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A-level Maths, Integration, Finding area between curves & Lines.

The diagram below shows a sketch of part of the curve with equation y = 2x³ 15x² + ax
where a is a constant. The curve has maximum and minimum turning points when x = 2
and x = 3 respectively.
a. Show that a = 36
b. Work out the area between the curve, the y-axis and the line y = 28

I keep getting a negative result when evaluating the definite integral. (-27/32). What am I doing wrong? I found the limits to be 7/2 and 2.
Reply 1
Original post by amaniix
The diagram below shows a sketch of part of the curve with equation y = 2x³ 15x² + ax
where a is a constant. The curve has maximum and minimum turning points when x = 2
and x = 3 respectively.
a. Show that a = 36
b. Work out the area between the curve, the y-axis and the line y = 28

I keep getting a negative result when evaluating the definite integral. (-27/32). What am I doing wrong? I found the limits to be 7/2 and 2.

Youd want to integrate from the y-axis (x=0) to x=7/2? Though it would help to see the diagram.
Aso did you integrate 28-y or ...?
(edited 4 months ago)
Reply 2
Original post by amaniix
Can you see this image? It's a screenshot of the diagram.

Nope, thats a file on your file system. Youd have to upload it to another site and link the url.

If there is a shaded area and it starts at the y-axis, then the lower limit is x=0. Then the upper limit would either be (probably) where the first maximum is (x=2) or when it crosses y=28 at x=7/2.

What was the expression you actually integrated y-28 or 28-y?
Reply 3
Original post by mqb2766
Nope, thats a file on your file system. Youd have to upload it to another site and link the url.

If there is a shaded area and it starts at the y-axis, then the lower limit is x=0. Then the upper limit would either be (probably) where the first maximum is (x=2) or when it crosses y=28 at x=7/2.

What was the expression you actually integrated y-28 or 28-y?

https://ibb.co/XWFmKs0 Ok does this link work haha
I set 2x^3 - 15x^2 + 36 and y=28 equal to each other to get 2x^3 - 15x^2 + 36x - 28. Then integrated that with upper limit of 7/2 and lower limit of 2. Should I have done the lower limit as 0?
Reply 4
Original post by amaniix
https://ibb.co/XWFmKs0 Ok does this link work haha
I set 2x^3 - 15x^2 + 36 and y=28 equal to each other to get 2x^3 - 15x^2 + 36x - 28. Then integrated that with upper limit of 7/2 and lower limit of 2. Should I have done the lower limit as 0?

If y is the cubic, then 28 > y for the relevant domain, and 28-y>0 so integrating that will give a positive area.
Integrating y-28 < 0 will give a negative area.

The limits look like they should be 0 (y-axis) to 7/2 (cubic cuts the line y=28)
(edited 4 months ago)
Reply 5
Original post by mqb2766
If y is the cubic, then 28 > y for the relevant domain, and 28-y>0 so integrating that will give a positive area.
Integrating y-28 < 0 will give a negative area.

The limits look like they should be 0 (y-axis) to 7/2 (cubic cuts the line y=28)

Oh okay thank you, I'll use those limits. I'm a little confused with the domain, how do I use the integral of 28-y>0 to find area.
Reply 6
Original post by amaniix
Oh okay thank you, I'll use those limits. I'm a little confused with the domain, how do I use the integral of 28-y>0 to find area.

If y is the cubic, then the integrand is
28-y
so
28 - (2x^3 - 15x^2 + 36x)
I was simply noting its positive so > 0, Its just the usual area between two curves where you form
upper - lower
and integrate over that (positive) area.
(edited 4 months ago)

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