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.
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 ...?
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?
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?
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)
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.
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.