Function approximation

**arze** · February 16th 2010, 09:12 PM

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.
$\int^{0.2}_{0.1} \sqrt[3]{1-x^2} dx$
series expansion:
$(1-x^2)^{\frac{1}{3}}=1-\frac{1}{3}x^2-\frac{1}{9}x^4$
so i have:
$\int^{0.2}_{0.1}(1-\frac{1}{3}x^2-\frac{1}{9}x^4) dx$
$=[x-\frac{1}{9}x^3-\frac{1}{45}x^5]^{0.2}_{0.1}$
=0.09921533
answer is 0.9921556
Where is my error?
Thanks

**Danny** · February 17th 2010, 06:10 AM

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.
$\int^{0.2}_{0.1} \sqrt[3]{1-x^2} dx$
series expansion:
$(1-x^2)^{\frac{1}{3}}=1-\frac{1}{3}x^2-\frac{1}{9}x^4$
so i have:
$\int^{0.2}_{0.1}(1-\frac{1}{3}x^2-\frac{1}{9}x^4) dx$
$=[x-\frac{1}{9}x^3-\frac{1}{45}x^5]^{0.2}_{0.1}$
=0.09921533
answer is 0.9921556
Where is my error?
Thanks

I don't see an error here.

**arze** · February 17th 2010, 03:40 PM

ok, then the other explanation is the answer is inaccurate?

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