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Hard Integration Question!

This is from an AS textbook but its pretty hard...
I just need help integrating the equations.
Screenshot 2016-09-02 at 14.12.59.png
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Avatar for B_9710
B_9710
18
To integrate each expression just expand all the brackets out.
How would you integrate the circle?


Original post by B_9710
To integrate each expression just expand all the brackets out.
Original post by TheAlphaParticle
How would you integrate the circle?


You don't need to integrate it. You can find the radius and the area is r2πr^2\pi
Original post by TheAlphaParticle
This is from an AS textbook but its pretty hard...
I just need help integrating the equations.
Screenshot 2016-09-02 at 14.12.59.png


area of a circle is πr2 \displaystyle \pi r^2 , you dont need to integrate for it

Spoiler

Original post by DylanJ42
area of a circle is πr2 \displaystyle \pi r^2 , you dont need to integrate for it

Spoiler




Original post by RDKGames
You don't need to integrate it. You can find the radius and the area is r2πr^2\pi


Original post by B_9710
To integrate each expression just expand all the brackets out.


I need help integrating the black curve, I don't know the limits of the curve so how am i meant to integrate it.
Original post by TheAlphaParticle
I need help integrating the black curve, I don't know the limits of the curve so how am i meant to integrate it.


Just count the squares. Each one is 1 in a given direction of either up/down or left/right. So the limits for the black curve are 9 and 14. But that will give area of everything below it. I'm sure you know how to get the actual black area from there.
(edited 7 years ago)
Original post by TheAlphaParticle
I need help integrating the black curve, I don't know the limits of the curve so how am i meant to integrate it.


just count the squares on the x axis until you reach the point directly under the furthest left point of the black shape, thats your lower limit

do the same but for the furthest right point of the black shape, this is your upper limit
Original post by RDKGames
Just count the squares. Each one is 1. So the limits for the black curve are 9 and 14. But that will give area of everything below it. I'm sure you know how to get the actual black area from there.


I've tried that and it gives 2.2222m^2 which means it needs one can of paint but the answer is 3 cans.
Original post by TheAlphaParticle
I've tried that and it gives 2.2222m^2 which means it needs one can of paint but the answer is 3 cans.


The area should be (8x5) - integral between 9 and 14.
(edited 7 years ago)
Original post by RDKGames
The area should be (8x5) - integral between 9 and 14.


IMG_20160902_150514.jpg Where did i go wrong?
Original post by TheAlphaParticle
Where did i go wrong?


(x−9)(x−15)3+8≠x2−24x+143\frac{(x-9)(x-15)}{3}+8 \not= x^2-24x+143

Not sure why you completely ignored the fact that the quadratic is being divided by 3.
Original post by RDKGames
(x−9)(x−15)3+8≠x2−24x+143\frac{(x-9)(x-15)}{3}+8 \not= x^2-24x+143

Not sure why you completely ignored the fact that the quadratic is being divided by 3.


It's being multiplied by 1/3 later on it makes it easier to integrate
Original post by TheAlphaParticle
It's being multiplied by 1/3 later on it makes it easier to integrate


You did it wrong then. If you want to be taking that route then you need to get the 8 under the same denominator and add them before factoring out the third altogether and place it outside the integral.
Original post by RDKGames
You did it wrong then. If you want to be taking that route then you need to get the 8 under the same denominator and add them before factoring out the third altogether and place it outside the integral.


Ahh thanks so much, I understand now, what is another route (out of curiosity?)
Original post by TheAlphaParticle
Ahh thanks so much, I understand now, what is another route (out of curiosity?)


Not a very different one. You just divide each term of the quadratic by 3 then integrate each one between the limits. It's neater the way you want to do it.
Original post by RDKGames
Not a very different one. You just divide each term of the quadratic by 3 then integrate each one between the limits. It's neater the way you want to do it.


It still doesn't work it gives 28.888 now which is still wrong
Original post by TheAlphaParticle
It still doesn't work it gives 28.888 now which is still wrong


Well of course it doesn't work when you consider it that way because when you integrate you find the area below the curve...
Original post by RDKGames
Well of course it doesn't work when you consider it that way because when you integrate you find the area below the curve...


Oh!! the area of Black and Red are equal, so are Green and Blue.

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