Answer
\[\int_{1}^{4}\sqrt{16t^6+4t^2-4t+1}dt\;\,\approx 255.376\]
Work Step by Step
\[x=t^2-t\;,\;y=t^4\;\;\,1\leq t\leq 4\]
Differentiate $x$ with respect to $t$:
\[\frac{dx}{dt}=2t-1\]
Differentiate $y$ with respect to $t$:
\[\frac{dy}{dt}=4t^3\]
Therefore the arc length is given by:
\[L=\int_{1}^{4}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]
\[L=\int_{1}^{4}\sqrt{(2t-1)^2+\left(4t^3\right)^2}dt\]
\[L=\int_{1}^{4}\sqrt{(4t^2+1-4t)+\left(16t^6\right)}dt\]
\[L=\int_{1}^{4}\sqrt{16t^6+4t^2-4t+1}dt\]
By using calculator,
\[L\approx 255.376\]
