Answer
$\frac{1}{27}(13^{3/2}-8)$
Work Step by Step
Given: $r(t)=i+t^2j+t^3k$ ; $0 \leq t \leq 1$
To calculate the length of the curve we will have to use the formula:
$L=\int_a^b |r'(t)| dt$
Thus,
$r'(t)=\lt 0,2t,3t^2\gt$
and $|r'(t)|=\sqrt {( 2t)^2+(3t^2)^2}dt$
$=\sqrt{ 4t^2+9t^4}$
$L=\int_{0}^1(\sqrt{ 4t^2+9t^4}) dt=\int_{0}^1t(\sqrt{ 4+9t^2}) dt$
$\implies L=\frac{1}{18}(\frac{2}{3}(4+9t^2)^{3/2}|_{0}^1$
$=\frac{1}{27}[(4+9(1)^2)-(4+9(0)^2)]^{3/2}$
$=\frac{1}{27}(13^{3/2}-8)$
Hence, $L=\frac{1}{27}(13^{3/2}-8)$
