Intergration calculating centroid of an area

TheBody

Hedgecutter wrote: »

intergration by parts?

No, regular substitution.

Hedgecutter

Am I on the right track?

TheBody

Good attempt but there are errors. You have

[latex]\int\frac{16x}{(4x^2-3)^2}dx=16\int\frac{x}{(4x^2-3)^2}dx[/latex].

Let [latex]u=4x^2-3[/latex]. Differentiating we get [latex]\frac{du}{dx}=8x[/latex]. If we rearrange this we can get:

[latex]\frac{du}{8}=xdx[/latex].

We can now re-write the original integral to get:

[latex]16\int\frac{1}{u^2} \frac{du}{8} =2\int u^{-2} du=2\frac{u^{-1}}{-1}+c=\frac{-2}{u}+c=-\frac{1}{4x^2-3}+c.[/latex]

I'm off to bed but I'll reply to any posts in the morning. Good luck!!

Hedgecutter

TheBody wrote: »

Good attempt but there are errors. You have

[latex]\int\frac{16x}{(4x^2-3)^2}dx=16\int\frac{x}{(4x^2-3)^2}dx[/latex].

Let [latex]u=4x^2-3[/latex]. Differentiating we get [latex]\frac{du}{dx}=8x[/latex]. If we rearrange this we can get:

[latex]\frac{du}{8}=xdx[/latex].

We can now re-write the original integral to get:

[latex]16\int\frac{1}{u^2} \frac{du}{8} =2\int u^{-2} du=2\frac{u^{-1}}{-1}+c=\frac{-2}{u}+c=-\frac{1}{4x^2-3}+c.[/latex]

I'm off to bed but I'll reply to any posts in the morning. Good luck!!

Thanks for your help. Ill finish this one and try one more.

Hedgecutter

TheBody wrote: »

Good attempt but there are errors. You have

[latex]\int\frac{16x}{(4x^2-3)^2}dx=16\int\frac{x}{(4x^2-3)^2}dx[/latex].

Let [latex]u=4x^2-3[/latex]. Differentiating we get [latex]\frac{du}{dx}=8x[/latex]. If we rearrange this we can get:

[latex]\frac{du}{8}=xdx[/latex].

We can now re-write the original integral to get:

[latex]16\int\frac{1}{u^2} \frac{du}{8} =2\int u^{-2} du=2\frac{u^{-1}}{-1}+c=\frac{-2}{u}+c=-\frac{1}{4x^2-3}+c.[/latex]

I'm off to bed but I'll reply to any posts in the morning. Good luck!!

I don't understand how you got -2/u+c in the second last step

TheBody

I presume you are ok to the third last step so.

In that step we have:

[latex]2\frac{u^{-1}}{-1}+c[/latex]. The positive 2 divided by the -1 gives -2. Then using the rules of indices, I can bring the [latex]u^{-1}[/latex] under the line by changing the sign on the power. So we get:

[latex]-\frac{2}{u}+c[/latex]

Hedgecutter

TheBody wrote: »

I presume you are ok to the third last step so.

In that step we have:

[latex]2\frac{u^{-1}}{-1}+c[/latex]. The positive 2 divided by the -1 gives -2. Then using the rules of indices, I can bring the [latex]u^{-1}[/latex] under the line by changing the sign on the power. So we get:

[latex]-\frac{2}{u}+c[/latex]

Ya I understand the third step and understand what your doing with indices rule.
Cheers.

Hedgecutter

This is the last one i was attempting last night. Struggled on the last line and not sure how to finish.

TheBody

I'm just up to my neck at work today. I'll will reply to your question later this evening though.

Hedgecutter

TheBody wrote: »

I'm just up to my neck at work today. I'll will reply to your question later this evening though.

Don't worry about, I'm back to thermodynamics this evening so won't be back to this till Friday.
Thanks

Hedgecutter

I got the lecture to have a look at that question. He said intergration by substition. Image below.

Also on the second image iv tried to find the area of a curve y=16-x^2 and X-axis.

TheBody

Hedgecutter wrote: »

I got the lecture to have a look at that question. He said intergration by substition. Image below.

Also on the second image iv tried to find the area of a curve y=16-x^2 and X-axis.

Did your lecturer write out that solution for you in the first image? It is incorrect. Look at step 4. How do the [latex]12x-5[/latex] cancel!!

Hedgecutter

TheBody wrote: »

Did your lecturer write out that solution for you in the first image? It is incorrect. Look at step 4. How do the [latex]12x-5[/latex] cancel!!

Cant remember the term he used but it was not cancel, I will clarify what he did next time I meet him.

