# How do you find the arc length of the curve y = 2x - 3, -2 ≤ x ≤ 1?

Jun 26, 2015

The arc length is $3 \sqrt{5} \setminus \approx 6.7082$

#### Explanation:

Since the graph of $y = f \left(x\right) = 2 x - 3$ is a straight line, there's actually no need to use calculus. Instead, just find the straight-line distance between the points $\left(- 2 , f \left(- 2\right)\right) = \left(- 2 , - 7\right)$ and $\left(1 , f \left(1\right)\right) = \left(1 , - 1\right)$. The answer, by the distance formula (Pythagorean theorem), is

$\sqrt{{\left(- 2 - 1\right)}^{2} + {\left(- 7 - \left(- 1\right)\right)}^{2}} = \sqrt{9 + 36} = \sqrt{45}$

$= \sqrt{9} \sqrt{5} = 3 \sqrt{5} \setminus \approx 6.7082$

If you want to confirm this with calculus, evaluate the integral ${\int}_{- 2}^{1} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \setminus \mathrm{dx} = {\int}_{- 2}^{1} \sqrt{1 + 4} \setminus \mathrm{dx}$

$= \sqrt{5} x {|}_{- 2}^{1} = \sqrt{5} \left(1 - \left(- 2\right)\right) = 3 \sqrt{5}$