Intergration calculating centroid of an area

Hedgecutter

TheBody wrote: »

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

Thanks. Make since now.

TheBody

No problem.

Hedgecutter

Wondered about how I finished these questions. Both the same question but not sure which if any is finished right.

TheBody

Hedgecutter wrote: »

Wondered about how I finished these questions. Both the same question but not sure which if any is finished right.

Looks ok to me. The only suggestion I would make is to use brackets instead of the "x" for multiplication. It can be confusing whether you mean multiplication or the variable x.

Hedgecutter

Cheers.
Going over some stuff I've done before. In 6(I) last step we had -2/u+c, how did that become -1/4x^2-3

Also in 6(ii) to intergrate (1/2 sin2x)(6dx) what does that become is it 1 1/2 cos 2x +c?

TheBody

Hedgecutter wrote: »

Cheers.
Going over some stuff I've done before. In 6(I) last step we had -2/u+c, how did that become -1/4x^2-3

Also in 6(ii) to intergrate (1/2 sin2x)(6dx) what does that become is it 1 1/2 cos 2x +c?

With regards your first question, your original problem had a variable of x in it. You then switched it to a "u" type integral. After you did the integration, you needed to switch back to "x" type stuff; so you just replaced back in for "u".


For your second question, you have:

[latex]\int \frac{1}{2}\sin(2x)(6dx)=3\int \sin(2x)dx=(3)(-\frac{1}{2})\cos(2x)+c =-\frac{3}{2}\cos(2x)+c[/latex]

Hedgecutter

TheBody wrote: »

With regards your first question, your original problem had a variable of x in it. You then switched it to a "u" type integral. After you did the integration, you needed to switch back to "x" type stuff; so you just replaced back in for "u".


For your second question, you have:

[latex]\int \frac{1}{2}\sin(2x)(6dx)=3\int \sin(2x)dx=(3)(-\frac{1}{2})\cos(2x)+c =-\frac{3}{2}\cos(2x)+c[/latex]

About the first one does the 1 replace the -2?

TheBody

Hedgecutter wrote: »

About the first one does the 1 replace the -2?

Oh yea , that's a mistake. Sorry, I missed that. There should be a 2 on top. You also need +c in EVERY bit after you do the integration.

Hedgecutter

TheBody wrote: »

Oh yea , that's a mistake. Sorry, I missed that. There should be a 2 on top. You also need +c in EVERY bit after you do the integration.

perfect thanks

Hedgecutter

Looking at my tutors solution on this intergration question and can't understand where he got the 6e ^x/2 from

dlouth15

Hedgecutter wrote: »

Looking at my tutors solution on this intergration question and can't understand where he got the 6e ^x/2 from

If you integrate e^(ax) you get

[latex]\displaystyle{\int e^{ax}\, dx=\frac{1}{a}e^{ax}+c}[/latex]

This is just a standard integral that can be looked up. The result is just e^(ax) divided by a.

Now [latex]\displaystile{e^{x/2}=e^{\frac{1}{2}x}}[/latex]. Here a=1/2. So dividing by 1/2 is the same as multiplying by 2.

So [latex]\displaystyle{\int e^{x/2}\, dx=\int e^{\frac{1}{2}x}\, dx=2e^{\frac{1}{2}x}+c.}[/latex]

What we want is just three times that. So

[latex]\displaystyle{3\int e^{x/2}\, dx=3\int e^{\frac{1}{2}x}\, dx=3\times2e^{\frac{1}{2}x}+c=6e^{\frac{1}{2}x}+c.}[/latex]

Hedgecutter

dlouth15 wrote: »

If you integrate e^(ax) you get

[latex]\displaystyle{\int e^{ax}\, dx=\frac{1}{a}e^{ax}+c}[/latex]

This is just a standard integral that can be looked up. The result is just e^(ax) divided by a.

Now [latex]\displaystile{e^{x/2}=e^{\frac{1}{2}x}}[/latex]. Here a=1/2. So dividing by 1/2 is the same as multiplying by 2.

So [latex]\displaystyle{\int e^{x/2}\, dx=\int e^{\frac{1}{2}x}\, dx=2e^{\frac{1}{2}x}+c.}[/latex]

What we want is just three times that. So

[latex]\displaystyle{3\int e^{x/2}\, dx=3\int e^{\frac{1}{2}x}\, dx=3\times2e^{\frac{1}{2}x}+c=6e^{\frac{1}{2}x}+c.}[/latex]

so you don't bring down 1/2 and multiply that by 3, you only bring down the 2?

dlouth15

Hedgecutter wrote: »

so you don't bring down 1/2 and multiply that by 3, you only bring down the 2?

