# How do you find the length of the curve y=3x-2, 0<=x<=4?

May 15, 2018

$L = {\int}_{0}^{4} \sqrt{1 + 9} \cdot \mathrm{dx} = {\left[\sqrt{10} x\right]}_{0}^{4} = 4 \sqrt{10}$

#### Explanation:

show below

$y = 3 x - 2$

$y ' = 3$

$L = {\int}_{a}^{b} \sqrt{1 + {\left(y '\right)}^{2}} \cdot \mathrm{dx} = {\int}_{0}^{4} \sqrt{1 + 9} \cdot \mathrm{dx}$

$= {\left[\sqrt{10} x\right]}_{0}^{4} = 4 \sqrt{10}$