Disclaimer. I do type things in a quite hasty manner, so I tend to make mistakes. If I do, you should discuss with me to figure out what the correct approach should be. If you merely copy what I say without understanding, it will backfire you on your exam!
For #1, there are notations that are not available in the reading. This is unfortunate, but for this, read the beginning of Section 7.5 of your book (more specifically, understand the notations $\mathrm{LEFT}(n)$ and $\mathrm{RIGHT}(n)$ on p.377). Then take a look at the problem again.
One thing you need to remember is that for POSITIVE function $f(x)$ on an interval $[a, b],$ we have the following phenomena.
(1) If $f(x)$ is increasing (i.e., $f'(x) \geq 0$ for all $x$) and positive (i.e., $f(x) > 0$ for all $x$), then $$\mathrm{LEFT}(n) \leq \int_{a}^{b} f(x) dx \leq \mathrm{RIGHT}(n).$$
(2) If $f(x)$ is decreasing (i.e., $f'(x) \leq 0$ for all $x$) positive (i.e., $f(x) > 0$ for all $x$), then $$\mathrm{RIGHT}(n) \leq \int_{a}^{b} f(x) dx \leq \mathrm{LEFT}(n).$$
Exercise. Explain why (1) and (2) should be true as stated (Hint: draw a picture).
Exercise. What happens to (1) and (2) if $f(x)$ is negative?
Hints for #1. For #1 (a) and (b), try to violate the statements by drawing a good picture. If you realize you cannot violate one at all, chances are that the statement is true. Otherwise, you will have a counterexample to say that the statement is false.
For #1 (c), I would think of $h(x) = g(x) - f(x)$ as a single function. Then
- $h(2) = g(2) - f(2) = 0$ and
- $h'(x) = g'(x) - f'(x) > 0$ for all $x,$ so $h(x)$ must be strictly increasing.
Our integrand is $xh(x)$ and its derivative is $h(x) + xh'(x)$ by the product rule.
For #1 (d), we have $h(t) = G(100 - t^{2}),$ for a suitable choice of $G(t).$ By chain rule, we have $$h'(t) = -2tG'(100 - t^{2}).$$ Use 2nd FTC to find $G'(t)$ and replace $t$ with $100 - t^{2}$ to compute $G'(100 - t^{2}).$
Hints for #2. For #2 (a), recall the definition of the average value. Since this appears in the team homework, maybe the definition will appear on the exam. (This is a mere guess, so please don't count on me.)
For #2 (b), we just try our guess: how about $$G(t) = \int_{0}^{t}g(x) dx.$$ Since 2nd FTC gives $G'(t) = g(t),$ you might feel like this is correct. No! We are supposed to have $G(0) = 18.4.$ (Do you see why?) How do we fix one thing about our wrong answer to make it correct? Think about it.
For #2 (c), let $c$ be the number (amount) of wasps that Conner brings each day (as I think the problem seems to suggest that he brings them back and forth--I could be wrong), so $200c$ grasshoppers will be killed each day. The total number of grasshoppers at the time $7 \leq t \leq 10$ will be $R(t) = G(t) - 200c(t - 7).$ (Can you see why?) We want $R(10) = 0,$ so using this (with the information given by the graph), deduce what $c$ ought to be.
For #2 (d), note that $R(t)$ is given for $7 \leq t \leq 10.$ Noting that Conner brings no wasps for the time $0 \leq t < 7,$ think about what $R(t)$ should be. (Hint: this should be easy to you.) For concavity, you might want to recall your knowledge about the second derivatives, which is available on p.196 of your book.
Hints for #3. First, you need to retain your Calculus 1 knowledge to pass your first Gateways, so say that's our motivation to stop Darth Integrator.
For #3 (a), note that $p(x)$ looks quite continuous, so it should be continuous at $x = 3/4.$
For #3 (b), let $T$ be the area of the triangle with the vertices $(0, 0), (c, 0), (c, p(c)).$ Then the problem is saying that $$\frac{1}{7}\int_{0}^{5/2}p(x)dx = T + \int_{c}^{5/2}p(x)dx.$$ Using the explicit descriptions about $p(x)$ given in #3 (a) (with your answer for specific $a$), you can compute the integrals above, and presumably it will give you an equation in $c.$ Then you should try to solve that equation to get $c,$ but since the problem is saying you should approximate, maybe it will be hard to do solve that equation in $c$ with your hand. (Then use the calculator.)
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