use finite approximations to estimate the area under the graph of the function f(x)=3 - x² - 2x between x =…

use finite approximations to estimate the area under the graph of the function f(x)=3 - x² - 2x between x = - 3 and x = 1 for each of the following cases. a. using a lower sum with two rectangles of equal width b. using a lower sum with four rectangles of equal width c. using an upper sum with two rectangles of equal width d. using an upper sum with four rectangles of equal width

use finite approximations to estimate the area under the graph of the function f(x)=3 - x² - 2x between x = - 3 and x = 1 for each of the following cases. a. using a lower sum with two rectangles of equal width b. using a lower sum with four rectangles of equal width c. using an upper sum with two rectangles of equal width d. using an upper sum with four rectangles of equal width

Answer

Explanation:

Step1: Calculate width of rectangles

The interval is $[a,b]=[-3,1]$. For $n$ rectangles, the width $\Delta x=\frac{b - a}{n}$.

Step2: For part a (lower - sum, $n = 2$)

$\Delta x=\frac{1-(-3)}{2}=2$. The sub - intervals are $[-3,-1]$ and $[-1,1]$. For a lower sum, we evaluate the function at the right - hand endpoints of each sub - interval. $f(-1)=3-(-1)^2-2(-1)=3 - 1+2 = 4$ and $f(1)=3-1^2-2(1)=3 - 1 - 2=0$. The lower sum $L_2=\sum_{i = 1}^{2}f(x_i)\Delta x=f(-1)\times2+f(1)\times2=4\times2+0\times2 = 8$.

Step3: For part b (lower - sum, $n = 4$)

$\Delta x=\frac{1-(-3)}{4}=1$. The sub - intervals are $[-3,-2],[-2,-1],[-1,0],[0,1]$. For a lower sum, we evaluate the function at the right - hand endpoints of each sub - interval. $f(-2)=3-(-2)^2-2(-2)=3 - 4 + 4=3$, $f(-1)=4$, $f(0)=3$, $f(1)=0$. $L_4=\sum_{i = 1}^{4}f(x_i)\Delta x=f(-2)\times1+f(-1)\times1+f(0)\times1+f(1)\times1=3 + 4+3+0 = 10$.

Step4: For part c (upper - sum, $n = 2$)

$\Delta x = 2$. The sub - intervals are $[-3,-1]$ and $[-1,1]$. For an upper sum, we evaluate the function at the left - hand endpoints of each sub - interval. $f(-3)=3-(-3)^2-2(-3)=3 - 9+6 = 0$ and $f(-1)=4$. The upper sum $U_2=\sum_{i = 1}^{2}f(x_i)\Delta x=f(-3)\times2+f(-1)\times2=0\times2+4\times2 = 8$.

Step5: For part d (upper - sum, $n = 4$)

$\Delta x = 1$. The sub - intervals are $[-3,-2],[-2,-1],[-1,0],[0,1]$. For an upper sum, we evaluate the function at the left - hand endpoints of each sub - interval. $f(-3)=0$, $f(-2)=3$, $f(-1)=4$, $f(0)=3$. $U_4=\sum_{i = 1}^{4}f(x_i)\Delta x=f(-3)\times1+f(-2)\times1+f(-1)\times1+f(0)\times1=0 + 3+4+3 = 10$.

Answer:

a. 8 b. 10 c. 8 d. 10