Answer
$\dfrac{\delta(10\sqrt {10}-1)}{54}$
Work Step by Step
$m_x=\int_m^n \sqrt {1+[y']^2} dx=\delta \times \int_0^1 (x^3) \sqrt {1+9x^4} dx$
Consider $1+9x^4=k$
Now, $m_x=\delta \int_0^1 (x^3) \sqrt {1+9x^4} dx\\=\dfrac{\delta}{36} \int_{1}^{10} \sqrt k \space dk \\=\dfrac{\delta(10\sqrt {10}-1)}{54}$
