Core 4 Integration Help??Watch

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#1
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??
0
2 years ago
#2
Partial fractions?
0
2 years ago
#3
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.
0
#4
(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
0
2 years ago
#5
(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
0
2 years ago
#6
(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 0
#7
(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 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!
0
2 years ago
#8
(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?
0
2 years ago
#9
it is not difficult to change the variable to u = 2x - 1....
0
2 years ago
#10
(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.
0
#11
(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
0
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