How Do You Calculate the Arc Length of y = ln(x) from x = 1 to x = e?

[Dylan6866]

∫

Homework Statement

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)

Homework Equations

∫sqrt(1+(f'(x))^2)

distance formula I'm guessing?

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?

 

[SammyS]

∫

Homework Statement

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)

Homework Equations

∫sqrt(1+(f'(x))^2) dx

distance formula I'm guessing?

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?

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

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

 

[Dylan6866]

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

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

Yes, so the distance formula would be right?

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

But, is the integration correct?

 

[SammyS]

Yes, so the distance formula would be right?

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

But, is the integration correct?

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