Mixing
graphical meaning of integration by parts of this function
$begingroup$
I have seen this and this. After looking at the first link I was pretty satisfied with the answer. Now I understand why
$int vdu=uv-int udv$
works graphically.
I have even worked out $int (sin^{-1}x)dx$ using integration by parts and understood it graphically.
But I am not able to visualize this $int (xsin x)dx$ in the graphical manner as I'm not able to find an inverse which is important for the graphical method to work. However, I could solve it mathematically.
Now my questions are
i) Is the graphical intuition only limited to functions which can be integrated this way?
$int f^{-1}(y)dy=xy-int f(x)dx$
ii)Please help me with a graphical intuition behind solving functions like $int (xsin x)dx$ using integration by parts (if any). I will try to do the remaining math myself.
Any suggestion is of great help to me. Thank you.
calculus integration definite-integrals indefinite-integrals
asked Jan 27 at 16:06
user8718165user8718165
146
$endgroup$
|
show 1 more comment
$begingroup$
I have seen this and this. After looking at the first link I was pretty satisfied with the answer. Now I understand why
$int vdu=uv-int udv$
works graphically.
I have even worked out $int (sin^{-1}x)dx$ using integration by parts and understood it graphically.
But I am not able to visualize this $int (xsin x)dx$ in the graphical manner as I'm not able to find an inverse which is important for the graphical method to work. However, I could solve it mathematically.
Now my questions are
i) Is the graphical intuition only limited to functions which can be integrated this way?
$int f^{-1}(y)dy=xy-int f(x)dx$
ii)Please help me with a graphical intuition behind solving functions like $int (xsin x)dx$ using integration by parts (if any). I will try to do the remaining math myself.
Any suggestion is of great help to me. Thank you.
calculus integration definite-integrals indefinite-integrals
asked Jan 27 at 16:06
user8718165user8718165
146
$endgroup$
$begingroup$
I know this does not answer your questions but is related.
$endgroup$
– Exp ikx
Jan 27 at 16:14
$begingroup$
can you please help me
$endgroup$
– user8718165
Jan 27 at 16:16
$begingroup$
This is not a direct answer to the question, but it could be (I hope) a nevertheless helpful comment. When we try to see things graphically, the aim is (usually) to give us some kind of insight into the solution. I am unaware of any such insightful graphical interpretation of IBP. The link you gave was kinda neat, but I confess I see no obvious practical application for it. Instead, I always view integration by parts as an algebraic exercise. How can we separate an expression so that if we integrate one part and differentiate the other, we obtain something we can integrate nicely?
$endgroup$
– Ben W
Jan 27 at 16:18
$begingroup$
I have tried to solve it algebraically and it worked. Thanks
$endgroup$
– user8718165
Jan 27 at 16:20
$begingroup$
...In your case, we have $xsin(x);dx$. There are only three ways to separate it: $u=x$, $dv=sin(x);dx$; or: $u=sin(x)$, $dv=x;dx$; or: $u=xsin(x)$, $dv=dx$. One of these choices will work, the others will not.
$endgroup$
– Ben W
Jan 27 at 16:20
|
show 1 more comment
$begingroup$
I have seen this and this. After looking at the first link I was pretty satisfied with the answer. Now I understand why
$int vdu=uv-int udv$
works graphically.
I have even worked out $int (sin^{-1}x)dx$ using integration by parts and understood it graphically.
But I am not able to visualize this $int (xsin x)dx$ in the graphical manner as I'm not able to find an inverse which is important for the graphical method to work. However, I could solve it mathematically.
Now my questions are
i) Is the graphical intuition only limited to functions which can be integrated this way?
$int f^{-1}(y)dy=xy-int f(x)dx$
ii)Please help me with a graphical intuition behind solving functions like $int (xsin x)dx$ using integration by parts (if any). I will try to do the remaining math myself.
Any suggestion is of great help to me. Thank you.
calculus integration definite-integrals indefinite-integrals
asked Jan 27 at 16:06
user8718165user8718165
146
$endgroup$
I have seen this and this. After looking at the first link I was pretty satisfied with the answer. Now I understand why
$int vdu=uv-int udv$
works graphically.
I have even worked out $int (sin^{-1}x)dx$ using integration by parts and understood it graphically.
But I am not able to visualize this $int (xsin x)dx$ in the graphical manner as I'm not able to find an inverse which is important for the graphical method to work. However, I could solve it mathematically.
