Question

use finite approximations to estimate the area under the graph of the function f(x) = 3 - x² - 2x be 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 a. the area under the graph of f(x) using a lower sum with two rectangles of equal width is \square. (t integer or a decimal.)

Explanation:

Step1: Find the width of each rectangle

The interval is from \( x = -3 \) to \( x = 1 \), so the length of the interval is \( 1 - (-3)=4 \). We are using 2 rectangles, so the width \( \Delta x=\frac{4}{2} = 2 \).

Step2: Determine the subintervals

The subintervals are \( [-3, -1] \) and \( [-1, 1] \).

Step3: Find the minimum value of \( f(x) \) on each subinterval

First, analyze the function \( f(x)=3 - x^{2}-2x=-(x^{2}+2x)+3=-(x + 1)^{2}+4 \). This is a parabola opening downwards with vertex at \( x=-1 \).

• For the subinterval \( [-3, -1] \): The function is increasing on \( [-3, -1] \) (since the vertex is at \( x = - 1 \) and it opens down), so the minimum value occurs at \( x=-3 \). \( f(-3)=3-(-3)^{2}-2\times(-3)=3 - 9 + 6=0 \).
• For the subinterval \( [-1, 1] \): The function is decreasing on \( [-1, 1] \), so the minimum value occurs at \( x = 1 \). \( f(1)=3-(1)^{2}-2\times1=3 - 1-2 = 0 \).

Step4: Calculate the lower sum

The lower sum \( L=\sum_{i = 1}^{n}f(x_{i})\Delta x \), where \( n = 2 \), \( \Delta x=2 \), \( f(x_1)=f(-3) = 0 \), \( f(x_2)=f(1)=0 \). So \( L=0\times2+0\times2 = 0 \). Wait, that seems wrong. Wait, let's recalculate \( f(-3) \): \( f(-3)=3-(-3)^2-2\times(-3)=3 - 9 + 6=0 \), \( f(-1)=3-(-1)^2-2\times(-1)=3 - 1 + 2=4 \), \( f(1)=3 - 1-2 = 0 \). Wait, on \( [-3,-1] \), the function goes from \( f(-3)=0 \) to \( f(-1)=4 \), so it's increasing, so minimum at \( x=-3 \) is 0. On \( [-1,1] \), function goes from \( f(-1)=4 \) to \( f(1)=0 \), so decreasing, minimum at \( x = 1 \) is 0. So the lower sum is \( (0\times2)+(0\times2)=0 \)? Wait, but maybe I made a mistake in the function analysis. Wait, \( f(x)=3 - x^{2}-2x \), let's find its derivative \( f^\prime(x)=-2x - 2=-2(x + 1) \). So when \( x < - 1 \), \( f^\prime(x)>0 \) (function increasing), when \( x > - 1 \), \( f^\prime(x)<0 \) (function decreasing). So on \( [-3,-1] \), increasing, so min at left endpoint \( x=-3 \), \( f(-3)=3 - 9 + 6=0 \). On \( [-1,1] \), decreasing, so min at right endpoint \( x = 1 \), \( f(1)=3 - 1-2 = 0 \). So the lower sum with two rectangles is \( (f(-3)\times2)+(f(1)\times2)=(0\times2)+(0\times2)=0 \).

Answer:

\( 0 \)