I'm running into some troubles while trying to evaluate $$\int_{0}^{2}(x^2 + x^{3/2} + x + x^{1/2})dx$$ using Simpson's Rule
Simpson's Rule states $$Q(f) = \frac{b-a}{6}(f(a) + 4f(\frac{a+b}{2}) +f(b))$$
In our case: $$f(x) = x^2 + x^{3/2} + x + x^{1/2}$$ $$a=0, b=2, f(a=0) = 0 $$ $$ f(b=2) = 4 + 2\sqrt{2} + 2 + \sqrt{2} = 3(2 + \sqrt{2})$$ $$f(\frac{a + b}{2}) = f(1) = 4$$ Thus we get
$$\frac{2-0}{6}(0 + 4 \cdot 4 + 3(2 + \sqrt{2})) = \frac{2(22 + 3\sqrt{2})}{6} = \frac{(22 + 3\sqrt{2})}{3} \approx 8.75$$
However, when evaluating the integral, Wolfram Alpha gives $\approx 8.815$.
Simpson's Rule has degree of accuracy three. Because we are dealing with a polynomial of degree two, shouldn't Simpson's Rule give the exact value of the integral back ? Or are those simply rounding errors ?
Thanks for your help !
This integral can easily be computed, using the following fact : $$ \left(\forall \alpha\in\mathbb{R}\setminus\left\lbrace -1\right\rbrace\right),\ \int{x^{\alpha}\,\mathrm{d}x}=\frac{x^{1+\alpha}}{1+\alpha}+C $$
Thus, $$ \int_{0}^{2}{\left(x^{2}+x^{\frac{3}{2}}+x+x^{\frac{1}{2}}\right)\mathrm{d}x}=\left[\frac{x^{3}}{3}+\frac{2x^{\frac{5}{2}}}{5}+\frac{x^{2}}{2}+\frac{2x^{\frac{3}{2}}}{3}\right]_{0}^{2}=\frac{2}{15}\left(35+22\sqrt{2}\right) $$