Now my questions are
i) Is the graphical intuition only limited to functions which can be integrated this way?
$int f^{-1}(y)dy=xy-int f(x)dx$
ii)Please help me with a graphical intuition behind solving functions like $int (xsin x)dx$ using integration by parts (if any). I will try to do the remaining math myself.
Any suggestion is of great help to me. Thank you.
calculus integration definite-integrals indefinite-integrals
calculus integration definite-integrals indefinite-integrals
asked Jan 27 at 16:06
user8718165user8718165
146
asked Jan 27 at 16:06
user8718165user8718165
146
edited Jan 27 at 16:56
user8718165
asked Jan 27 at 16:06
user8718165user8718165
146
asked Jan 27 at 16:06
user8718165user8718165
146
asked Jan 27 at 16:06
user8718165user8718165
146
146
$begingroup$
I know this does not answer your questions but is related.
$endgroup$
– Exp ikx
Jan 27 at 16:14
$begingroup$
can you please help me
$endgroup$
– user8718165
Jan 27 at 16:16
$begingroup$
This is not a direct answer to the question, but it could be (I hope) a nevertheless helpful comment. When we try to see things graphically, the aim is (usually) to give us some kind of insight into the solution. I am unaware of any such insightful graphical interpretation of IBP. The link you gave was kinda neat, but I confess I see no obvious practical application for it. Instead, I always view integration by parts as an algebraic exercise. How can we separate an expression so that if we integrate one part and differentiate the other, we obtain something we can integrate nicely?
$endgroup$
– Ben W
Jan 27 at 16:18
$begingroup$
I have tried to solve it algebraically and it worked. Thanks
$endgroup$
– user8718165
Jan 27 at 16:20
$begingroup$
...In your case, we have $xsin(x);dx$. There are only three ways to separate it: $u=x$, $dv=sin(x);dx$; or: $u=sin(x)$, $dv=x;dx$; or: $u=xsin(x)$, $dv=dx$. One of these choices will work, the others will not.
$endgroup$
– Ben W
Jan 27 at 16:20
|
show 1 more comment
$begingroup$
I know this does not answer your questions but is related.
$endgroup$
– Exp ikx
Jan 27 at 16:14
$begingroup$
can you please help me
$endgroup$
– user8718165
Jan 27 at 16:16
$begingroup$
This is not a direct answer to the question, but it could be (I hope) a nevertheless helpful comment. When we try to see things graphically, the aim is (usually) to give us some kind of insight into the solution. I am unaware of any such insightful graphical interpretation of IBP. The link you gave was kinda neat, but I confess I see no obvious practical application for it. Instead, I always view integration by parts as an algebraic exercise. How can we separate an expression so that if we integrate one part and differentiate the other, we obtain something we can integrate nicely?
$endgroup$
– Ben W
Jan 27 at 16:18
$begingroup$
I have tried to solve it algebraically and it worked. Thanks
$endgroup$
– user8718165
Jan 27 at 16:20
$begingroup$
...In your case, we have $xsin(x);dx$. There are only three ways to separate it: $u=x$, $dv=sin(x);dx$; or: $u=sin(x)$, $dv=x;dx$; or: $u=xsin(x)$, $dv=dx$. One of these choices will work, the others will not.
$endgroup$
– Ben W
Jan 27 at 16:20
$begingroup$
I know this does not answer your questions but is related.
$endgroup$
– Exp ikx
Jan 27 at 16:14
$begingroup$
I know this does not answer your questions but is related.
$endgroup$
– Exp ikx
Jan 27 at 16:14
$begingroup$
can you please help me
$endgroup$
– user8718165
Jan 27 at 16:16
$begingroup$
can you please help me
$endgroup$
– user8718165
Jan 27 at 16:16
$begingroup$
This is not a direct answer to the question, but it could be (I hope) a nevertheless helpful comment. When we try to see things graphically, the aim is (usually) to give us some kind of insight into the solution. I am unaware of any such insightful graphical interpretation of IBP. The link you gave was kinda neat, but I confess I see no obvious practical application for it. Instead, I always view integration by parts as an algebraic exercise. How can we separate an expression so that if we integrate one part and differentiate the other, we obtain something we can integrate nicely?
