## Calculus 8th Edition

$\frac{1}{27}(13^{3/2}-8)$
As we are given that $r(t)=i+t^2j+t^3k$ ; $0 \leq t \leq 1$ Length of the curve can be obtained by using formula, such as $L=\int_a^b |r'(t)| dt$ Now, $r'(t)=\lt 0,2t,3t^2\gt$ ; $|r'(t)|=\sqrt {( 2t)^2+(3t^2)^2}dt$ or, $=\sqrt{ 4t^2+9t^4}$ Since, $L=\int_{0}^1(\sqrt{ 4t^2+9t^4}) dt=\int_{0}^1t(\sqrt{ 4+9t^2}) dt$ This yields, $L=\frac{1}{18}(\frac{2}{3}(4+9t^2)^{3/2}|_{0}^1=\frac{1}{27}[(4+9(1)^2)-(4+9(0)^2)]^{3/2}=\frac{1}{27}(13^{3/2}-8)$ After simplifications, we get $L=\frac{1}{27}(13^{3/2}-8)$