I have $$\int_0^2 \frac{1}{1+x^6} dx$$
We vaguely went over this in class today and I understand how to work Simpson's rule but only when I have $n$ already given to me. How can I determine $n$ when it isn't given? Thank you for your time and help.
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I'm not clear about your reference to $n$: Perhaps this formula will be of help. Simpson's Rule (numerical approximation of definite integrals): $\displaystyle \int_a^b f(x)dx \approx \frac{b-a}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right) + f(b)\right] $.
If you want the composite Simpson's Rule: Use $$\int_a^b f(x) \, dx\approx \frac{h}{3}\bigg[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\cdots+4f(x_{n-1})+f(x_n)\bigg]$$ where $h = \dfrac{b-a}{n}$ Note: Usually -- in an exercise -- "$n$" will be explicitly given to you. In real-life scenarios, keep doubling the number of $x_i$ until your answer stabilizes. It turns out that the computation for $n = 2k$ (with $k$ an integer) can be incorporated into the answer for $n = 2(2k)$, so you don't need to start from scratch when doubling the number of $x_i$. Choose $n$ sufficiently small to bound the error term. The error term is given by the following: $$\frac{h^4}{180}(b-a) \max_{\xi\in[a,b]} |f^{(4)}(\xi)|$$ |
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Using $$\int_a^b f(x) \, dx\approx \frac{h}{3}\bigg[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\cdots+4f(x_{n-1})+f(x_n)\bigg]$$ the error term is given by $$\frac{h^4}{180}(b-a) \max_{\xi\in[a,b]} |f^{(4)}(\xi)|$$ Choose $h$ so that the error term is sufficiently small. |
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