$endgroup$
– Ben W
Jan 27 at 16:18
$begingroup$
This is not a direct answer to the question, but it could be (I hope) a nevertheless helpful comment. When we try to see things graphically, the aim is (usually) to give us some kind of insight into the solution. I am unaware of any such insightful graphical interpretation of IBP. The link you gave was kinda neat, but I confess I see no obvious practical application for it. Instead, I always view integration by parts as an algebraic exercise. How can we separate an expression so that if we integrate one part and differentiate the other, we obtain something we can integrate nicely?
$endgroup$
– Ben W
Jan 27 at 16:18
$begingroup$
I have tried to solve it algebraically and it worked. Thanks
$endgroup$
– user8718165
Jan 27 at 16:20
$begingroup$
I have tried to solve it algebraically and it worked. Thanks
$endgroup$
– user8718165
Jan 27 at 16:20
$begingroup$
...In your case, we have $xsin(x);dx$. There are only three ways to separate it: $u=x$, $dv=sin(x);dx$; or: $u=sin(x)$, $dv=x;dx$; or: $u=xsin(x)$, $dv=dx$. One of these choices will work, the others will not.
$endgroup$
– Ben W
Jan 27 at 16:20
$begingroup$
...In your case, we have $xsin(x);dx$. There are only three ways to separate it: $u=x$, $dv=sin(x);dx$; or: $u=sin(x)$, $dv=x;dx$; or: $u=xsin(x)$, $dv=dx$. One of these choices will work, the others will not.
$endgroup$
– Ben W
Jan 27 at 16:20
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
I am not sure if this helps, but I'll show you a proof for integration by parts then use it to evaluate $int xsin x,dx$. I don't know how to really think of integration by parts graphically, most of the visualization for me comes from seeing the algebra.
Let $u(x)$ and $v(x)$ be differentiable, continuous functions. Then we know that
$$frac{d}{dx}left[u(x)v(x)right]=u'(x)v(x)+u(x)v'(x)$$
And if we integrate both sides, we have that
$$int frac{d}{dx}left[u(x)v(x)right] dx=int u'(x)v(x)dx+int u(x)v'(x)dx$$
$$u(x)v(x)=int u'(x)v(x)dx+int u(x)v'(x)dx$$
And after some simple re-arrangement,
$$int u(x)v'(x)dx=u(x)v(x)-int u'(x)v(x)dx$$
Which is usually just abbreviated as
$$int udv=uv-int vdu$$
Which completes our proof.
Next, we evaluate some examples.
Ex 1: $$I=int xsin(x)dx$$
For this, I do not recommend trying to visualize the integration, because that could just make things more complicated. Instead integrate by parts with
$$u=xRightarrow du=dx\ dv=sin(x)dxRightarrow v=-cos(x)$$
Hence,
$$I=-xcos(x)+intcos(x)dx$$
$$I=-xcos(x)+sin(x)+C$$
Which is rather easy: no visualization required.
Ex. 2:
$$I=int xtan^{-1}(x)dx$$
I chose this example because it is rather hard to visualize, but can be computed with integration by parts as follows:
$$u=tan^{-1}(x)Rightarrow du=frac{dx}{1+x^2}\ dv=xdxRightarrow v=frac{x^2}2$$
Hence
$$I=frac{x^2}2tan^{-1}(x)-frac12intfrac{x^2}{1+x^2}dx$$
This final integral is easier than it looks, and it can be computed by noting that
$$begin{align}
intfrac{x^2}{x^2+1}dx&=intfrac{x^2+1-1}{x^2+1}dx\
&=intfrac{x^2+1}{x^2+1}dx-intfrac{dx}{x^2+1}\
&=int dx-intfrac{dx}{x^2+1}\
&=x-tan^{-1}(x)\
end{align}$$
So, without having to visualize anything,
$$I=frac{x^2+1}2tan^{-1}(x)-frac{x}2+C$$
Tell me if there's anything I can do to improve my answer if it is not what you needed.
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
$endgroup$
$begingroup$
Thank you so much for answering!!
$endgroup$
– user8718165
Jan 28 at 0:13
$begingroup$
I am not able to upvote you because don't have enough reputation.
$endgroup$
– user8718165
Jan 28 at 0:19
$begingroup$
@user8718165 You are very welcome. You do have enough reputation to accept answers though. Just click the check-mark next to the voting button if my answer helped :)
$endgroup$
– clathratus
Jan 28 at 0:52
add a comment |
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$begingroup$
I am not sure if this helps, but I'll show you a proof for integration by parts then use it to evaluate $int xsin x,dx$. I don't know how to really think of integration by parts graphically, most of the visualization for me comes from seeing the algebra.
