You are Here: Home

# C3 integration Tweet

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. C3 integration
2 Questions:

Now I know I could multiply out brackets and integrate but I'm wondering if I've missed a more straigtforward method. The chapter I'm on is about how to differentiate and integrate functions of the form , but this integral doesn't seem to be in this form.

I get the integral as , giving me a final answer of , but according to my textbook the answer is . Can anyone tell me where I've gone wrong?
Last edited by Julii92; 31-07-2012 at 10:17.
2. Re: C3 integration
If you wrote "(3x-7)^3 (3x-1)" as "(3x-7)^4", you would have decreased "(3x-1)" to "(3x-7)" (which is -6), so your integral's value would have been changed by "-6(3x-7)". Therefore, could you not write the function as "(3x-7)^4 + 6(3x-7)"?

That's what I'd try, but I haven't done integration for ages so sorry if I'm totally wrong..
3. Re: C3 integration
(Original post by M^2012)
If you wrote "(3x-7)^3 (3x-1)" as "(3x-7)^4", you would have decreased "(3x-1)" to "(3x-7)" (which is -6), so your integral's value would have been changed by "-6(3x-7)". Therefore, could you not write the function as "(3x-7)^4 + 6(3x-7)"?

That's what I'd try, but I haven't done integration for ages so sorry if I'm totally wrong..
My mistake, it's actually (x-1), not (3x-1). I've fixed that now.
4. Re: C3 integration
You can still use M^2012's idea if you write it as .
5. Re: C3 integration
(Original post by Julii92)

I get the integral as , giving me a final answer of , but according to my textbook the answer is . Can anyone tell me where I've gone wrong?
Are you sure you've posted this integral correctly?

The definite integral that you've posted does not converge i.e. it cannot be expressed as a number (since dividing by 0 is involved).
6. Re: C3 integration
(Original post by notnek)
Are you sure you've posted this integral correctly?

The definite integral that you've posted does not converge i.e. it cannot be expressed as a number (since dividing by 0 is involved).
I'm given two curves: and The shaded region is the area enclosed by these two curves, x=3, and the point where the first curve crosses the x-axis ie.
7. Re: C3 integration
(Original post by Julii92)
I'm given two curves: and The shaded region is the area enclosed by these two curves, x=3, and the point where the first curve crosses the x-axis ie.
Your working would've been correct if the region was enclosed by only the two curves but here the region is enclosed by the two curves, y=0 and x=3.

Split the region into two by drawing a vertical line through the point where the curves meet. Then you can find the area of the two regions separately and then add them together.
Last edited by notnek; 31-07-2012 at 11:48.
8. Re: C3 integration
(Original post by M^2012)
If you wrote "(3x-7)^3 (3x-1)" as "(3x-7)^4", you would have decreased "(3x-1)" to "(3x-7)" (which is -6), so your integral's value would have been changed by "-6(3x-7)". Therefore, could you not write the function as "(3x-7)^4 + 6(3x-7)"?

That's what I'd try, but I haven't done integration for ages so sorry if I'm totally wrong..

(Original post by BabyMaths)
You can still use M^2012's idea if you write it as .

(Original post by notnek)
Your working would've been correct if the region was enclosed by only the two curves but here the region is enclosed by the two curves, y=0 and x=3.

Split the region into two by drawing a vertical line through the point where the curves meet. Then you can find the area of the two regions separately and then add them together.
Thanks guys!