TheBody

Hedgecutter wrote: »

Cant remember the term he used but it was not cancel, I will clarify what he did next time I meet him.

He is talking rubbish. I'm just working through your other problem here. I'll post it in a minute for you.

Hedgecutter

TheBody wrote: »

He is talking rubbish. I'm just working throught your other problem here. I'll post it in a minute for you.

That's very worrying, I will question him

TheBody

With regards to the second image, you have a bunch of stuff early on that I don't really understand.

It's a good idea to begin with drawing a sketch of the curve. [latex]y=16-x^2[/latex]. It looks like an "n" shaped (maybe sad face is a better description). It crosses the x-axis at -4 and 4 and crosses the y-axis at y=16. The reason a sketch is useful is because you want to know when the area you are interested in is above or below the x-axis. Any areas below the x axis will give you a negative value which you have to make positive as a negative area makes no sense.

In this case, all the region in question is above the x-axis so it's all good.

[latex]\int_{-4}^4 (16-x^2) dx= (16x-\frac{x^3}{3})|_{-4}^{4}=((16(4)-\frac{(4)^3}{3})-(16(-4)-\frac{(-4)^3}{3})=(\frac{128}{3})-(-\frac{128}{3})=\frac{256}{3}[/latex]units squared.

TheBody

Hedgecutter wrote: »

That's very worrying, I will question him

Don't get me wrong, the integral can be done but it would require more advanced techniques than you seem to have covered.

Hedgecutter

TheBody wrote: »

With regards to the second image, you have a bunch of stuff early on that I don't really understand.

It's a good idea to begin with drawing a sketch of the curve. [latex]y=16-x^2[/latex]. It looks like an "n" shaped (maybe sad face is a better description). It crosses the x-axis at -4 and 4 and crosses the y-axis at y=16. The reason a sketch is useful is because you want to know when the area you are interested in is above or below the x-axis. Any areas below the x axis will give you a negative value which you have to make positive as a negative area makes no sense.

In this case, all the region in question is above the x-axis so it's all good.

[latex]\int_{-4}^4 (16-x^2) dx= (16x-\frac{x^4}{4})|_{-4}^{4}=((16(4)-\frac{(4)^4}{4})-(16(-4)-\frac{(-4)^4}{4})=(0)-(-128)=128 units squared.[/latex]

I had it drawn but couldn't fit it on the image.

TheBody

Hedgecutter wrote: »

I had it drawn but couldn't fit it on the image.

Notice how I didn't have to split up the integral as it is all above the x axis.

Hedgecutter

TheBody wrote: »

Notice how I didn't have to split up the integral as it is all above the x axis.

First thing I do is draw Y=16-X^2 and work from there?

TheBody

Hedgecutter wrote: »

First thing I do is draw Y=16-X^2 and work from there?

I would always recommend drawing a sketch first. It is a big help when you can clearly see the region in question. Of course this means that you need to know the shapes of some basic curves in the first place! How would you be fixed in that regard?

Hedgecutter

TheBody wrote: »

I would always recommend drawing a sketch first. It is a big help when you can clearly see the region in question. Of course this means that you need to know the shapes of some basic curves in the first place! How would you be fixed in that regard?

To plot the graph it would be (0,-4) (16,0) (0,4)

TheBody

Hedgecutter wrote: »

To plot the graph it would be (0,-4) (16,0) (0,4)

Yes, a "sad face" curve through those points. There is a great free package called geogebra you can download if you want to see is done perfectly.

You can download it here: https://www.geogebra.org/

Hedgecutter

TheBody wrote: »

Yes, a "sad face" curve through those points. There is a great free package called geogebra you can download if you want to see is done perfectly.

You can download it here: https://www.geogebra.org/

Perfect, so if I let x=0 y=16, where as I had x=1 y=15

TheBody

Hedgecutter wrote: »

To plot the graph it would be (0,-4) (16,0) (0,4)

Sorry, I just noticed a small mistake here. (-4,0) (0,16) and (4,0) are the points you need.

TheBody

Hang on!!!! I made big mistake in my integration above. Give me a minute to fix it!!

Hedgecutter

TheBody wrote: »

Hang on!!!! I made big mistake in my integration above. Give me a minute to fix it!!

Im downloading geogeba so no rush,

TheBody

Ok, post 107 is changed now. Sorry about that. I dunno what I was thinking when I did the integration!

Hedgecutter

TheBody wrote: »

Ok, post 107 is changed now. Sorry about that. I dunno what I was thinking when I did the integration!

Thanks, time to call it a night. Thanks again for the help.

TheBody

Hedgecutter wrote: »

Thanks, time to call it a night. Thanks again for the help.

No problem. Keep up the good work. It will all pay off come exam time.