Let $u(x)$ and $v(x)$ be differentiable, continuous functions. Then we know that
$$frac{d}{dx}left[u(x)v(x)right]=u'(x)v(x)+u(x)v'(x)$$
And if we integrate both sides, we have that
$$int frac{d}{dx}left[u(x)v(x)right] dx=int u'(x)v(x)dx+int u(x)v'(x)dx$$
$$u(x)v(x)=int u'(x)v(x)dx+int u(x)v'(x)dx$$
And after some simple re-arrangement,
$$int u(x)v'(x)dx=u(x)v(x)-int u'(x)v(x)dx$$
Which is usually just abbreviated as
$$int udv=uv-int vdu$$
Which completes our proof.
Next, we evaluate some examples.
Ex 1: $$I=int xsin(x)dx$$
For this, I do not recommend trying to visualize the integration, because that could just make things more complicated. Instead integrate by parts with
$$u=xRightarrow du=dx\ dv=sin(x)dxRightarrow v=-cos(x)$$
Hence,
$$I=-xcos(x)+intcos(x)dx$$
$$I=-xcos(x)+sin(x)+C$$
Which is rather easy: no visualization required.
Ex. 2:
$$I=int xtan^{-1}(x)dx$$
I chose this example because it is rather hard to visualize, but can be computed with integration by parts as follows:
$$u=tan^{-1}(x)Rightarrow du=frac{dx}{1+x^2}\ dv=xdxRightarrow v=frac{x^2}2$$
Hence
$$I=frac{x^2}2tan^{-1}(x)-frac12intfrac{x^2}{1+x^2}dx$$
This final integral is easier than it looks, and it can be computed by noting that
$$begin{align}
intfrac{x^2}{x^2+1}dx&=intfrac{x^2+1-1}{x^2+1}dx\
&=intfrac{x^2+1}{x^2+1}dx-intfrac{dx}{x^2+1}\
&=int dx-intfrac{dx}{x^2+1}\
&=x-tan^{-1}(x)\
end{align}$$
So, without having to visualize anything,
$$I=frac{x^2+1}2tan^{-1}(x)-frac{x}2+C$$
Tell me if there's anything I can do to improve my answer if it is not what you needed.
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
$endgroup$
$begingroup$
Thank you so much for answering!!
$endgroup$
– user8718165
Jan 28 at 0:13
$begingroup$
I am not able to upvote you because don't have enough reputation.
$endgroup$
– user8718165
Jan 28 at 0:19
$begingroup$
@user8718165 You are very welcome. You do have enough reputation to accept answers though. Just click the check-mark next to the voting button if my answer helped :)
$endgroup$
– clathratus
Jan 28 at 0:52
add a comment |
$begingroup$
I am not sure if this helps, but I'll show you a proof for integration by parts then use it to evaluate $int xsin x,dx$. I don't know how to really think of integration by parts graphically, most of the visualization for me comes from seeing the algebra.
Let $u(x)$ and $v(x)$ be differentiable, continuous functions. Then we know that
$$frac{d}{dx}left[u(x)v(x)right]=u'(x)v(x)+u(x)v'(x)$$
And if we integrate both sides, we have that
$$int frac{d}{dx}left[u(x)v(x)right] dx=int u'(x)v(x)dx+int u(x)v'(x)dx$$
$$u(x)v(x)=int u'(x)v(x)dx+int u(x)v'(x)dx$$
And after some simple re-arrangement,
$$int u(x)v'(x)dx=u(x)v(x)-int u'(x)v(x)dx$$
Which is usually just abbreviated as
$$int udv=uv-int vdu$$
Which completes our proof.
Next, we evaluate some examples.
Ex 1: $$I=int xsin(x)dx$$
For this, I do not recommend trying to visualize the integration, because that could just make things more complicated. Instead integrate by parts with
$$u=xRightarrow du=dx\ dv=sin(x)dxRightarrow v=-cos(x)$$
Hence,
$$I=-xcos(x)+intcos(x)dx$$
$$I=-xcos(x)+sin(x)+C$$
Which is rather easy: no visualization required.
