# Show that the trapezoid rule is more precise than Simpson rule for $x^{5} - \alpha \cdot x^{4}$

I need to show that for $\int_{0}^{1}(x^{5}-\alpha \cdot x^{4})dx$, the trapezoid rule is more precise than Simpson when $\frac{15}{14} < \alpha < \frac{85}{74}$

What I have done : Trapezoid is more precise means $$Error^{trapezoid}_{s}(f,0,1) < Error^{Simpson}_{s}(f,0,1)$$

$$\leftrightarrow \int_{0}^{1}f(x)dx - \frac{1}{2}(f(0) + f(1)) < \int_{0}^{1}f(x)dx - \frac{1}{6}(f(0) +4\cdot f(\frac{1}{2})+ f(1))$$ $$\leftrightarrow\frac{1}{2}(f(0) + f(1)) < \frac{1}{6}(f(0) +4\cdot f(\frac{1}{2})+ f(1))$$ $$\ldots$$ $$\frac{15}{14} < \alpha$$

But from where does the condition $\alpha < \frac{85}{74}$ appear ?

The error is the absolute value of your first inequalities, so you are asking $$|\int_{0}^{1}f(x)dx - \frac{1}{2}(f(0) + f(1))| < |\int_{0}^{1}f(x)dx - \frac{1}{6}(f(0) +4\cdot f(\frac{1}{2})+ f(1))|$$ Let us work with the left side $$|\int_{0}^{1}f(x)dx - \frac{1}{2}(f(0) + f(1))| =|\int_0^1 (x^5-\alpha x^4)dx-\frac 12(1-\alpha)|\\=|\frac 16-\frac \alpha 5-\frac 12+\frac \alpha 2|\\ =|\frac {3\alpha}{10}-\frac 13|$$ You can evaluate the right side similarly, then find the $\alpha$s that make the inequality true.
• Although the error can be negative, when one is speaking of "more precise" you need the absolute value bars. Here our integral is $\frac 16- \frac \alpha 5$. For $\alpha=1$, for example, this is $\frac {-1}{30}$. If the trapezoid gave $1$ whle Simpson gave $0$ the version without absolute values would prefer trapezoid. – Ross Millikan Nov 15 '17 at 18:47