1. **Problem statement:** Find the area under the curve $y=4-x^2$ on the interval $[-2,2]$ using rectangles. 2. **Formula and explanation:** The area under a curve can be approximated by Riemann sums. For $n$ rectangles, the width of each rectangle is $\Delta x=\frac{b-a}{n}$ where $a=-2$ and $b=2$. - **Lower sum:** Use the minimum function value on each subinterval. - **Upper sum:** Use the maximum function value on each subinterval. 3. **Calculate for 2 rectangles:** - Width $\Delta x=\frac{2-(-2)}{2}=2$ - Subintervals: $[-2,0]$, $[0,2]$ Lower sum (minimum on each interval): - On $[-2,0]$, $y$ decreases from $0$ at $x=-2$ to $4$ at $x=0$, minimum is $0$ - On $[0,2]$, $y$ decreases from $4$ at $x=0$ to $0$ at $x=2$, minimum is $0$ - Area $= 2\times0 + 2\times0=0$ Upper sum (maximum on each interval): - On $[-2,0]$, maximum is $4$ at $x=0$ - On $[0,2]$, maximum is $4$ at $x=0$ - Area $= 2\times4 + 2\times4=16$ 4. **Calculate for 4 rectangles:** - Width $\Delta x=\frac{4}{4}=1$ - Subintervals: $[-2,-1]$, $[-1,0]$, $[0,1]$, $[1,2]$ Lower sum (minimum on each interval): - $[-2,-1]$: min at $x=-2$, $y=0$ - $[-1,0]$: min at $x=-1$, $y=3$ - $[0,1]$: min at $x=1$, $y=3$ - $[1,2]$: min at $x=2$, $y=0$ - Area $= 1\times0 + 1\times3 + 1\times3 + 1\times0=6$ Upper sum (maximum on each interval): - $[-2,-1]$: max at $x=-1$, $y=3$ - $[-1,0]$: max at $x=0$, $y=4$ - $[0,1]$: max at $x=0$, $y=4$ - $[1,2]$: max at $x=1$, $y=3$ - Area $= 1\times3 + 1\times4 + 1\times4 + 1\times3=14$ 5. **Summary:** - 2 rectangles lower sum: $0$ - 2 rectangles upper sum: $16$ - 4 rectangles lower sum: $6$ - 4 rectangles upper sum: $14$