Intergration calculating centroid of an area

TheBody · 31-12-2014 6:17pm

Hedgecutter wrote: »

=3/2 LN +c

=3/2 LN [1+x^2]+c

Brackets are incorrect but all I can find.

Well done. That's it all done.

Hedgecutter · 31-12-2014 6:53pm

Lovely. Thanks again for your help.

I have another turning point question think I might be close.

calculate the turning points and max and min value on the curve

Y=x^3-3x^3-9x+10.

Hedgecutter · 31-12-2014 9:24pm

using integration find the area between curves y=x2+1 and y=7-x

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x=-3 x=2

wondering if im on the right track

TheBody · 31-12-2014 10:13pm

I'm out on the sauce for New Year's Eve but I will reply to your posts tomorrow if nobody gets there before me. Happy new year!

Hedgecutter · 31-12-2014 10:18pm

TheBody wrote: »

I'm out on the sauce for New Year's Eve but I will reply to your posts tomorrow if nobody gets there before me. Happy new year!

Enjoy, cracking open a can here myself

Happy New Year

TheBody · 01-01-2015 1:44pm

Hedgecutter wrote: »

Lovely. Thanks again for your help.

I have another turning point question think I might be close.

calculate the turning points and max and min value on the curve

Y=x^3-3x^3-9x+10.

Everything looks good in that one. The ony comment I'd make is that after testing the points in the second derivative to determine which point is the local max and min, you should write down which point is which. i.e say that (-1,15) is the local max and (3,-17) is the local min.

TheBody · 01-01-2015 1:48pm

Hedgecutter wrote: »

using integration find the area between curves y=x2+1 and y=7-x

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x=-3 x=2

wondering if im on the right track

That is almost correct. You made a small error with the signs of the factors. It should be [latex](x+3)(x-2)=0[/latex]. Giving you that x=-3 or x=2.

You have found the x coordinates where the two curves meet.

When finding the area betwen two curves, it's really useful to draw a rough sketch of the curves first. You need to know which curve is "on top" and which curve is on "the bottom".

Keep going and see how you get on.

Hedgecutter · 01-01-2015 2:39pm

TheBody wrote: »

This is fine so far. You have found the x coordinates where the two curves meet.

When finding the area betwen two curves, it's really useful to draw a rough sketch of the curves first. You need to know which curve is "on top" and which curve is on "the bottom".

Keep going and see how you get on.

Should I have x=0

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
x(x-3)(x+2)
x=0,x=-3, x=2

Looking at my notes I noticed that we sketched the curved to see which was positive and which was negative.

I'm I right in saying I need y coordinates?

X=1
Y=X^2+X-6
Y=(1)^2+(1)-6
Y=1+1-6
Y=-4

(1,-4)

I'm a bit confused why we need to know this

TheBody · 01-01-2015 4:23pm

I have just changed post #38. I missed a small mistake.

X=0 is NOT a root of the quadratic equation. What you had done the first time (apart from the small error pointed out in post #38 was correct.

With regards finding the y values, the ony use they serve is in helping you draw a more accurate sketch. They won't play any part in the actual integration.

Also in the post above, you ran x=1 through the function. I don't know why you did this. Run the x values found earlier through one of the functions. It doesn't matter which one you use.

Hedgecutter · 01-01-2015 4:46pm

TheBody wrote: »

I have just changed post #38. I missed a small mistake.

X=0 is NOT a root of the quadratic equation. What you had done the first time (apart from the small error pointed out in post #38 was correct.

With regards finding the y values, the ony use they serve is in helping you draw a more accurate sketch. They won't play any part in the actual integration.

Also in the post above, you ran x=1 through the function. I don't know why you did this. Run the x values found earlier through one of the functions. It doesn't matter which one you use.

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x-3=0,x=3
x+2=0,x=-2
x=3, x=-2

On the formula I need to plug in the coordinates,

What i'm not sure of is where to get the other two coordinates.

Ill use two parts of the area formula, on that formula I have a,b,b,c where I plug in the coordinates are the other two coordinates 0?

TheBody · 01-01-2015 4:52pm

Hedgecutter wrote: »

On the formula I need to plug in the coordinates,

What i'm not sure of is where to get the other two coordinates.

Ill use two parts of the area formula, on that formula I have a,b,b,c where I plug in the coordinates are the other two coordinates 0?

I'm not sure what you mean by this.

What you need to do next is draw a sketch of curves. Then to find the area you work out:

[latex]\int_{-3}^2[/latex](upper curve)-(lower curve) dx

Hedgecutter · 01-01-2015 5:16pm

Sketched the graph and nothing below the line

Still confused to the importance of knowing this.

TheBody · 01-01-2015 5:22pm

Hedgecutter wrote: »

Sketched the graph and nothing below the line

Still confused to the importance of knowing this.

Sketch looks ok. You need to do this to find out which curve was "on top". Now you know the [latex]7-x[/latex] is on top and the [latex]x^2+1[/latex] is on the bottom (at least over the interval from -3 to 2).