If you go back to the standard integral

[latex]\displaystyle{\int e^{ax}\, dx=\frac{1}{a}e^{ax}+c}[/latex]

you see that the the expression is divided by a in the result. But here a=1/2. Dividing by 1/2 is the same as multplying by 2.

Then you multiply 3 by the 2.

So you don't multiply 3 by 1/2, you divide 3 by 1/2. Which is the same as multplying 3 by 2.

Hedgecutter

dlouth15 wrote: »

If you go back to the standard integral

[latex]\displaystyle{\int e^{ax}\, dx=\frac{1}{a}e^{ax}+c}[/latex]

you see that the the expression is divided by a in the result. But here a=1/2. Dividing by 1/2 is the same as multplying by 2.

Then you multiply 3 by the 2.

So you don't multiply 3 by 1/2, you divide 3 by 1/2. Which is the same as multplying 3 by 2.

ya I was making stupid mistake, dividing 1/2 by 1/2 instead of 1 divided by 1/2.

cheers

Hedgecutter

Have I drawn my graph wrong. the black and white solution is my tutors.

dlouth15

Hedgecutter wrote: »

Have I drawn my graph wrong. the black and white solution is my tutors.

I think your tutor may have used y=9 as the straight line when drawing the graph and thereby getting the limits of integration wrong rather than y=3x. Following through, his integration is OK but his evaluation with the limits -3 to 3 is incorrect.

Second last line on his solution should be -18 not 18:

[latex]\displaystyle{\frac{27}{2}-9-\left(\frac{27}{2}+9\right)=-18}[/latex].

But -18 is wrong also because the limits of integration are wrong.

In your solution you have correctly established the limits of integration and done the integration properly. But you have made a mistake on the third last line where you evaluate it. Have a look at the previous line to that and see if you can spot the error.

Also I would advise sticking with fractions until the very end or leave the answer as a simple fraction if that is permitted.

Hedgecutter

dlouth15 wrote: »

I think your tutor may have used y=9 as the straight line when drawing the graph and thereby getting the limits of integration wrong rather than y=3x. Following through, his integration is OK but his evaluation with the limits -3 to 3 is incorrect.

Second last line on his solution should be -18 not 18:

[latex]\displaystyle{\frac{27}{2}-9-\left(\frac{27}{2}+9\right)=-18}[/latex].

But -18 is wrong also because the limits of integration are wrong.

In your solution you have correctly established the limits of integration and done the integration properly. But you have made a mistake on the third last line where you evaluate it. Have a look at the previous line to that and see if you can spot the error.

Also I would advise sticking with fractions until the very end or leave the answer as a simple fraction if that is permitted.

Answer should be 4 1/2 ?

dlouth15

Hedgecutter wrote: »

Answer should be 4 1/2 ?

Yes.

Hedgecutter

Trying to fine tune my work, wondering if this is correct

TheBody

Hedgecutter wrote: »

Trying to fine tune my work, wondering if this is correct

I'm afraid you made a mistake in the second last line.

What you need is:

[latex]\int 4x \cos(2x)dx=2x\sin(2x)-2 \int \sin(2x)dx=2x \sin(2x)-2(-\frac{1}{2} \cos(2x))+c=2x \sin(2x)+ \cos(2x)+c [/latex]

Hedgecutter

TheBody wrote: »

I'm afraid you made a mistake in the second last line.

What you need is:

[latex]\int 4x \cos(2x)dx=2x\sin(2x)-2 \int \sin(2x)dx=2x \sin(2x)-2(-\frac{1}{2} \cos(2x))+c=2x \sin(2x)+ \cos(2x)+c [/latex]

Thanks, I'm doing tonnes of practice to see if I can tighten up on areas like that. Understanding a lot more.

TheBody

I presume you have exams coming up soon?

Hedgecutter

TheBody wrote: »

I presume you have exams coming up soon?

June :eek:

TheBody

Hedgecutter wrote: »

June :eek:

You are doing lots of work. It will pay off in the end.

Hedgecutter

Can't remember what happened in the circled area. How does 2e become 4e?

TheBody

It's coming because all that stuff in the bracket is to be squared. i.e. 2^2=4.

TheBody

On a seperate issue, it would be usual to not keep repeating, for example in this case, "rms" all the way down the parge. You just need to write it in at the top and use the "=" from then on.

Hedgecutter

TheBody wrote: »

It's coming because all that stuff in the bracket is to be squared. i.e. 2^2=4.

ya but what happens to the x/2?

TheBody

Hedgecutter wrote: »

ya but what happens to the x/2?

Oh sorry, I though you were just wondering about the 2.

The rules of indices are doing their thing here. The rule you need is

[latex](a^x)^y=a^{xy} [/latex]

In other words, we need to multiply the two powers.

