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Integration

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Ahh ok

So when Ive used this method in the past its only been when the line interests the curve and doesnt involve the x axis - which you mentioned earlier.

Thats why Im getting them both a little muddled up.

Thanks to you all - that visual really helped me understand Zacken, as well.

Honestly, the best way to know when to apply which method is by just sketching lines and shading areas under the curve. If you integrate from here to here, mark the area on the diagram with pencil and see if it helps. If you draw a line from here to here and split the area into a triangle, how can you find the other remaining area? Does integration get you that? Experiment and find out by sketching and shading.

Reply 21

LewisClothier

Original post by Zacken

The area that you have found is the blue area plus the green area (they will have opposite signs and cancel each other out a little).

As you can see, the area that you have found (the area between the curve and the line) is nothing close to the area that you need to find.

Thanks Zacken!! :biggrin:

#wouldrepifIcould

Reply 22

TSR2016

Original post by christinajane

Just doing a past paper from June 2006 C2

https://c4a3f001dcd45afe69d0ceec83003f9fbc283b2c.googledrive.com/host/0B1ZiqBksUHNYSGtmV3dzVVVvNTg/June%202006%20QP%20-%20C2%20Edexcel.pdf

Question 10

I got the equation of the line AN y=-12x+40

So I done took the equation of the line away from the equation of the curve to get

x^3 - 8x^2 + 32x - 40

I then integrated this to get

1/4x^4 - 8/3x^3 +16x^2 - 40x

with the limits 10/3 and 0

BUT I cant seem to get the right answer using this method :-(

Can someone just show me where I have gone wrong coz I think Ive gone blind to any mistakes now....

I know that there are two ways of doing it - finding the area of the triangle which ive done etc. But just outta curiosity I tried this way and no matter what I DO I cant get the same answer....