<p>
</p>
<p>You can do it with the stated domain without looking at the graph; the graph only supports what the domain says.</p>
<p>(Why exactly do you want to do it without the graph, though?)</p>
<p>On the m/c there are no graphs</p>
<p>
</p>
<p>As far as I can tell, it’s impossible to do this problem if you have neither the domain nor the graph. At the very least, with the domain you can imagine a rough facsimile of the graph (it starts at 0 and ends at pi, so logically the integral should be 0 and pi.)</p>
<p>But like I said earlier, usually on MC the graph is something relatively simple, like cosθ. The formula for this area given in MC will usually involve symmetry, so make sure to keep that in mind.</p>
<p>does anyone remember the formula for the disk, washer, and shell methods?</p>
<p>i know the disk and washer is pi intergral (R^2-r^2), but is the shell method 2 pi integral R times h?</p>
<p>also what are the domains for the polar graphs? the circle things, cardioids and roses?</p>
<p>(2pi)fnInt[(x)(height)]</p>
<p>x or y is used interchangeably from the axis of rotation.</p>
<p>I’m pretty sure shells aren’t required on the AP exam any more…?</p>
<p>they’re not but its helpful to use them sometimes</p>
<p>All right, so I’ve been working on these few problems, and it’s rather late, but if someone can just walk me through any of these… I know the answer, I’m just looking for explanation. </p>
<ol>
<li><p>What is the value of (-2/3) ^ n summed from n = 0 to n = infinity?
The answer is 3/5</p></li>
<li><p>The function f is twice differentiable, and the graph of f has no points of inflection. If f(6) = 3, f’(6) = -1/2, and f’'(6) = -2, what is f(7)?
Ans: 2</p></li>
<li><p>The radius of convergence for the power series [(x-3) ^ (2n)]/n from n = 1 to n = infinity is equal to one. What is the interval of convergence?</p></li>
</ol>
<p>2 < x < 4</p>
<ol>
<li>How do you know if any of these converge:
(All of them are the sums from n = 1 to n = infinity):
1 / sqrt of n </li>
</ol>
<p>(3^n) / n!</p>
<p>(e/pi) ^ n</p>
<p>The last two converge…</p>
<p>Oh man, I am not ready for this test.</p>
<h1>1 is geometric, so you can use the equation s = a/(1-r)</h1>
<p>a= the first term
r= (-2/3)</p>
<p>
</p>
<ol>
<li>Key for this is recognizing that (-2/3) ^ n from 0 to infinity is a geometric series. From that, you remember that the formula for the value of a geometric series is k/(1-(x)) for a series in format x^n, where k is the first term of the series. Therefore, this series = 1/(1+(2/3)) or 3/5.</li>
</ol>
<p>For 3, just use the Ratio Test.</p>
<p>4)
1/sqrt of n is 1 / n^(1/2)
Use p-series test. Since 1/2 = p, series diverges.</p>
<p>{(3^n)(3)/(n+1!)}/{(3^n)/(n!)} < 1
This is the Ratio Test and it is = to 3 if you take the limit n–> inf so it must diverge…? **Am I doing something wrong?</p>
<p>(e/pi)^n
e/pi is < 1 so it converges(geometric)</p>
<ol>
<li>1/(n^1/2) is a pseries p=1/2 which is less than one, thus it diverges.
I think you need to use the ratio test for (3^n)/n!
e/pi is geometric, and it converges because e<pi, so r is less than 1</li>
</ol>
<ol>
<li>For 1/sqrt of n, use the p series where p is 1/2. since p is less than 1, the series diverges.</li>
</ol>
<p>For 3^n/n! use the ratio test and find that the value is less than 1, so it converges.</p>
<p>For e/pi realize that it is a constant to the n power, and the constant is less than (2.7183/3.14159265) than 1, so by the geometric series, it converges.</p>
<p>Hey guys, sorry it’s last moment, but does anyone now how to do 2008 question 86?</p>
<p>
</p>
<p>The answer choices are 6,7,8,9, and 10.</p>
<p>
</p>
<p>Well, the first statement basically means that dy/dx = 2x. Therefore, we can integrate that to get our y in terms of x: y = x^2 + c. To find c just plug in the point we’ve been given, (2,3). This gives us 3 = 4 + c, so c = -1. Thus the final equation is y = x^2 - 1. To find f(3) then just substitute 3 for x, to get 8?</p>
<p>Big thanks, guys!</p>
<p>Okay so … .dy/dx = slope = 2x which which means that we get x^2 + C and plug in to get C = -1 so the equation is x^2 - 1. If x = 3, 9-1 = 8 got it.</p>
<p>heinochus = you need x squared not 2x.</p>
<p>8 is indeed right. I was missing the part about putting dy/dx back in an equation form.</p>
<p>(3^n) / n!</p>
<p>Can someone find whether this is divergent or convergent using the Ratio Test? I’m getting 3 which tells me divergent, but the answer is convergent apparently…</p>
<p>Hotinpursuit: I get 3/(n+1) at the end which is zero. Convergent.</p>
<p>EDIT: Be sure the n! / (n+1)! simplifies to (n+1).</p>
<p>Using the ratio test, you get lim as n approaches infinity of 3/n+1, which gradually becomes zero. According to the ratio test, this is <1 and converges.</p>