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Core 4 Integration Help??

It's a parametrics question but it came to have the cartesian equation:

y=x/(2x-1)

and the question:

The finite region between the curve C and the x-axis, bounded by the lines with equations x = 2/3 and x = 1, is shown shaded in the figure above.
(c) Calculate the exact value of the area of this region, giving your answer in the form a + b ln c, where a, b and c are constants.

I know to integrate with the limits 1 and 2/3 but I just don't know how to integrate x/(2x-1)

help??
Reply 1
Avatar for Aethrell
Aethrell
2
Partial fractions?
x=1/2(2x-1)+1/2

so x/(2x-1) = [1/2(2x-1)+1/2 ] / (2x-1) = 1/2 + 1/2(2x-1)

Now integrate.

If you do not understand the technique above then use long division.
Reply 3
Avatar for geohan
geohan
OP
1
Original post by Math12345
x=1/2(2x-1)+1/2

so x/(2x-1) = [1/2(2x-1)+1/2 ] / (2x-1) = 1/2 + 1/2(2x-1)

Now integrate.

If you do not understand the technique above then use long division.


I'm still stuck.. I understand where that's came from but I just don't know where to go from there
Original post by geohan
I'm still stuck.. I understand where that's came from but I just don't know where to go from there


Give us a link to the paper
Original post by geohan
It's a parametrics question but it came to have the cartesian equation:

y=x/(2x-1)

and the question:

The finite region between the curve C and the x-axis, bounded by the lines with equations x = 2/3 and x = 1, is shown shaded in the figure above.
(c) Calculate the exact value of the area of this region, giving your answer in the form a + b ln c, where a, b and c are constants.

I know to integrate with the limits 1 and 2/3 but I just don't know how to integrate x/(2x-1)

help??


have you tried maybe doing it as parametric? Differentiate x, then multiply the differential of x by y then integrate that with respect to t.
Make sure you adapt the bounds from x to t and you should be all gravy :smile:
Reply 6
Avatar for geohan
geohan
OP
1
Original post by Math12345
Give us a link to the paper


Original post by k.russell
have you tried maybe doing it as parametric? Differentiate x, then multiply the differential of x by y then integrate that with respect to t.
Make sure you adapt the bounds from x to t and you should be all gravy :smile:


It's all good now I think, thanks anyway! I followed what you said before about making x=1/2(2x-1)+1/2 etc etc
and I took a factor of 1/4 out, so it was:

1/4 into the integral of 2 + 2/(2x-1)

once that was integrated and expanded the quarter, I got [1/2x+1/4ln(2x-1)] and ended up with 1/6+1/4ln(3) as the answer!
Original post by geohan
It's all good now I think, thanks anyway! I followed what you said before about making x=1/2(2x-1)+1/2 etc etc
and I took a factor of 1/4 out, so it was:

1/4 into the integral of 2 + 2/(2x-1)

once that was integrated and expanded the quarter, I got [1/2x+1/4ln(2x-1)] and ended up with 1/6+1/4ln(3) as the answer!


Was there a reason you were integrating with cartesian not parametric? Normally in parametric integration questions, don't they just want you to integrate parametrically?
it is not difficult to change the variable to u = 2x - 1....
Original post by the bear
it is not difficult to change the variable to u = 2x - 1....


Yeah this is what I would've suggested. The breaking up the fraction trick is also ok.
Reply 10
Avatar for geohan
geohan
OP
1
Original post by k.russell
Was there a reason you were integrating with cartesian not parametric? Normally in parametric integration questions, don't they just want you to integrate parametrically?


They gave the cartesian equation in the previous part of the question, and it seemed a lot more straight forward to follow on with that equation to integrate. I think I just had a mind blank to be honest, first time I've looked at core 4 in weeks hahaha

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