Answer
$$L = \int_{1}^{4} \sqrt{(16t^6 + 4t^2 -4t + 1 )} \space dt \approx 255.3756$$
Work Step by Step
1. Write the formula for the length "$L$"of a curve:
$$L = \int_{\alpha}^{\beta} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(t^2 - t)}{dt} = 2t -1$
$\frac{dy}{dt} = \frac{d(t^4)}{dt} = 4t^3$
3. Substitute the values into the definite integral:
$L = \int_{\alpha}^{\beta} \sqrt{(2t -1)^2 + (4t^3)^2} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(4t^2 -4t + 1) + (16t^6)} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(16t^6 + 4t^2 -4t + 1 )} \space dt$
- Since $t$ goes from 1 to 4:
$L = \int_{1}^{4} \sqrt{(16t^6 + 4t^2 -4t + 1 )} \space dt$
4. Write this integral on your calculator, and get the result.
$L \approx 255.3756$
