer would then be $frac13(pi^3-sin(pi^3))$.
– herb steinberg
Jul 25 at 15:46
Changing the limit to $pi$ looks strange. The final answer would then be $frac13(pi^3-sin(pi^3))$.
– herb steinberg
Jul 25 at 15:46
add a comment |Â
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2
Is your real question how to integrate things of the form $x^4sin (ax^2)$ with respect to $x$? Because that's the hard part.
– The Count
Jul 24 at 15:03
The limits of integration can't involve the differential. $x$ is a limit of your $dx$ integral.
– B. Goddard
Jul 24 at 17:18
@TheCount yes but instead of a constant $a$ I have a variable $y$
– K.M.
Jul 24 at 17:52
From the point of view of the first integral as you had originally written it, $y$ is a constant. That's a good conceptual point to grasp. It is now a totally different question....
– The Count
Jul 25 at 1:35