C3 Question

I did CCEA Mathematics C3 today and we got a question that involved integrating ln(x).
Integration of ln(x) is not on our specification, but I had a go at it anyway.

If anyone is able to do this question, please do. So I can see how close I was.

Anyway the question went a bit like.

(i) Sketch a graph of y=ln(2x+1)               (2)

(ii) Find the EXACT area bounded by graph of y=ln(2x+1) and the line y=1.              (8)

I got the first part right I think. Did I? ---> clicky

But I would be surprised if I got the 2nd part correct.
Can't remember exactly how I did it now, just remember that i got 0.3something.

Reply 1

dvs

The graph of y=ln(2x+1) actually lies above the graph of y=ln(x), and they do not intersect.

I don't think you were supposed to integrate ln(2x+1) for the second part, because the question is asking you for the area bounded between its graph and the line y=1, i.e. if y = ln(2x+1), then x = (e^y - 1)/2, which is what you want to integrate.

Reply 2

blunty

dvs

The graph of y=ln(2x+1) actually lies above the graph of y=ln(x), and they do not intersect.

Ah crap, one of my many stupid mistakes. Need to cut down on those.

I don't think you were supposed to integrate ln(2x+1) for the second part, because the question is asking you for the area bounded between its graph and the line y=1, i.e. if y = ln(2x+1), then x = (e^y - 1)/2, which is what you want to integrate.

hmmmmm.

Reply 3

Aired

You can integrate lnx by writing it as 1.lnx and integrating by parts

∫1.lnx dx, let u = lnx and dv/dx = 1.

du/dx = 1/x and v = x.

= xlnx - ∫1 dx

= xlnx - x + c

Reply 4

dvs

blunty

hmmmmm.

I believe they were looking for the area bounded between y=ln(2x+1), the y-axis, and the line y=1. So you'd have to integrate something of the form x=f(y) with respect to y.

Reply 5

Aired

₀¹∫ 1.ln(2x+1) dx

Let u = ln(2x+1) and dv/dx = 1.

u = ln(2x+1), let a = 2x+1, u = lna.

da/dx = 2 and du/da = 1/a.

.`. du/dx = 2/2x+1 and v = x.

₀¹∫ 1.ln(2x+1) dx = [xln(2x+1)]₀¹ - ∫(2x)/(2x+1) dx

I got to there with part ii, no idea if it's right :s-smilie:

Reply 6

blunty

dvs

I believe they were looking for the area bounded between y=ln(2x+1), the y-axis, and the line y=1. So you'd have to integrate something of the form x=f(y) with respect to y.