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2 years ago

Why is it important that goals be measurable?

Advanced Placement (AP)

1 answer:

Natalija [7]2 years ago

5 0

Answer:

Explanation:

If you have a specific goal in mind you'll be more motivated and know exactly what you're working towards instead of blindly doing something without having a vision for what you want it to look like in the end, and it will make you want to do it more because you know exactly what you need to do to get to that end goal.

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3 years ago

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3 years ago

HEYY I NEED HELP WITH AN AP CALC ASSIGNMENT ASAP

Vladimir79 [104]

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 $c=\frac\pi2$ .

5. This question seems to be incomplete...

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4 years ago