Ex. 2:
$$I=int xtan^{-1}(x)dx$$
I chose this example because it is rather hard to visualize, but can be computed with integration by parts as follows:
$$u=tan^{-1}(x)Rightarrow du=frac{dx}{1+x^2}\ dv=xdxRightarrow v=frac{x^2}2$$
Hence
$$I=frac{x^2}2tan^{-1}(x)-frac12intfrac{x^2}{1+x^2}dx$$
This final integral is easier than it looks, and it can be computed by noting that
$$begin{align}
intfrac{x^2}{x^2+1}dx&=intfrac{x^2+1-1}{x^2+1}dx\
&=intfrac{x^2+1}{x^2+1}dx-intfrac{dx}{x^2+1}\
&=int dx-intfrac{dx}{x^2+1}\
&=x-tan^{-1}(x)\
end{align}$$
So, without having to visualize anything,
$$I=frac{x^2+1}2tan^{-1}(x)-frac{x}2+C$$
Tell me if there's anything I can do to improve my answer if it is not what you needed.
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
$endgroup$
$begingroup$
Thank you so much for answering!!
$endgroup$
– user8718165
Jan 28 at 0:13
$begingroup$
I am not able to upvote you because don't have enough reputation.
$endgroup$
– user8718165
Jan 28 at 0:19
$begingroup$
@user8718165 You are very welcome. You do have enough reputation to accept answers though. Just click the check-mark next to the voting button if my answer helped :)
$endgroup$
– clathratus
Jan 28 at 0:52
add a comment |
$begingroup$
I am not sure if this helps, but I'll show you a proof for integration by parts then use it to evaluate $int xsin x,dx$. I don't know how to really think of integration by parts graphically, most of the visualization for me comes from seeing the algebra.
Let $u(x)$ and $v(x)$ be differentiable, continuous functions. Then we know that
$$frac{d}{dx}left[u(x)v(x)right]=u'(x)v(x)+u(x)v'(x)$$
And if we integrate both sides, we have that
$$int frac{d}{dx}left[u(x)v(x)right] dx=int u'(x)v(x)dx+int u(x)v'(x)dx$$
$$u(x)v(x)=int u'(x)v(x)dx+int u(x)v'(x)dx$$
And after some simple re-arrangement,
$$int u(x)v'(x)dx=u(x)v(x)-int u'(x)v(x)dx$$
Which is usually just abbreviated as
$$int udv=uv-int vdu$$
Which completes our proof.
Next, we evaluate some examples.
Ex 1: $$I=int xsin(x)dx$$
For this, I do not recommend trying to visualize the integration, because that could just make things more complicated. Instead integrate by parts with
$$u=xRightarrow du=dx\ dv=sin(x)dxRightarrow v=-cos(x)$$
Hence,
$$I=-xcos(x)+intcos(x)dx$$
$$I=-xcos(x)+sin(x)+C$$
Which is rather easy: no visualization required.
Ex. 2:
$$I=int xtan^{-1}(x)dx$$
I chose this example because it is rather hard to visualize, but can be computed with integration by parts as follows:
$$u=tan^{-1}(x)Rightarrow du=frac{dx}{1+x^2}\ dv=xdxRightarrow v=frac{x^2}2$$
Hence
$$I=frac{x^2}2tan^{-1}(x)-frac12intfrac{x^2}{1+x^2}dx$$
This final integral is easier than it looks, and it can be computed by noting that
$$begin{align}
intfrac{x^2}{x^2+1}dx&=intfrac{x^2+1-1}{x^2+1}dx\
&=intfrac{x^2+1}{x^2+1}dx-intfrac{dx}{x^2+1}\
&=int dx-intfrac{dx}{x^2+1}\
&=x-tan^{-1}(x)\
end{align}$$
So, without having to visualize anything,
$$I=frac{x^2+1}2tan^{-1}(x)-frac{x}2+C$$
Tell me if there's anything I can do to improve my answer if it is not what you needed.
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
$endgroup$
I am not sure if this helps, but I'll show you a proof for integration by parts then use it to evaluate $int xsin x,dx$. I don't know how to really think of integration by parts graphically, most of the visualization for me comes from seeing the algebra.
Let $u(x)$ and $v(x)$ be differentiable, continuous functions. Then we know that
$$frac{d}{dx}left[u(x)v(x)right]=u'(x)v(x)+u(x)v'(x)$$
And if we integrate both sides, we have that
$$int frac{d}{dx}left[u(x)v(x)right] dx=int u'(x)v(x)dx+int u(x)v'(x)dx$$
$$u(x)v(x)=int u'(x)v(x)dx+int u(x)v'(x)dx$$
And after some simple re-arrangement,
$$int u(x)v'(x)dx=u(x)v(x)-int u'(x)v(x)dx$$
Which is usually just abbreviated as
$$int udv=uv-int vdu$$
Which completes our proof.
