Question

(1 point) use the midpoint rule to approximate (int_{-2.5}^{5.5}x^{3}dx) with (n = 8).

Explanation:

Step1: Calculate the width of sub - intervals

The width of each sub - interval $\Delta x=\frac{b - a}{n}$, where $a=-2.5$, $b = 5.5$, and $n = 8$. So $\Delta x=\frac{5.5-( - 2.5)}{8}=\frac{8}{8}=1$.

Step2: Determine the mid - points of sub - intervals

The mid - points $x_i$ of the sub - intervals are: $x_1=-2$, $x_2=-1$, $x_3 = 0$, $x_4=1$, $x_5 = 2$, $x_6=3$, $x_7 = 4$, $x_8=5$.

Step3: Evaluate the function at mid - points

We have $f(x)=x^{3}$. Calculate $f(x_i)$ for each $i$:
$f(x_1)=(-2)^{3}=-8$, $f(x_2)=(-1)^{3}=-1$, $f(x_3)=0^{3}=0$, $f(x_4)=1^{3}=1$, $f(x_5)=2^{3}=8$, $f(x_6)=3^{3}=27$, $f(x_7)=4^{3}=64$, $f(x_8)=5^{3}=125$.

Step4: Apply the Midpoint Rule

The Midpoint Rule formula is $M_n=\Delta x\sum_{i = 1}^{n}f(x_i)$.
$M_8=1\times(-8-1 + 0+1+8+27+64+125)=216$.

Answer:

$216$