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zepelin [54]
3 years ago
11

c) Choose one person, event, or movement from an earlier period of American history (prior to 1820) and briefly explain how that

person, event, or movement did or did not fit what Emerson said of "the former generations."
Advanced Placement (AP)
1 answer:
Free_Kalibri [48]3 years ago
3 0

Answer:

One person could be Emerson related to Alice Cary, both were writers and poets who have lot of influence in the historynof the United States. He was the leader of the Trascendentalist thinking.

Explanation:

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HEYY I NEED HELP WITH AN AP CALC ASSIGNMENT ASAP
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CALCULATOR PART

1. The area of R + S is unsigned, meaning you want to find

\displaystyle\int_a^b\left|f(x)-g(x)\right|\,\mathrm dx

where [a,b] is the interval between the leftmost and rightmost intersections of f and g.

First use your calculator to find these intersections:

\cos x=\dfrac{x+1}3\implies x\approx-3.64,x\approx-1.86,x\approx0.889

so that a=-3.64 and b=0.889. Now compute the integral using your calculator:

\displaystyle\int_a^b\left|f(x)-g(x)|\,\mathrm dx\approx1.662

2. The volume, using the washer method, is given by the integral

\displaystyle\pi\int_{-1.86}^{0.889}(|2-g(x)|^2-|2-f(x)|^2)\,\mathrm dx\approx12.078

3. A circle of radius r has area \pi r^2; a semicircle with the same radius thus has area \frac{\pi r^2}2. Each cross section of this solid is a semicircle whose diameter is the vertical distance between f(x) and g(x), or |f(x)-g(x)|. In terms of the diameter d=2r, the area of each semicircle would be \frac{\pi d^2}8. Then the volume of the solid is

\displaystyle\frac\pi8\int_{-3.64}^{-1.86}|f(x)-g(x)|^2\,\mathrm dx\approx0.0425

NON-CALCULATOR PART

4. The mean value theorem says that for a function F continuous on an interval [a,b] and differentiable on (a,b), there is some c\in(a,b) such that

F'(c)=\dfrac{F(b)-F(a)}{b-a}

If this F happens to be an antiderivative of f, then we end up with

f(c)=\displaystyle\frac1{b-a}\int_a^bf(x)\,\mathrm dx

\cos x is continuous and differentiable everywhere, so the MVT applies. We have F'(x)=f(x)=\cos x, so the MVT tells us there is some c\in[0,\pi such that

\cos c=\dfrac{\sin\pi-\sin0}{\pi-0}=0

That is, the average value of f(x) on [0,\pi] is 0. The MVT says there is some c in the interval such that the function takes on the average value itself; this happens for c=\frac\pi2.

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