Homework Help: Exact arc length of ln x

1. Feb 28, 2012

Dylan6866

1. The problem statement, all variables and given/known data

Use the integration tables to find the exact arc length of the curve
f(x)=ln x 1≤x≤e

Reference the table number formula used

Then approx. your answer and compare that to the approx. "straight line distance" between 2 points

coordinates of two points (1,0) (e,1)

2. Relevant equations
∫sqrt(1+(f'(x))^2)

distance formula I'm guessing?

3. The attempt at a solution
y = ln(x)
y' = (1/x)

L = ∫ (1 to e) [ sqrt(1 + (1/x)^2) ] dx
L = ∫ (1 to e) [ sqrt(1 + (1/x)^2) ] dx

= ∫(1 to e) [ sqrt(1 + (1/x^2)) ] dx
= ∫ (1 to e) [ sqrt( (x^2 + 1) / x^2) ) ] dx
=∫ (1 to e) [(1/x)*sqrt(x^2 + 1) ] dx

Integral # 28
∫ [(1/u)sqrt(a^2 + u^2)] du
= sqrt(a^2 + u^2) - a*ln | [a + sqrt(a^2 + u^2)] / u] | + C

In this integral, a = 1 and u = x

Int (1 to e) [(1/x)*sqrt(x^2 + 1) ] dx =
sqrt(1 + x^2) - ln | [1 + sqrt(1 + x^2)] / x] | (1 to e)

=
sqrt(1 + e^2) - ln | [1 + sqrt(1 + e^2)] / e] |
=
sqrt(2) - ln | [1 + sqrt(2)]] |

0.53

straight line =1.98

Is this right?

2. Feb 28, 2012

SammyS

Staff Emeritus
Isn't a straight line the shortest distance between two pints?

You only evaluated the anti-derivative at x = e, not at x = 1.

3. Feb 28, 2012

Dylan6866

Yes, so the distance formula would be right?

And I evauluated it again, the answer should be ≈ 2.

But, is the integration correct?

4. Feb 28, 2012

SammyS

Staff Emeritus
It is if the integral table is correct. You appear to have used it correctly.