Next, we evaluate some examples.
Ex 1: $$I=int xsin(x)dx$$
For this, I do not recommend trying to visualize the integration, because that could just make things more complicated. Instead integrate by parts with
$$u=xRightarrow du=dx\ dv=sin(x)dxRightarrow v=-cos(x)$$
Hence,
$$I=-xcos(x)+intcos(x)dx$$
$$I=-xcos(x)+sin(x)+C$$
Which is rather easy: no visualization required.
Ex. 2:
$$I=int xtan^{-1}(x)dx$$
I chose this example because it is rather hard to visualize, but can be computed with integration by parts as follows:
$$u=tan^{-1}(x)Rightarrow du=frac{dx}{1+x^2}\ dv=xdxRightarrow v=frac{x^2}2$$
Hence
$$I=frac{x^2}2tan^{-1}(x)-frac12intfrac{x^2}{1+x^2}dx$$
This final integral is easier than it looks, and it can be computed by noting that
$$begin{align}
intfrac{x^2}{x^2+1}dx&=intfrac{x^2+1-1}{x^2+1}dx\
&=intfrac{x^2+1}{x^2+1}dx-intfrac{dx}{x^2+1}\
&=int dx-intfrac{dx}{x^2+1}\
&=x-tan^{-1}(x)\
end{align}$$
So, without having to visualize anything,
$$I=frac{x^2+1}2tan^{-1}(x)-frac{x}2+C$$
Tell me if there's anything I can do to improve my answer if it is not what you needed.
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
answered Jan 27 at 21:23
clathratusclathratus
5,2291438
5,2291438
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Thank you so much for answering!!
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– user8718165
Jan 28 at 0:13
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I am not able to upvote you because don't have enough reputation.
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– user8718165
Jan 28 at 0:19
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@user8718165 You are very welcome. You do have enough reputation to accept answers though. Just click the check-mark next to the voting button if my answer helped :)
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– clathratus
Jan 28 at 0:52
add a comment |
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Thank you so much for answering!!
$endgroup$
– user8718165
Jan 28 at 0:13
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I am not able to upvote you because don't have enough reputation.
$endgroup$
– user8718165
Jan 28 at 0:19
$begingroup$
@user8718165 You are very welcome. You do have enough reputation to accept answers though. Just click the check-mark next to the voting button if my answer helped :)
$endgroup$
– clathratus
Jan 28 at 0:52
$begingroup$
Thank you so much for answering!!
$endgroup$
– user8718165
Jan 28 at 0:13
$begingroup$
Thank you so much for answering!!
$endgroup$
– user8718165
Jan 28 at 0:13
$begingroup$
I am not able to upvote you because don't have enough reputation.
$endgroup$
– user8718165
Jan 28 at 0:19
$begingroup$
I am not able to upvote you because don't have enough reputation.
$endgroup$
– user8718165
Jan 28 at 0:19
$begingroup$
@user8718165 You are very welcome. You do have enough reputation to accept answers though. Just click the check-mark next to the voting button if my answer helped :)
$endgroup$
– clathratus
Jan 28 at 0:52
$begingroup$
@user8718165 You are very welcome. You do have enough reputation to accept answers though. Just click the check-mark next to the voting button if my answer helped :)
$endgroup$
– clathratus
Jan 28 at 0:52
add a comment |
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I know this does not answer your questions but is related.
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– Exp ikx
Jan 27 at 16:14
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can you please help me
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– user8718165
Jan 27 at 16:16
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This is not a direct answer to the question, but it could be (I hope) a nevertheless helpful comment. When we try to see things graphically, the aim is (usually) to give us some kind of insight into the solution. I am unaware of any such insightful graphical interpretation of IBP. The link you gave was kinda neat, but I confess I see no obvious practical application for it. Instead, I always view integration by parts as an algebraic exercise. How can we separate an expression so that if we integrate one part and differentiate the other, we obtain something we can integrate nicely?
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– Ben W
Jan 27 at 16:18
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I have tried to solve it algebraically and it worked. Thanks
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– user8718165
Jan 27 at 16:20
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...In your case, we have $xsin(x);dx$. There are only three ways to separate it: $u=x$, $dv=sin(x);dx$; or: $u=sin(x)$, $dv=x;dx$; or: $u=xsin(x)$, $dv=dx$. One of these choices will work, the others will not.
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– Ben W
Jan 27 at 16:20