- #1

mathmari

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Calculate using the Simpson's Rule the integral $\int_0^1\sqrt{1+x^4}\, dx$ approximately such that the error is less that $0,5\cdot 10^{-3}$. Which has to be $h$ ?

So we use here the composite Simpson's rule, right?

An upper bound of the error of that rule is defined as $$\frac{h^4}{180}(b-a)\max_{\xi \in [a,b]}|f^{(4)}(\xi)|$$ and we have to set this equal to $0,5\cdot 10^{-3}$, right?

So we get $$\frac{h^4}{180}\max_{\xi \in [a,b]}|f^{(4)}(\xi)|=0,5\cdot 10^{-3} $$

The $4$.th derivative of $f(x)=\sqrt{1+x^4}$ is $\frac{12 (1 - 14 x^4 + 5 x^8)}{(1 + x^4)^{7/2}}$. If we check if this is increasing or decreasing on $[0,1]$ we see that the maximum on that interval is at $x=0$ and it is equal to $12$.

So we get $$\frac{h^4}{180}\cdot 12=0,5\cdot 10^{-3} \Rightarrow h^4=\frac{180}{12}0,5\cdot 10^{-3} \Rightarrow h^4=7.5\cdot 10^{-3} \Rightarrow h\approx 0.294283$$

Is that correct?