Ups and downs

Accumulating Wind Power

\[ \newcommand{\dnorm}{\text{dnorm}} \newcommand{\pnorm}{\text{pnorm}} \newcommand{\recip}{\text{recip}} \]

Exercise 1 As you know, \[\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx\\ \text{and}\\\int_a^c f(x) dx = - \int_c^a f(x) dx\ .\]

Here are some definite integrals for which, without stating anything more about the function, we give you the numerical result.


$\int_{2}^{7} f(x) \,dx = -8$	$\int_{-6}^{-2} g(x) \,dx = 3$
$\int_{2}^{12} f(x) \,dx = -14$	$\int_{0}^{2} g(x) \,dx = 1$
$\int_{2}^{7} h(x) \,dx = 5$	$\int_{0}^{2} h(x) \,dx = 6$

Consider these the facts you have to work with when answering the following questions:

$\int_{2}^{7} 3f(x) \,dx =$?

$\int_{7}^{12} f(x) \,dx =$?

$\int_{2}^{7} f(x) + g(x) \,dx =$?

$\int_{2}^{2} r(x) \,dx =$?

$\int_{-6}^{-2} \left[\strut g(x)+3\right] \,dx =$?

$\int_{12}^{7} f(x) \,dx =$

Exercise 6 The (so-called) “First Fundamental Theorem of Calculus” says:

\[\partial_t \int_a^t f(x) dx \ = \ f(t)\]

Consider this new quantity: \[\partial_t \int_t^a f(x) dx\] Which of the following is a valid simplification of the quantity?

Exercise 9 Which of the following is NOT equivalent to \[\int_1^4\frac{1}{x}dx\ ?\]

$\ln|x| {\Large|_{\tiny 1}^{\tiny 4}}$

$1.386294$

$\ln|4|-\ln|1|$

$\frac {x^0}{0} {\Large|_{\tiny 1}^{\tiny 4}}$

question id: ape-lie-coat-1

Exercise 12 Active R chunk lst-tent-area-2 gives the function $A(x, y)$ for the differential $A(x,y)\,dy\,dx$ of the surface area of the tent at point (x,y).

Active R chunk 2

This is more than 4 times the ground area of the domain $-5 < x < 5, \ -5 < y < 5$ over which the tent has been raised. You can get some feel for why this is by looking at the several graphs of slices of the tent. The “ground length” in each of these plots is 10 (since they run from $y=-5$ to $5$.) The length of the tent material itself is given by the length of the black curve in each plot.

Modify the code in Active R chunk lst-tent-area-2 to find the surface area of the tent over the domain $3 < x < 5, \ 0 < y < 3$.

1. What is the area (in square meters) of the smaller domain itself? (As opposed to the surface area of the tent raised over that domain. You don't need the codebox for this question.)

```{mcq}
#| label: tent-area-1
#| inline: true
#| show_hints: true
1. 3 [ hint: Area is length times width. ]
2. 6 [ correct hint: Correct The x extent is 2, the y extent is 3, so the area is 6 ]
3. 9 [ hint: Area is length times width. ]
4. 12 [ hint: Area is length times width. ]
```

2. What is the surface area (in square meters) of the part of the tent raised over the small area?

```{mcq}
#| label: tent-area-2
#| inline: true
#| show_hints: true
1. 0 
2. 13 [ correct hint: Right  ]
3. 27 
4. 99 
```

Essay: The ratio of tent surface area to ground area in the small region of the tent is only about 2. For for the tent as a whole, the ratio is much bigger: about 4. Look at the contour plot of the tent height and explain why there is such a big difference in ratios.

question id: tent-area-essay

Activities

Exercise 13

Figure 2: Power produced by the author’s rooftop photovoltaic array in June 4 (solid line) and June 3 (bars), 2022.”

Figure fig-solar-june-c3, previously used in Exercise exr-goldfish-begin-pantry and Exercise exr-fox-sit-kayak, comes from the monitoring app for a photovoltaic (solar electricity) array on two consecutive days in June 2022. The vertical axis has units of kilowatts and dimension [power] L² M / T².

Electric utilities don’t charge for power, they charge for energy, dimension L² M / T, which is typically denominated in units of kilowatt-hours. The local utility charges $0.14 per kilowatt-hour. For each of the two days, what was the money value of the energy produced by the solar array. (Your answer can be rough, but aim to be within $$10%.)

June 3:
June 4:

Hint: Find the value for June 4 first. That will make the June 3 estimate easier.

No answers yet collected


\(\int_{2}^{7} f(x) \,dx = -8\)	\(\int_{-6}^{-2} g(x) \,dx = 3\)
\(\int_{2}^{12} f(x) \,dx = -14\)	\(\int_{0}^{2} g(x) \,dx = 1\)
\(\int_{2}^{7} h(x) \,dx = 5\)	\(\int_{0}^{2} h(x) \,dx = 6\)