Answer
$$\approx 0.3546472$$ with an error of about $10^{-5}$.
Work Step by Step
We integrate the integral with respect to $ x $ and take a lower limit as $0$ and upper limit as $0.35$.
$\int_0^{0.35}\sqrt [3] {1+x^2} dx=\int_0^{0.35} [1+\dfrac{x^2}{3}-\dfrac{x^{4}}{9}+...] dx $
or, $=0.35-\dfrac{(0.35)^3}{9}-\dfrac{(0.35)^5}{45}+\dfrac{5(0.35)^7}{567}....$
But $\dfrac{5(0.35)^7}{567} \approx 5.67 \times 10^{-6} $ and the proceeding term is greater than $10^{-5}$.
So, the first three terms of the series would give the accuracy.
$\int_0^{0.35}\sqrt [3] {1+x^2} dx \approx 0.3546472$ with an error of about $10^{-5}$.
