Hedgecutter wrote: » I'm revising at the moment and I have hit a wall with one question. I have to find the centroid of the area of a curve Y=3x^2 the x axis and ordinates x=0,y=2 looking over my notes and I dont understand how we found y bar. I don't understand how we went from 3x^2 to 9x^4, if we integrate it should be 3x^3/3? I have attached my notes for y bar any help would be appreciated
TheBody wrote: » Hi there. On the top of the second line, you just forgot the square when y got repalced. It should read [latex]\frac{1}{2}\int_0^2(3x^2)^2dx [/latex]
Hedgecutter wrote: » Cheers. so its 3*3=9 and add the multiply the powers 2*2=4 to give 9^4 [latex]9x^4[/latex]
TheBody wrote: » FYP a bit!!
Hedgecutter wrote: » thanks on another note how do you use superscript on boards?
TheBody wrote: » I'm afraid you have made a number of mistakes. To begin, this time you are dealing with a function of t. Therefore, we have [latex]\overline{x}=\frac{\int_0^4(t)(4t-t^2)dt}{\int_0^44t-t^2dt}[/latex]. Notice that I don't have a mixture of x and t. In general, you NEVER want a mixture of variables in an integral. There are a number of errors in the [latex]\overline{y}[/latex] bit too. To get you going, you should have: [latex]\overline{y}=\frac{\frac{1}{2}\int_0^4(4t-t^2)^2dt}{\int_0^44t-t^2dt}=\frac{\frac{1}{2}\int_0^4t^4-8t^3+16t^2dt}{\int_0^44t-t^2dt}[/latex]. See if you can finish it off from there.
Hedgecutter wrote: » Ya I have a bad habit of mixing those up. Could you write out long hand how you worked this out please. [latex]\overline{y}=\frac{\frac{1}{2}\int_0^4(4t-t^2)^2dt}{\int_0^44t-t^2dt}=\frac{\frac{1}{2}\int_0^4t^4-8t^3+16t^2dt}{\int_0^44t-t^2dt}[/latex].
TheBody wrote: » What you missed was that [latex](4t-t^2)^2=(4t-t^2)(4t-t^2)=t^4-8t^3+16t^2[/latex]. Notice that: [latex](4t-t^2)^2\not=16t^2-t^4[/latex]
Hedgecutter wrote: » Basic stuff really. Once I get a few more under my belt I should be fine. thanks again, ill post the finished solution tomorrow.
TheBody wrote: » It'll all come to you with time. Keep your head up and you will do fine. What course are you doing? (if you don't mind me asking?)
Hedgecutter wrote: » Question when I worked out this I got 16t^2-8t^3+t^4 does this make a difference?
TheBody wrote: » You got the same thing as me, only mine is in a different order. Well done!
Hedgecutter wrote: » Do I need to do something else with it before I integrate?
TheBody wrote: » Sorry for the delay in getting back to you. You made a small mistake at the very beginning. You have an [latex]-18t^3[/latex] term when is should be an [latex]-8t^3[/latex] term. You had it correct earlier; I guess you just wrote it down wrong. Other than that I think it looks ok. Just note the once yu do the integrating bit you should drop the integral symbol and the dt. i.e from the second line onwards. Almost there!! P.S. I got [latex]\overline{y}=\frac{8}{5}[/latex] as the final answer.
Hedgecutter wrote: » Ya that's what I get for trying to do it with kids around. I'll have a go again in the morning. Part "b" looks interesting indefinite integral. Factorising could catch me again.
Hedgecutter wrote: » Ya 8/5.
TheBody wrote: » Well done! Have you exams in January?
Hedgecutter wrote: » (B) next one I have. We usually factorise what's under the line but not sure how to factorise 1+x^2
TheBody wrote: » You can't factorise that. You don't need to anyway. This is just regular integration using substitution. Let u=1+x^2. Give it a go and see how you get on.
Hedgecutter wrote: » Lovely I was making it hard for myself.
TheBody wrote: » The idea here is to turn an integration of [latex]x[/latex] into an integration depending totally on [latex]u[/latex]. As I said before, you never want to have a mixture of variables. To get you started, we want: [latex]\int\frac{3x}{1+x^2}dx=3\int\frac{x}{1+x^2}dx[/latex]. Remember we are trying to exchange all the x stuff. We already know that [latex]u=1+x^2[/latex], so we need to replace the [latex]xdx[/latex] bit. As you did, [latex]\frac{du}{dx}=2x[/latex], therefore [latex]xdx}=\frac{du}{2}[/latex] We can now replace all the x stuff to get: [latex]3\int\frac{1}{u}\frac{du}{2}=\frac{3}{2}\int\frac{1}{u}du[/latex]. Can you finish it from there?
