Answer
(a) A lower sum with two rectangles of equal width: $8$
(b) A lower sum with four rectangles of equal width: $10$
(c) An upper sum with two rectangles of equal width: $8$
(d) An upper sum with four rectangles of equal width: $10$
Work Step by Step
Dividing the interval [-2, 2] into subintervals of equal width and approximate the area under the curve using rectangles. Let's calculate the lower and upper sums for different numbers of rectangles.
(a) A lower sum with two rectangles of equal width:
First, we need to calculate the width of each rectangle:
Width = $\frac{(2 - (-2)) }{2} = \frac{4}{2} = 2$
Next, we calculate the height of each rectangle using the left endpoint of each subinterval:
For the first rectangle, the left endpoint is $-2$, so the height is $f(-2) = 4 - (-2)^{2}= 4 - 4 = 0$.
For the second rectangle, the left endpoint is $0$ (midpoint of the interval), so the height is $f(0) = 4 - 0^{2} = 4$.
Finally, we calculate the area of each rectangle and sum them up:
Area of rectangle $1$ = Width * Height$ = 2 \times 0 = 0$
Area of rectangle $2$ = Width * Height $= 2 \times 4 = 8$
The lower sum with two rectangles is $0 + 8 = 8$.
(b) A lower sum with four rectangles of equal width:
Now we divide the interval into four subintervals of equal width:
Width = $\frac{(2 - (-2)) }{4} = 1$
The left endpoints of the subintervals are: $-2, -1, 0, 1$.
For each rectangle, we calculate the height using the left endpoint of each subinterval:
Height of rectangle $1$: $f(-2) = 4 - (-2)^{2} = 4 - 4 = 0$
Height of rectangle $2$: $f(-1) = 4 - (-1)^{2} = 4 - 1 = 3$
Height of rectangle $3$: $f(0) = 4 - (0)^{2}= 4 - 0 = 4$
Height of rectangle $4$: $f(1) = 4 - (1)^{2} = 4 - 1 = 3$
The areas of the rectangles are:
Area of rectangle $1$ = Width $\times$Height $= 1 \times 0 = 0$
Area of rectangle $2$ = Width $\times$ Height$ = 1 \times 3 = 3$
Area of rectangle $3$ = Width $\times$ Height$ = 1 \times 4 = 4$
Area of rectangle $4$ = Width $\times$ Height $= 1 \times 3 = 3$
The lower sum with four rectangles is $0 + 3 + 4 + 3 = 10$.
(c) An upper sum with two rectangles of equal width:
Using the same intervals and widths as in part (a), we calculate the heights using the right endpoint of each subinterval:
Height of rectangle $1$: $f(0) = 4 - 0^{2} = 4 - 0 = 4$
Height of rectangle $2$: $f(2) = 4 - 2^{2} = 4 - 4 = 0$
The areas of the rectangles are:
Area of rectangle $1$ = Width $\times$ Height$ = 2 \times 4 = 8$
Area of rectangle $2$ = Width $\times$ Height$ = 2 \times 0 = 0$
The upper sum with two rectangles is $8 + 0 = 8$.
(d) An upper sum with four rectangles of equal width:
Using the same intervals and widths as in part (b), we calculate the heights using the right endpoint of each subinterval:
Height of rectangle $1$: $f(-1) = 4 - (-1)^{2}= 4 - 1 = 3$
Height of rectangle $2$: $f(0) = 4 - 0^{2} = 4 - 0 = 4$
Height of rectangle $3$: $f(1) = 4 - 1^{2} = 4 - 1 = 3$
Height of rectangle $4$: $f(2) = 4 - 2^{2}= 4 - 4 = 0$
The areas of the rectangles are:
Area of rectangle $1$ = Width $\times$ Height$ = 1 \times 3 = 3$
Area of rectangle $2$ = Width $\times$ Height$ = 1 \times 4 = 4$
Area of rectangle $3$ = Width $\times$ Height$ = 1 \times 3 = 3$
Area of rectangle $4$ = Width $\times$ Height$ = 1 \times 0 = 0$
The upper sum with four rectangles is $3 + 4 + 3 + 0 = 10$.
