# What is the arc length of teh curve given by f(x)=3x^6 + 4x in the interval x in [-2,184]?

Apr 15, 2017

$L = 1.16 \times {10}^{14}$

#### Explanation:

To find arc length, we use the equation:

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

Take the derivative of $f \left(x\right)$:

$f ' \left(x\right) = 18 {x}^{5} + 4$

Now plug into the formula with the interval:

${\int}_{-} {2}^{184} \sqrt{1 + {\left(18 {x}^{5} + 4\right)}^{2}} \mathrm{dx}$

Solve inside the parentheses:

${\int}_{-} {2}^{184} \sqrt{1 + 324 {x}^{10} + 144 {x}^{5} + 16} \mathrm{dx}$

Simplify inside the square root:

${\int}_{-} {2}^{184} \sqrt{324 {x}^{10} + 144 {x}^{5} + 17} \mathrm{dx}$

Solve the integral:

$L = 1.16 \times {10}^{14}$