# Thread: Area under graph question. Integral of ln.

1. ## Area under graph question. Integral of ln.

Can't seem to solve this question:

Sketch the graph of ln(x-3), then, without evaluating the integral, prove that

3/2ln10 < Integral of lnlx-3l, upper limit = 8, lower limit = 5 < ln125

Sorry I can't type the inequalities out in proper form ><, and by lx-3l I mean the modulus/absolute value.

This question falls under the topic of 'area under curve by integration', so I know it has something to do with that, but I just can't figure it out.

This is the graph with y = ln(x-3), y = 3/2ln10 and y = ln124

2. ## Re: Area under graph question. Integral of ln.

Hint:

Area of trapezoid: $A_T=\frac{1}{2}\left(B+b \right)h=\frac{3}{2}\left(\ln(2)+\ln(5) \right)=\frac{3}{2}\ln(10)$

Area of rectangle: $A_R=bh=3\ln(5)=\ln(125)$

3. ## Re: Area under graph question. Integral of ln.

Originally Posted by MarkFL2
Hint:

Area of trapezoid: $A_T=\frac{1}{2}\left(B+b \right)h=\frac{3}{2}\left(\ln(2)+\ln(5) \right)=\frac{3}{2}\ln(10)$

Area of rectangle: $A_R=bh=3\ln(5)=\ln(125)$
Woah, how did you see that so quickly

Ahh got it after drawing the whole graph out on paper. Thanks a lot

4. ## Re: Area under graph question. Integral of ln.

Glad to help. Rectangles and trapezoids are commonly used to approximate definite integrals, so those were the shapes I considered first.