So the [latex]\frac{x}{2}[/latex] gets multiplied by the outer 2 to give x.

Hedgecutter

Think my part looks ok but no idea how to do part 2

Hedgecutter

Be handy if I put up an image

Think part 1 is ok but can't figure out part 2.

TheBody

In the first part, you should have [latex]\frac{d\theta}{dt} [/latex] instead of [latex]\frac{dx}{dt} [/latex].

With regards to the second part, in the first part you found a formula for the angular acceleration to be: [latex]\frac{d^2\theta}{dt^2}=20-12t [/latex].

So just let this equal 0 and solve for t. In other words solve:

[latex]20-12t=0 [/latex] for t.

Hedgecutter

TheBody wrote: »

In the first part, you should have [latex]\frac{d\theta}{dt} [/latex] instead of [latex]\frac{dx}{dt} [/latex].

With regards to the second part, in the first part you found a formula for the angular acceleration to be: [latex]\frac{d^2\theta}{dt^2}=20-12t [/latex].

So just let this equal 0 and solve for t. In other words solve:

[latex]20-12t=0 [/latex] for t.

Handy question if it comes up. Cheers.

TheBody

Hi Hedgecutter.

Just wondering if you passed your exams in the end?

Hedgecutter

TheBody wrote: »

Hi Hedgecutter.

Just wondering if you passed your exams in the end?

Still waiting for results. Should get them this week. Found exam very tough. If I don't pass I'll give it another go.

Scuid Mhór

Hedgecutter I think you may owe TheBody a pint or two when you find out you've passed...

TheBody

Hedgecutter wrote: »

Still waiting for results. Should get them this week. Found exam very tough. If I don't pass I'll give it another go.

Hopefully you passed. Let us know how you got on when you get the results.

TheBody

Scuid Mhór wrote: »

Hedgecutter I think you may owe TheBody a pint or two when you find out you've passed...

I was planning to send him an invoice when he passes!!!

Hedgecutter

A sun holiday I'd say ;D

TheBody

So did you pass?

Hedgecutter

Still no results. Ridiculous at this stage, nearly a month late.

TheBody

Hedgecutter wrote: »

Still no results. Ridiculous at this stage, nearly a month late.

That's very odd. Sure college is starting back soon!! Are you sure it's not just yours that are missing?

Hedgecutter

TheBody wrote: »

That's very odd. Sure college is starting back soon!! Are you sure it's not just yours that are missing?

Everyone in the same boat. Email afew of the lads in my class, they have received nothing. Guys at work who are a year behind have received some but not all. I email the exam office and they said that the delay is at city and guilds side.

Part time course I'm due to start begins in a couple of weeks. I have no idea if i qualify yet. Crazy.

Hedgecutter

Good news passed my exams. Starting day course next week. Maths module Friday morning, so I'm sure it won't be long before I'm back on here looking for help

TheBody

Ah that's great news!! Well done. Congrats!!

Hedgecutter

Stuck with these two turning points questions. Can't even start.

(I) find dy/dx then find X, once you have X you can find Y, then your max and min points.

I can't get of the mark to find X

Hedgecutter

Got number 6 sorted but still stuck on 8



I know this is basic algebra but cant remember how to find X, Is my work correct so far?

Finding turning points

Y=5X-2lnX

dy/dx=5-2/x

dy/dx=5-2x^-1
dydx=0

5-2x^-1=0
-2x^-1=-5

TheBody

Hedgecutter wrote: »

Stuck with these two turning points questions. Can't even start.

(I) find dy/dx then find X, once you have X you can find Y, then your max and min points.

I can't get of the mark to find X

Can you elaborate a little on what yor problem is? With regards the first one:

Rather than using the product rule to differentiate, I'd begin by expanding the brackets.

[latex]x=\theta(6-\theta)=6\theta-\theta^2 [/latex].

You can just differentiate this as normal then. In other words,

[latex]\frac{dx}{d\theta}=6-2\theta [/latex].

Can you finish it from there?

With regrads the second:

[latex]y=5x-2\ln(x)[/latex]. Differentiating we get:

[latex]\frac{dy}{dx}=5-\frac{2}{x}[/latex]. Can you finish it from here?

TheBody

I see you replied while I was writing the previous post!! We have:

[latex]5-\frac{2}{x}=0[/latex]
[latex]\implies\frac{2}{x}=5[/latex]
[latex]\implies 5x=2[/latex]
[latex]x=\frac{2}{5}[/latex].

Hedgecutter

Do I bring X above the line like I did in post 198?

Find the square root of 5/2 using second function on calculator putting a -1 before root symbol?

I got 0.4