C3 integration
Maths and statistics discussion, revision, exam and homework help.
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C3 integration2 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. -
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Re: C3 integrationIf 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.. -
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Re: C3 integrationMy mistake, it's actually (x-1), not (3x-1). I've fixed that now.(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.. -
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Re: C3 integrationAre you sure you've posted this integral correctly?(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?
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). -
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Re: C3 integrationI'm given two curves:(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).
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. 
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Re: C3 integrationYour 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.(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.
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. -
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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..
Thanks guys!(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.
