# What is the arclength of f(x)=1/e^(3x) on x in [1,2]?

##### 1 Answer
Apr 10, 2016

$\frac{1}{3} \cdot \ln \left(\frac{\sqrt{{e}^{12} + 9} + {e}^{6}}{\sqrt{{e}^{6} + 9} + {e}^{3}}\right) + \frac{1}{3} \left(\frac{\sqrt{{e}^{6} + 9}}{e} ^ 3 - \frac{\sqrt{{e}^{12} + 9}}{e} ^ 6\right) \cong 1.001850$

#### Explanation:

$f \left(x\right) = {e}^{- 3 x}$
$f \text{'} \left(x\right) = - 3 {e}^{- 3 x}$

$L = {\int}_{a}^{b} \sqrt{1 + {\left[f \text{'} \left(x\right)\right]}^{2}} . \mathrm{dx}$

$F \left(x\right) = \int \sqrt{1 + 9 {e}^{- 6 x}} . \mathrm{dx}$
$F \left(x\right) = \int \sqrt{{e}^{6 x} + 9} \cdot {e}^{- 3 x} \cdot \mathrm{dx}$
Making
${e}^{3 x} = 3 \tan y$
$\cancel{3} \cdot {e}^{3 x} \cdot \mathrm{dx} = \cancel{3} {\sec}^{2} y \cdot \mathrm{dy}$
${\left({e}^{- 3 x}\right)}^{2} \cdot {e}^{3 x} \cdot \mathrm{dx} = \frac{{\sec}^{2} y \cdot \mathrm{dy}}{3 \tan y} ^ 2$
${e}^{- 3 x} \cdot \mathrm{dx} = \frac{{\sec}^{2} y \cdot \mathrm{dy}}{9 {\tan}^{2} y}$

So
$F \left(y\right) = \int 3 \sec y \cdot \frac{{\sec}^{2} y \cdot \mathrm{dy}}{9 {\tan}^{2} y} = \frac{1}{3} \int \left(\frac{1}{\cos} ^ 3 y\right) \cdot \left({\cos}^{2} \frac{y}{\sin} ^ 2 y\right) \cdot \mathrm{dy}$
$F \left(y\right) = \frac{1}{3} \int \frac{\mathrm{dy}}{\cos y . {\sin}^{2} y}$

But
$\frac{1}{\cos y . {\sin}^{2} y} = \frac{1}{\cos} y + \cos \frac{y}{\sin} ^ 2 y$

Then
$F \left(y\right) = \frac{1}{3} \int \sec y \cdot \mathrm{dy} + \frac{1}{3} \int \cos \frac{y}{\sin} ^ 2 y \cdot \mathrm{dy}$

Solving the last term of the expression above
$\frac{1}{3} \int \cos \frac{y}{\sin} ^ 2 y \cdot \mathrm{dy}$
Making
$\sin y = u$ => $\cos y \cdot \mathrm{dy} = \mathrm{du}$
Resulting in
$= \frac{1}{3} \int \frac{\mathrm{du}}{u} ^ 2 = - \frac{1}{3} \cdot \frac{1}{u} = - \frac{1}{3} \cdot \frac{1}{\sin} y$

Back to the main expression
$F \left(y\right) = \frac{1}{3} \cdot \ln | \sec y + \tan y | - \frac{1}{3} \cdot \frac{1}{\sin} y$
But
$\tan y = {e}^{3 x} / 3$ => $\sin y = {e}^{3 x} / 3 \cdot \cos y$ => $\left({e}^{6 x} / 9 + 1\right) {\cos}^{2} y = 1$ => $\cos y = \frac{3}{\sqrt{{e}^{6 x} + 9}}$
$\to \sin y = {e}^{3 x} / \sqrt{{e}^{6 x} + 9}$

Therefore
$F \left(x\right) = \frac{1}{3} \cdot \ln | \frac{\sqrt{{e}^{6 x} + 9} + {e}^{3 x}}{3} | - \frac{1}{3} \cdot \frac{\sqrt{{e}^{6 x} + 9}}{e} ^ \left(3 x\right) + c o n s t .$

Finally
$L = F \left(x = 2\right) - F \left(x = 1\right)$
$L = \frac{1}{3} \cdot \ln \left(\frac{\sqrt{{e}^{12} + 9} + {e}^{6}}{3}\right) - \frac{1}{3} \cdot \frac{\sqrt{{e}^{12} + 9}}{e} ^ 6 - \left[\frac{1}{3} \cdot \ln \left(\frac{\sqrt{{e}^{6} + 9} + {e}^{3}}{3}\right) - \frac{1}{3} \cdot \frac{\sqrt{{e}^{6} + 9}}{e} ^ 3\right]$
$L = \frac{1}{3} \cdot \ln \left(\frac{\sqrt{{e}^{12} + 9} + {e}^{6}}{\sqrt{{e}^{6} + 9} + {e}^{3}}\right) + \frac{1}{3} \left(\frac{\sqrt{{e}^{6} + 9}}{e} ^ 3 - \frac{\sqrt{{e}^{12} + 9}}{e} ^ 6\right)$