use finite approximations to estimate the area under the graph of the function f(x)=15 - x² + 2x between x =…

Explanation:

Step1: Calculate the width of rectangles

The interval is from $x = - 3$ to $x = 5$, so the length of the interval $\Delta x=\frac{5 - (-3)}{n}$, where $n$ is the number of rectangles.

Step2: Find the sub - intervals

For $n = 2$, $\Delta x=\frac{5-(-3)}{2}=4$. The sub - intervals are $[-3,1]$ and $[1,5]$. For $n = 4$, $\Delta x=\frac{5-(-3)}{4}=2$. The sub - intervals are $[-3,-1]$, $[-1,1]$, $[1,3]$, $[3,5]$.

Step3: Evaluate the function at key points

The function is $f(x)=15 - x^{2}+2x$.

a. Lower sum with two rectangles

The lower sum $L_2$: On $[-3,1]$, the minimum of $f(x)$ occurs at $x=-3$ or $x = 1$. $f(-3)=15-9 - 6=0$, $f(1)=15 - 1+2=16$. The minimum on $[-3,1]$ is $0$. On $[1,5]$, $f(1)=16$, $f(5)=15 - 25 + 10=0$. The minimum on $[1,5]$ is $0$. $L_2=\sum_{i = 1}^{2}m_i\Delta x$, where $m_i$ is the minimum value of $f(x)$ on the $i$ - th sub - interval. $L_2=0\times4+0\times4 = 0$.

b. Lower sum with four rectangles

On $[-3,-1]$: $f(-3)=0$, $f(-1)=15 - 1-2 = 12$, minimum is $0$. On $[-1,1]$: $f(-1)=12$, $f(1)=16$, minimum is $12$. On $[1,3]$: $f(1)=16$, $f(3)=15 - 9+6 = 12$, minimum is $12$. On $[3,5]$: $f(3)=12$, $f(5)=0$, minimum is $0$. $L_4=0\times2+12\times2+12\times2+0\times2=48$.

c. Upper sum with two rectangles

On $[-3,1]$, the maximum of $f(x)$ occurs at $x=-3$ or $x = 1$. The maximum on $[-3,1]$ is $16$. On $[1,5]$, the maximum on $[1,5]$ is $16$. $U_2=\sum_{i = 1}^{2}M_i\Delta x$, where $M_i$ is the maximum value of $f(x)$ on the $i$ - th sub - interval. $U_2=16\times4+16\times4=128$.

d. Upper sum with four rectangles

On $[-3,-1]$: $f(-3)=0$, $f(-1)=12$, maximum is $12$. On $[-1,1]$: $f(-1)=12$, $f(1)=16$, maximum is $16$. On $[1,3]$: $f(1)=16$, $f(3)=12$, maximum is $16$. On $[3,5]$: $f(3)=12$, $f(5)=0$, maximum is $12$. $U_4=12\times2+16\times2+16\times2+12\times2 = 112$.

Answer:

a. $0$ b. $48$ c. $128$ d. $112$