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# C4 Integration!!!!!!!!! Watch

1. Q1.
2. So what have you done so far on part d)?

Ok, I guess you have a diagram?

Try working out the area of the triangle (you know the base and the height) and then subtract the sector of the ellipse.
4. The limits 0 and pi/4 are given to you in the question. 0 is the x axis and pi/4 is the point where the tangent meets the curve.
5. (Original post by rainbow1)
but how do you know that one of the limit is 0?
Because the angle 0 represents the positive x axis (as y = 4 sin t, you can see an argument of t=0 will give y=0).
6. (Original post by rainbow1)
Actually to be specific I don't know how to get the limits 0 and pi/4

Pls show me the workings!!

Much appreciated!!
Is this the diagram ?

7. (Original post by rainbow1)
but cant t=pi as well?? it also gives y=0

thank you once again..
Yes but that would be the negative x axis.

Do they provide a diagram? If not can you sketch one?

Draw the line y=x on your diagram (this is where t=pi/4).

Now the tangent, the line y=x and the line x=0 form a triangle.

Work out the area of this triangle.

Now work out the area of one eighth of the ellipse (either by integration or by knowing the formula for the area of an ellipse) and subtract it.
8. (Original post by rainbow1)
diagram in post 9..,
9. (Original post by rainbow1)
got that! have not thought about the negative value!!

thanks for your great help Mr M!
Glad that has moved you on - wasn't sure you were understanding.
10. sorry to bother you all but i am stuck on 6L question 2, part c.

i have brought up the solutionbank and still don't understand why the area labelled 'T2' on the solutionbank is added on to the area R.

if anyone could help, i would much appreciate it.

thanks.
11. (Original post by redmouse)
sorry to bother you all but i am stuck on 6L question 2, part c.

i have brought up the solutionbank and still don't understand why the area labelled 'T2' on the solutionbank is added on to the area R.

if anyone could help, i would much appreciate it.

thanks.
Possibly start your own thread about it rather than hijacking another one? Also, don't assume everyone has solutionbank - they don't - so you really need to post the question.
12. okay sorry i will do. thanks.

Updated: April 7, 2012
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