MA005: Calculus I

Practice 1: area = $3(6)+\frac{1}{2}(4)(3)=24$ and area = $\frac{1}{2}(3)(10)+\frac{1}{2}(6)(3)=24 \text {. }$

Practice 2: outside rectangular area = $(1)(1)+(1)\left(\frac{1}{2}\right)=1.5$

Practice 3: Using rectangles with base = $1 / 2$. Inside: area = $\frac{1}{2}\left(\frac{2}{3}+\frac{1}{2}+\frac{2}{5}+\frac{1}{3}\right)=\frac{57}{60} \approx 0.95$

Outside: area = $\frac{1}{2}\left(1+\frac{2}{3}+\frac{1}{2}+\frac{2}{5}\right)=\frac{72}{60} \approx 1.2$
The area of the region is between $0.95$ and $1.2$.

Practice 4: The area of the leaf is larger than the area of the dark rectangles, $32 \mathrm{~cm}^{2}$.

Practice 5:

$\mathrm{y}=\mathrm{B}(\mathrm{x})=2 \mathrm{x}$ is a line with slope 2 so it is parallel to $y=A(x)=2 x-2$.

Practice 7: See Fig. 34. Distance = area of shaded region = $\frac{1}{2}$( base )( height )
$=\frac{1}{2}(60 \text { seconds })(66 \text { feet } / \text { second })=1980 \text { feet. }$

Practice 8:

(a) At 2 pm both are walking at the same velocity. You are ahead.
(b) At 3 pm your friend is walking faster than you, but you are still ahead. (The "area" under your velocity curve is larger than the "area" under your friend's).
(c) You and your friend will be together on the trail when the "areas" (distances) under the two velocity graphs are equal.

Practice 9: Total calls = "area" under rate curve from 9 am to 11 am ≈ 300 calls.