Originally Posted by

**arze** Use the first three terms of the series expansion as an approximation for the function to be integrated. Estimate the value of the definite integral.

$\displaystyle \int^{0.2}_{0.1} \sqrt[3]{1-x^2} dx$

series expansion:

$\displaystyle (1-x^2)^{\frac{1}{3}}=1-\frac{1}{3}x^2-\frac{1}{9}x^4$

so i have:

$\displaystyle \int^{0.2}_{0.1}(1-\frac{1}{3}x^2-\frac{1}{9}x^4) dx$

$\displaystyle =[x-\frac{1}{9}x^3-\frac{1}{45}x^5]^{0.2}_{0.1}$

=0.09921533

answer is 0.9921556

Where is my error?

Thanks