3.
Changing the parameter of the curve to x x x we get the expression
y = ln x , x ∈ [ 1 , e ] . y = \ln x,\; x \in [1,e]. y = ln x , x ∈ [ 1 , e ] .
Applying the formula of function graph arc length we get
L = ∫ 1 e 1 + ( ln ′ ( x ) ) 2 d x = ∫ 1 e 1 + 1 x 2 d x = ∫ 1 e 1 + x 2 x d x . L = \int_1^e \sqrt{1 + (\ln'(x))^2} dx = \int_1^e \sqrt{1 + \frac{1}{x^2}} dx \\=
\int_1^e \frac{\sqrt{1+x^2}}{x} dx. L = ∫ 1 e 1 + ( ln ′ ( x ) ) 2 d x = ∫ 1 e 1 + x 2 1 d x = ∫ 1 e x 1 + x 2 d x .
Answer : L = ∫ 1 e 1 + x 2 x d x . L = \int_1^e \frac{\sqrt{1+x^2}}{x} dx. L = ∫ 1 e x 1 + x 2 d x .
4.
Similarly, let us change the curve parameter to y y y
x = e y , y ∈ [ 0 , 1 ] . x = e^y,\;y \in [0,1]. x = e y , y ∈ [ 0 , 1 ] .
Then,
L = ∫ 0 1 1 + ( d d y e y ) 2 d y = ∫ 0 1 1 + e 2 y d y . L = \int_0^1 \sqrt{1 + (\frac{d}{dy}e^y)^2}dy = \int_0^1 \sqrt{1 + e^{2y}}dy. L = ∫ 0 1 1 + ( d y d e y ) 2 d y = ∫ 0 1 1 + e 2 y d y .
Answer : L = ∫ 0 1 1 + e 2 y d y . L = \int_0^1 \sqrt{1 + e^{2y}}dy. L = ∫ 0 1 1 + e 2 y d y .
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