Answer
$\int_0^{0.35}\sqrt [3] {1+x^2} dx \approx 0.3546472$ (Error of about $10^{-5})$
Work Step by Step
Integrate the integral with respect to $ x $ as follows:
$\int_0^{0.35}\sqrt [3] {1+x^2} dx=\int_0^{0.35} [1+\dfrac{x^2}{3}-\dfrac{x^{4}}{9}+...] dx \\=0.35-\dfrac{(0.35)^3}{9}-\dfrac{(0.35)^5}{45}+\dfrac{5(0.35)^7}{567}....$
$\dfrac{5(0.35)^7}{567} \approx 5.67 \times 10^{-6} $ and this means that the proceeding term is greater than $10^{-5}$ and the first five terms of the series yield the accuracy.
Thus, $\int_0^{0.35}\sqrt [3] {1+x^2} dx \approx 0.3546472$ (Error of about $10^{-5})$
