Exact arc length of ln x

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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?
 

Answers and Replies

  • #2
SammyS
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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.
 
  • #3
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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?
 
  • #4
SammyS
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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.
 

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