Hedgecutter wrote: » intergration by parts?
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!!
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]
TheBody wrote: » I'm just up to my neck at work today. I'll will reply to your question later this evening though.
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.
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!!
Hedgecutter wrote: » Cant remember the term he used but it was not cancel, I will clarify what he did next time I meet him.
TheBody wrote: » He is talking rubbish. I'm just working throught your other problem here. I'll post it in a minute for you.
Hedgecutter wrote: » That's very worrying, I will question him
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]
Hedgecutter wrote: » I had it drawn but couldn't fit it on the image.
TheBody wrote: » Notice how I didn't have to split up the integral as it is all above the x axis.
Hedgecutter wrote: » First thing I do is draw Y=16-X^2 and work from there?
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?
Hedgecutter wrote: » To plot the graph it would be (0,-4) (16,0) (0,4)
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/
TheBody wrote: » Hang on!!!! I made big mistake in my integration above. Give me a minute to fix it!!
TheBody wrote: » Ok, post 107 is changed now. Sorry about that. I dunno what I was thinking when I did the integration!
Hedgecutter wrote: » Thanks, time to call it a night. Thanks again for the help.
