use the midpoint rule with n = 3 to approximate the integral \n\\(\\int_{-2}^{5}(-6x + 10x^{2})dx\\)

Explanation:

Step1: Calculate the width of sub - intervals

The formula for the width of sub - intervals $\Delta x=\frac{b - a}{n}$, where $a=-2$, $b = 5$ and $n = 3$. $\Delta x=\frac{5-(-2)}{3}=\frac{7}{3}$

Step2: Determine the sub - intervals and mid - points

The sub - intervals are $[-2,-2+\frac{7}{3}]=[-2,\frac{1}{3}]$, $[\frac{1}{3},\frac{1}{3}+\frac{7}{3}]=[\frac{1}{3},\frac{8}{3}]$, $[\frac{8}{3},\frac{8}{3}+\frac{7}{3}]=[\frac{8}{3},5]$. The mid - points are $x_1=\frac{-2+\frac{1}{3}}{2}=-\frac{5}{6}$, $x_2=\frac{\frac{1}{3}+\frac{8}{3}}{2}=\frac{3}{2}$, $x_3=\frac{\frac{8}{3}+5}{2}=\frac{23}{6}$

Step3: Evaluate the function at mid - points

Let $f(x)=-6x + 10x^{2}$. $f(x_1)=-6\times(-\frac{5}{6})+10\times(-\frac{5}{6})^{2}=5 + 10\times\frac{25}{36}=5+\frac{125}{18}=\frac{90 + 125}{18}=\frac{215}{18}$ $f(x_2)=-6\times\frac{3}{2}+10\times(\frac{3}{2})^{2}=-9 + 10\times\frac{9}{4}=-9+\frac{45}{2}=\frac{-18 + 45}{2}=\frac{27}{2}$ $f(x_3)=-6\times\frac{23}{6}+10\times(\frac{23}{6})^{2}=-23+10\times\frac{529}{36}=-23+\frac{2645}{18}=\frac{-414+2645}{18}=\frac{2231}{18}$

Step4: Apply the Midpoint Rule

The Midpoint Rule is $\int_{a}^{b}f(x)dx\approx\Delta x\sum_{i = 1}^{n}f(x_i)$. $\int_{-2}^{5}(-6x + 10x^{2})dx\approx\frac{7}{3}(f(x_1)+f(x_2)+f(x_3))$ $=\frac{7}{3}(\frac{215}{18}+\frac{27}{2}+\frac{2231}{18})$ $=\frac{7}{3}(\frac{215 + 243+2231}{18})$ $=\frac{7}{3}\times\frac{2689}{18}=\frac{18823}{54}\approx348.57$

Answer:

$\frac{18823}{54}\approx348.57$