Answer
$L=\int_1^2\sqrt{1+(2x+3x^2)^2}dx=10.0556$
Work Step by Step
Approximately 10.2, line length, $\sqrt{4+100}=10.2$
$L=\int_a^b\sqrt{1+\Big(\frac{dy}{dx}\Big)^2}dx$
$y=x^2+x^3$
$\frac{dy}{dx}=2x+3x^2$
$a=1$, $b=2$
$L=\int_1^2\sqrt{1+(2x+3x^2)^2}dx=10.0556$