At any rate now you find the area by working out:

[latex]\int_{-3}^2(7-x)-(x^2+1)dx[/latex]

Hedgecutter · 01-01-2015 5:28pm

TheBody wrote: »

Sketch looks ok. You need to do this to find out which curve was "on top". Now you know the [latex]7-x[/latex] is on top and the [latex]x^2+1[/latex] is on the bottom (at least over the interval from -3 to 2).

At any rate now you find the area by working out:

[latex]\int_{-3}^2(7-x)-(x^2+1)dx[/latex]

so because 7-x is on top it's plugged in to the formula first?

TheBody · 01-01-2015 5:29pm

Hedgecutter wrote: »

so because 7-x is on top it's plugged in to the formula first?

Yes, that's exactly it. The best way to see that is with a sketch.

Hedgecutter · 01-01-2015 5:35pm

TheBody wrote: »

Yes, that's exactly it. The best way to see that is with a sketch.

how do you know where the coordinates go a=-3 b=2 is it because -3 is first on the table working out y+7-x so -3=a?

Hedgecutter · 01-01-2015 5:51pm

Had a go at finishing

Hedgecutter · 01-01-2015 5:53pm

Had a go at finishing it

TheBody · 01-01-2015 10:33pm

Very close!! You made a small mistake near the beginning when you were removing the brackets. You should have,

[latex]\int_{-3}^2(7-x)-(x^2+1) dx=\int_{-3}^2 7-x-x^2-1 dx=\int_{-3}^2-x^2-x+6 dx[/latex].

Other than that your method was correct. Also, you should have equal signs as you progress from step to step.

TheBody · 01-01-2015 10:39pm

Hedgecutter wrote: »

how do you know where the coordinates go a=-3 b=2 is it because -3 is first on the table working out y+7-x so -3=a?

If you look at the sketch you made of the region, the most left x value is -3 and it runs up as far as x=2. Thats why I used those values.

Not sure if that makes sense??

Hedgecutter · 01-01-2015 10:39pm

TheBody wrote: »

Very close!! You made a small mistake near the beginning when you were removing the brackets. You should have,

[latex]\int_{-3}^2(7-x)-(x^2+1) dx=\int_{-3}^2 7-x-x^2-1 dx=\int_{-3}^2-x^2-x+6 dx[/latex].

Other than that your method was correct. Also, you should have equal signs as you progress from step to step.

Why did the sign change?

[latex]\int_{-3}^2(7-x)-(x^2+1) dx=\int_{-3}^2 7-x-x^2-1 dx=\int_{-3}^2-x^2-x+6 dx[/latex].

is it because -+=-

TheBody · 01-01-2015 10:41pm

Hedgecutter wrote: »

Why did the sign change?

[latex]\int_{-3}^2(7-x)-(x^2+1) dx=\int_{-3}^2 7-x-x^2-1 dx=\int_{-3}^2-x^2-x+6 dx[/latex].

is it because -+=-

Yes, the minus sign before the second bracket causes a sign change of everything inside the bracket.

Hedgecutter · 01-01-2015 10:42pm

TheBody wrote: »

Yes, the minus sign before the second bracket causes a sign change of everything inside the bracket.

Perfect.

TheBody · 01-01-2015 10:43pm

It's something you always need to be keeping an eye out for.

What year are in?

Hedgecutter · 01-01-2015 10:45pm

The equivalent of second year. If I pass this year I'll have a level 6 higher certificate.

Hedgecutter · 01-01-2015 10:46pm

What method do I use for part (1)

TheBody · 01-01-2015 10:47pm

Hedgecutter wrote: »

The equivalent of second year. If I pass this year I'll have a level 6 higher certificate.

Fair play. Keep working hard and you will do fine.

TheBody · 01-01-2015 10:52pm

Hedgecutter wrote: »

What method do I use for part (1)

When it comes to deciding if an integral is to be done using regular substitution, you ask yourself, is there anything there that if DIFFERENTIATED, would give you the other bit, and your not too bothered about constants? You have to deal with any constants but they don't cause any trouble. If there is, then you let that bit be "u".

So, With that in mind and looking at your problem, do you notice anything between the top and bottom?

Hedgecutter · 01-01-2015 11:08pm

TheBody wrote: »

When it comes to deciding if an integral is to be done using regular substitution, you ask yourself, is there anything there that if DIFFERENTIATED, would give you the other bit, and your not too bothered about constants? You have to deal with any constants but they don't cause any trouble. If there is, then you let that bit be "u".

So, With that in mind and looking at your problem, do you notice anything between the top and bottom?

do we use partial fractions method. factorise the top line?

but you can only factorise the denominator

TheBody · 01-01-2015 11:14pm

Hedgecutter wrote: »

do we use partial fractions method. factorise the top line?

but you can only factorise the denominator

No, it's not partial fractions. DIFFERENTIATE the top line and see if you notice anything......

Intergration calculating centroid of an area

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