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# C2 intergration, need answer watch

1. I know how to do it but I keep getting the wrong answer. Could someone please post what you get . y=2/x^2 +x. The points A and B have coordinates (0.5) and (2) , these are x-coordinates. Find the area between line AB and the curve. The answer is 27/8 and I got 10.25 ...
2. Show us your working and we can then see where you've made your mistake.
3. (Original post by SophieL1996)
I know how to do it but I keep getting the wrong answer. Could someone please post what you get . y=2/x^2 +x. The points A and B have coordinates (0.5) and (2) , these are x-coordinates. Find the area between line AB and the curve. The answer is 27/8 and I got 10.25 ...
What are the y coordinates of A and B? It is impossible to answer without them!
4. Ok so I found the area by: area of trapezium - integration of curve between limits of 2 and 0.5 area of trapezium = 0.5 x 1.5 (height) x (a+b) . a= 8.5 and b= 2.5. I found this by using the y= equation and sub in x coordinates. this gave an area of 8.25 I intergrated the curve equation to -2x^-1+2x which is -2/x + 2x I subbed in 2 and 0.5 which gave 3-5= -2 8.25- (-2)= 10.25
5. (Original post by Mr M)
What are the y coordinates of A and B? It is impossible to answer without them!
sub x values into equation.... so I got 8.5 and 2.5
6. (Original post by SophieL1996)
sub x values into equation.... so I got 8.5 and 2.5
What makes you think they are points of intersection?
7. (Original post by SophieL1996)
Ok so I found the area by: area of trapezium - integration of curve between limits of 2 and 0.5 area of trapezium = 0.5 x 1.5 (height) x (a+b) . a= 8.5 and b= 2.5. I found this by using the y= equation and sub in x coordinates. this gave an area of 8.25 I intergrated the curve equation to -2x^-1+2x which is -2/x + 2x I subbed in 2 and 0.5 which gave 3-5= -2 8.25- (-2)= 10.25
(Original post by SophieL1996)
y=2/x^2 +x.
The integral of x is not 2x . . .
8. it shows you that on the sketch.... and how else are you going to work it out? As they give you little information.
9. (Original post by joostan)
The integral of x is not 2x . . .
Ah yes thank you I seemed to have written it down wrong when working out as I wrote +2 which intergrated to 2x
10. Ive just tried it and got close (not got correct answer tho)
=2x^-2+x
=S -(2x^-1)/-1 + (x^2)/2
when X=2 S=1
and when X=0.5 S= -27/8
Where to go from there if it is right
11. The answer of 27/8 is correct.
12. (Original post by SophieL1996)
sub x values into equation.... so I got 8.5 and 2.5
That is only true if the question states that A and B lie on the curve
13. (Original post by Mr M)
The answer of 27/8 is correct.
No I intergrated what should be x to 2x as I wrote it wrong....
14. (Original post by kvohra)
That is only true if the question states that A and B lie on the curve
Yes, the sketch shows that...
15. (Original post by SophieL1996)
it shows you that on the sketch.... and how else are you going to work it out? As they give you little information.
When posting questions you need to give us all the information. It is frustrating to be expected to guess.
16. Boom I have it for ya!
=2x^-2+x
=S -(2x^-1)/-1 + (x^2)/2
when X=2 S=1
and when X=0.5 S= -27/8
1-(-)27/8= 35/8
Area of trapezium= 8.25
8.25-35/8= 27/8
Ill take a fiver for that
17. (Original post by Mr M)
When posting questions you need to give us all the information. It is frustrating to be expected to guess.
I thought it was obvious when I said find the area between AB and the curve, sorry I should have made it clearer.
18. (Original post by calm down)
Boom I have it for ya!
=2x^-2+x
=S -(2x^-1)/-1 + (x^2)/2
when X=2 S=1
and when X=0.5 S= -27/8
1-(-)27/8= 35/8
Area of trapezium= 8.25
8.25-35/8= 27/8
Ill take a fiver for that
Haha thanks and yes you deserve it!
19. (Original post by calm down)
Boom I have it for ya!
=2x^-2+x
=S -(2x^-1)/-1 + (x^2)/2
when X=2 S=1
and when X=0.5 S= -27/8
1-(-)27/8= 35/8
Area of trapezium= 8.25
8.25-35/8= 27/8
Ill take a fiver for that
8.25 - 35/8 =/= 27/8 . . .
20. For future reference you can do it all in one step:

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