# How do you find the definite integrals for the lengths of the curves, but do not evaluate the integrals for y=x^3, 0<=x<=1?

Mar 14, 2018

$A = {\int}_{0}^{1} \sqrt{1 + 9 {x}^{4}} \mathrm{dx}$

#### Explanation:

Recall that the arc length of a curve is given by

$A = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Here this would be

$A = {\int}_{0}^{1} \sqrt{1 + {\left(3 {x}^{2}\right)}^{2}} \mathrm{dx}$

$A = {\int}_{0}^{1} \sqrt{1 + 9 {x}^{4}} \mathrm{dx}$

Hopefully this helps!