MA005: Calculus I

Practice 1: Area = $\lim _{\text {mesh } \rightarrow 0}\left(\sum_{k=1}^{n} \sin \left(c_{k}\right) \Delta x_{k}\right)=\int_{0}^{\pi} \sin (\mathrm{x}) \mathrm{d} x$

Practice 2: $\lim _{m e s h \rightarrow 0}\left(\sum_{k=1}^{n}\left(2 c_{k}\right) \Delta x_{k}\right)=$ Shaded area in Fig. 18 = 8

$\int_{3}^{8} 4 d x$ = shaded area in Fig. 19 = 20 .

Practice 3:

(a) Total distance = 12.5 feet forward and 2.5 feet backward = 15 feet total travel.

(b) The bug ends up 10 feet forward of it's starting position at x = 12 so the bug's final location is at x = 22.

Practice 4: Between $\mathrm{x}=0$ and $\mathrm{x}=2 \pi$, the graph of $\mathrm{y}=\sin (\mathrm{x})$ (Fig. 20) has the same area above the x–axis as below the x–axis so the definite integral is $0: \int_{0}^{2 \pi} \sin (x) \mathrm{d} x=0$.

Practice 5:

(a) 20 miles west (from noon to 2 pm) plus 60 miles east (from 2 to 6 pm) is a total travel distance of 80 miles. (At 4 pm the driver is back at the starting position after driving 40 miles = 20 miles west and then 20 miles east.)

(b) The car is 40 miles east of the starting location. (East is the "negative" of west.)

Practice 6: $\Delta \mathrm{x}=\frac{2-0}{\mathrm{n}}=\frac{2}{\mathrm{n}} \cdot \mathrm{M}_{\mathrm{i}}=\frac{2}{\mathrm{n}} \mathrm{i} \text { so } \mathrm{f}\left(\mathrm{M}_{\mathrm{i}}\right)=\left\{\frac{2}{\mathrm{n}} \mathrm{i}\right\}^{2}=\frac{4}{\mathrm{n}^{2}} \mathrm{i}^{2}$ Then

$\mathrm{US}=\sum_{i=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{M}_{\mathrm{i}}\right) \Delta \mathrm{x}=\sum_{\mathrm{i}=1}^{\mathrm{n}} \frac{4}{\mathrm{n}^{2}} \mathrm{i}^{2} \frac{2}{\mathrm{n}}=\frac{8}{\mathrm{n}^{3}}\left\{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{i}^{2}\right\}$

$=\frac{8}{\mathrm{n}^{3}}\left\{\frac{1}{3} \mathrm{n}^{3}+\frac{1}{2} \mathrm{n}^{2}+\frac{2}{12} \mathrm{n}\right\}=\frac{8}{3}+\frac{4}{\mathrm{n}}+\frac{16}{12} \frac{1}{\mathrm{n}^{2}} \longrightarrow \frac{8}{3}$ as $n$ approaches infinity.