Intergration calculating centroid of an area

Hedgecutter

Having another go at this one struggling to finish.

TheBody

Long division is what I had in mind for that problem but it seemed a little complicated given the other problems you have done.

Everything looks ok so far.

In terms of integrating the hard bit, you have made a good start at it.

[latex]\int \frac{47}{24(12x-5)}dx=\frac{47}{24}\int \frac{1}{12x-5}dx [/latex]. As you said, we let [latex]u=12x-5[/latex] which will eventually give us that [latex]\frac{du}{12}=dx[/latex]. We can now replace all the x stuff with u stuff. We get:

[latex]\frac{47}{24}\int \frac{1}{12x-5}dx=\frac{47}{24}\int \frac{1}{u} \frac{du}{12} =\frac{47}{288}\int \frac{1}{u} du=\frac{47}{288}\ln|u|+c=\frac{47}{288}\ln|12x-5|+c[/latex].

Don't forget to integrate the other bits and put it all together!!






4,000 posts! Go me!!

Hedgecutter

Ya difficult one compared to what I have being doing, what to have it fully completed, correct and also understood before I show it to my lecture.

Hedgecutter

Latest attempt

TheBody

Hedgecutter wrote: »

Latest attempt

Almost correct!!

In the very last line, [latex]\int \frac{x}{2}dx=\frac{x^2}{4} [/latex]. Also in the [latex]\ln[/latex] bit you need absolute value signs instead of round brackets like I do in the post above.

Apart from that it looks ok. Your exposition could do with a little improving but overall it's not too bad.

Hedgecutter

TheBody wrote: »

Almost correct!!

In the very last line, [latex]\int \frac{x}{2}dx=\frac{x^2}{4} [/latex]. Also in the [latex]\ln[/latex] bit you need absolute value signs instead of round brackets like I do in the post above.

Apart from that it looks ok. Your exposition could do with a little improving but overall it's not too bad.

Thanks spotted that x^2/4 of course, silly.

Hedgecutter

Calculate root mean square for y=2e^x/2 Range x=0 x=2

TheBody

Hedgecutter wrote: »

Calculate root mean square for y=2e^x/2 Range x=0 x=2

To start, I have no idea what the formula for the root mean square is, so I can only check the calculations you have on the image.

You made a small mistake. You halved the 4 to get 2 but kept the half. Also, it's considered poor exposition to drop the square root symbol and then put it back in near the end. Keep it in all the time until you actually take the square root. Don't forget the = signs too!!

Hedgecutter

TheBody wrote: »

To start, I have no idea what the formula for the root mean square is, so I can only check the calculations you have on the image.

You made a small mistake. You halved the 4 to get 2 but kept the half. Also, it's considered poor exposition to drop the square root symbol and then put it back in near the end. Keep it in all the time until you actually take the square root. Don't forget the = signs too!!

Fair enough ill rewrite it with formula. need to get good habits going. cheers

TheBody

Hedgecutter wrote: »

Fair enough ill rewrite it with formula. need to get good habits going. cheers

I'm a bit of a maths nazi when it comes to exposition. I compare it to reading a book without any full stops or comma, etc. You could probably figure out what they are trying to say but it doesn't look very good.

Also, for yourself, it is much easier to fix any problems if your work is laid out properly.

Don't mean to bust your chops over it!!

Hedgecutter

TheBody wrote: »

I'm a bit of a maths nazi when it comes to exposition. I compare it to reading a book without any full stops or comma, etc. You could probably figure out what they are trying to say but it doesn't look very good.

Also, for yourself, it is much easier to fix any problems if your work is laid out properly.

Don't mean to bust your chops over it!!

Maths nazi:D

Hedgecutter

Hope this is better.

TheBody

Hedgecutter wrote: »

Hope this is better.

Lovely! I don't have a calculator to hand to check the number bits but everything else is ok.

Hedgecutter

TheBody wrote: »

Lovely! I don't have a calculator to hand to check the number bits but everything else is ok.

Time for a beer so;) Thanks again.

TheBody

Hedgecutter wrote: »

Time for a beer so;) Thanks again.

No problem. Glad I could help.

Hedgecutter

Had a go at this one this evening, similar to one I did the other day but what I don't fully understand is choosing the limits.

here we have 0,3, are the limits always the starting coordinate and the last coordinate on the axis that the area is on?

TheBody

Hedgecutter wrote: »

Had a go at this one this evening, similar to one I did the other day but what I don't fully understand is choosing the limits.

here we have 0,3, are the limits always the starting coordinate and the last coordinate on the axis that the area is on?

Solution looks good.

Yes, the limits just the x-values from start to finish of the region.

Hedgecutter

TheBody wrote: »

Solution looks good.

Yes, the limits just the x-values from start to finish of the region.

Cheers nearly six years of past papers covered in regards calculus. If I can keep going over them I should be ok.

TheBody

Hedgecutter wrote: »

Cheers nearly six years of past papers covered in regards calculus. If I can keep going over them I should be ok.

That's good going. It's a lot of work but it does pay off.

Hedgecutter

TheBody wrote: »

That's good going. It's a lot of work but it does pay off.

Ya, very happy with it, probability next. not too bad in class but can be tricky I think.

TheBody

Hedgecutter wrote: »

Ya, very happy with it, probability next. not too bad in class but can be tricky I think.

It can be very tricky. For me, it is very boring too. I just never had an interest probability or statistics for that matter.

Hedgecutter

TheBody wrote: »

It can be very tricky. For me, it is very boring too. I just never had an interest probability or statistics for that matter.

Lector says the exact same. Lining up a few maths grinds to get me over the line. might spend a night or two on probability, worth 10%.

TheBody

It's worth spending some time on it. Even if it turns out your not great at it, you may pick up a few marks. They all add up!

Hedgecutter

Wondering if I'm right with this turning point question ?

TheBody

Hedgecutter wrote: »

Wondering if I'm right with this turning point question ?

Looks good to me. The only comment I would make is you are lacking any explanation about what you are doing. For example, you used the second derivative to decide which points were local max/min but you didn't actually say which is which!

Hedgecutter

TheBody wrote: »

Looks good to me. The only comment I would make is you are lacking any explanation about what you are doing. For example, you used the second derivative to decide which points were local max/min but you didn't actually say which is which!

Fair comment I'll sort that out.

Hedgecutter

Not sure if I posted this intergration question before but I can't remember how I did the areas I circled in red. Where did 6/4 come from and the same in the last step (6/4)(3/2 e^2x+c)

TheBody

You are using the fact that for k a constant, [latex]\int e^{kx}dx=\frac{1}{k}e^{kx}+c [/latex]

The bit circled isn't qutie right. Looking at the bit directly above the circled bit:

[latex]\int \frac{1}{2}e^{2x} 6 dx=3\int e^{2x}dx=\frac{3}{2}e^{2x}+c.[/latex]

Hedgecutter

Find the area by intergration I'm curious why the sign changed to a minus in the part I've circled red near the bottom. 7-x-x^2-1dx. Previous line its a plus one. Can't remember why this happened.
Is it a mistake ?

TheBody

Hedgecutter wrote: »

Find the area by intergration I'm curious why the sign changed to a minus in the part I've circled red near the bottom. 7-x-x^2-1dx. Previous line its a plus one. Can't remember why this happened.
Is it a mistake ?

It changed sign because you multiplied the stuff in the brackets by the minus sign.