# What is the arc length of f(x)=2-3x on x in [-2,1]?

Jan 26, 2016

$3 \sqrt{10}$

#### Explanation:

$f ' \left(x\right) = - 3$

$\mathrm{dL} = \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

$= \sqrt{1 + {\left(- 3\right)}^{2}} \mathrm{dx}$

$= \sqrt{10} \mathrm{dx}$

$L = {\int}_{- 2}^{1} \sqrt{10} \mathrm{dx}$

$= 3 \sqrt{10}$

Alternatively, since this is a straight line, we can use Pythagoras' Theorem.

The horizontal is $1 - \left(- 2\right) = 3$ units.
The vertical is $f \left(1\right) - f \left(- 2\right) = \left(- 1\right) - 11 = - 12$ units.

Using Pythagoras' Theorem, we have the length of the diagonal as

$\sqrt{{3}^{2} + {12}^{2}} = 3 \sqrt{10}$