Question

Consider the arc length of the curve
y = ln x on 1<= x <= 2.

a) First, set up, but do not evaluate, an integral that represents the exact arc length.
b) Next, apply the u-substitution, u = 1/x, to your integral in part (a). Leave your integral in
terms of u only without evaluating the integral.
c) Now, apply the trigonometric substitution
u = tan(theta) to your integral in part (b), and find the arc length.

Answer 1

arc length = \(\large \mathbb{\int_1^2 \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx}\)

Answer 2

find the derivative and plugin

Answer 3

I get \[\int\limits_{1}^{2} \frac{ 1 }{ x } \sqrt{x^2+1}dx\] for (a). Im suppose to u-sub u=1/x for (b) but I get confused here

Answer 4

just do watever it is asking to do..
u not going to evaluate it anyways

Answer 5

I get u = 1/x \[du = \frac{ -1 }{ x^2 }\] I don't know where to plug in the du

Answer 6

sub \( u = \frac{1}{x}\)

\(du = \frac{-1}{x^2} dx\)
\(-xdu = \frac{1}{x} dx\)

\(-\frac{1}{u}du = \frac{1}{x} dx\)

Answer 7

as x->1, u->1
as x->2, u->1/2

Answer 8

so the integral becomes :
\(\large \int\limits_{1}^{\frac{1}{2}} \sqrt{(\frac{1}{u})^2+1} (\frac{-du}{u}) \)

Answer 9

leave it like that...

Answer 10

Then I have to use trig sub

Answer 11

u = tan(theta)

Answer 12

ya go ahead