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Calculate the value of the integral $$ \int_0^2 \sqrt[3]{x^2 + 2x - 1} \,dx $$ with measurement uncertainty not larger than $10^{-3}$.

I know we can evaluate integration using the "trapezoidal rule" or "Simpson's rule". But if we want to calculate the uncertainty, using the first rule, we have to calculate $\max_{x \in [0, 2]} |f''(x)|$, which does not exist ($|f''(x)| = {2 \over 9}|{x^2 + 2x + 7 \over (x^2 + 2x - 1)^{5/3}}| $, which has limit $+\infty$ when $x \rightarrow \sqrt{2} - 1$). Using the second rule, we need to calculate $\max_{x \in [0, 2]} |f''''(x)|$, which doesn't exist either. So, how can we evaluate this integral? Can someone give me a suggestion? Thanks.

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    $\begingroup$ The integral has two parts, for each part, you take $y^3 = x^2 + 2x - 1$, solve $x$ in terms of $y$, which is definitely differentiable. And rewrite your problem in $\int y (dx/dy) dy$ $\endgroup$ – Yimin May 12 '15 at 2:06
  • $\begingroup$ $\int_0^2{\sqrt[3]{x^2 + 2x - 1}}dx = \int_1^3{\sqrt[3]{x^2 - 2}}dx$ then Taylor for $(1-\frac{x^2}{2})^\frac{1}{3}$ and integrate. $\endgroup$ – Alexey Burdin May 12 '15 at 2:18
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    $\begingroup$ i think Yimin meant that now the derivatives of the integrand are bounded so that you can bound the error. $\endgroup$ – abel May 12 '15 at 2:29
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    $\begingroup$ @abel you are right! $\endgroup$ – Yimin May 12 '15 at 2:32
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    $\begingroup$ @Yimin, you see i can read your mind! $\endgroup$ – abel May 12 '15 at 2:34
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An important point to take account is that the function $x^2+2x-1$ is negative on a part of the range of integration. In order to avoid the problems with the square root of negative numbers in the numerical calculus, it is advised to split the integral into two parts and compute separately : $$ \int_{\sqrt 2-1}^2 \sqrt[3]{x^2 + 2x - 1} \,dx $$ and $$ \int_0^{\sqrt 2-1} \sqrt[3]{-(x^2 + 2x - 1)} \,dx $$ Since I suppose that the goal is to find the numerical value for the integration on the real path, you should obtain approximately $1.83052$

Are you sure that there is no typo in the wording of the problem ? Because it is surprising that a school problem of this kind involves complex range of integration. If there is no typo, one could compare the computed result to an analytical calculus. But it would be arduous because involving special functions of higher level.

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  • $\begingroup$ Yes, I'm quite pretty sure that the problem is right in typo. What you said is what I mind mostly, in time of 30', I don't think I can do those very complicated task and get perfect result. But thanks anyway, @JJacquelin $\endgroup$ – le duc quang May 12 '15 at 14:28

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