<p>For that particular question, you can’t actually answer a midpoint Riemann sum with the information given, because you don’t have information about the midpoints of the intervals [1, 2], [2, 3], [3, 5], [5, 8], and [8, 13]. You’d need to know what f(1.5), f(2.5), f(4), f(6.5), and f(10.5) are to answer the question.</p>
<p>It’s one of the reasons why they typically won’t ask a midpoint Riemann sum question using a table.</p>
<p>So apparently this is a thread were we ask TheMathProf Calculus questions and he explains to us, right?</p>
<p>Can you please explain to me what approach I should use for solving volume questions (volumes of revolution / cross section volume questions)? I find these questions really tough without the help of a graphing calculator. Like, how do you know the limits? How do you know what to subtract from what in the integral?</p>
<p>I guess that’s what this is turning into. :)</p>
<p>For cross-section volume questions, the key is to know the area formulas for basic shapes and to be able to then accumulate those areas across the interval. Typically, they’ll give you one piece (i.e. cross-sections perpendicular to the x-axis are diagonals of a square) and then you have to figure out how to use that information to come up with the area.</p>
<p>The limits of integration are going to depend significantly from question-to-question, so some general tidbits to look for:</p>
<ul>
<li>Do they give you an equation of a horizontal or a vertical line? If so, this is either a boundary and/or one of your limits of integration.</li>
<li>In many cases, finding where two curves intersect will give you one of the limits of integration.</li>
<li>Being able to make a quick sketch by hand is immensely helpful.</li>
<li>Keep in mind that with the disc/washer method, when you’re revolving around a horizontal line, your limits of integration have to do with the variable x; when you’re revolving around a vertical line, your limits of integration have to do with the variable y.</li>
</ul>
<p>As far as the volumes of revolution goes, the biggest key is to know which curve is furthest from the axis of revolution? You’ll use that curve to determine R and the other curve (if there is one) to determine r. Keep in mind that you’ll have to adjust in cases where the axes of revolution don’t fall on the x- or y-axes.</p>
<p>If a question gives you a velocity function, and tells you to find the average velocity (say, from 0 to 5), why do you have to take the integral? Shouldn’t it be (v(5) - v(0)) / 5 instead of (s(5) - s(0)) / 5?</p>
<p>A function f has y = 4 as a horizontal asymptote. Why must [the limit of f as x approaches infinity equals to 4] be true? I can picture it as an asymptote with f approaching it as x approaches negative infinity. And why can f(x) = 4?</p>
<p>Does ANYONE here have a T1-89? I need desperate help.</p>
<p>The limit as x–>infinity doesn’t necessarily have to be 4 if it has a horizontal asymptote of y=4. You’re right in that it could be that the function goes to negative infinity. f(x) can equal four. Think of the function with the limit as x–>negative infinity as 4 (the asymptote). Going right, it increases slightly, reaching say, a maximum value of 7 at x=1. Then it decreases so that the limit as x–>positive infinity is negative infinity. Between 1 and infinity, there will be a point c where f(c)=4.</p>
<p>Thanks MathProf, I’ll be referring to that post as I go through volume problems. Thanks again!</p>
<p>And ryanxing, I have a titanium. What exactly do you want help with?</p>
<p>can someone explain the trapezoidal rule?</p>
<p>@ Malfunction: my T1-89 can’t take the indefinite integral of ln(2^x + 1). Why is that?</p>
<p>Also, let’s say I have a velocity function. I want to find the value of the position function at a certain point but only given the velocity function. Is there any faster way to find that value than take the integral of the velocity function and then plug in that x value in the position function? That take an awfully long time. Any way to do both of them in 1 step?</p>
<p>Unrelated to calculator:</p>
<p>I have a velocity function, let’s say v(t) = (x^2 + 1)(cos x)</p>
<p>I am asked to find the greatest distance between particle and origin at a given interval, let’s say between 0 and 5. How do I do that?</p>
<p>You graph v(t). It should be above the x-axis for a certain period of time, and then should cross under the x-axis (almost always does for these types of problems). Find the zero on the right side.</p>
<p>The theory behind this is that the distance a particle travels is the integration of v(t) on a certain interval [0, b]. You need to find the b-value that maximizes the integral value. In the case above, it should just be the zero.</p>
<p>Can someone explain to me how to do 2d and 4d of the 2003 Free Response? I don’t get them at all.</p>
<p>For 2d:</p>
<p>What you really want to do is find where x(t), the position function, has an absolute maximum and absolute minimum, then find the what x is at the absolute min and max (using a similar “strategy” as shown below), then take the absolute value of each of those answers, and the bigger one is where x is furthest from the origin.</p>
<p>For 4d:
The key to this is that they give us f(0) = 3. So…</p>
<p>INT(from -3 to 0) f’(x)dx = f(0) - f(-3)</p>
<p>Your only unknown here is f(-3), since we know f(0) and can calculate the area under the curve using basic math. So all you have to do is rearrange and solve for f(-3). f(4) is found similarly.</p>
<p>I’m still confused about 4d. Could you please elaborate?</p>
<p>You’re basically setting up a definite integral such that you’re only unknown is the value you’re trying to find. If we take the integral from -3 to 0 of f’(x), then</p>
<p>1) We know the value of the integral (since it’s just area under the curve)</p>
<p>2) We know that the integral = f(0) - f(-3), by fundamental theorem of calc</p>
<p>3) We know f(0) because it’s given in the problem</p>
<p>So we know everything needed to solve for f(-3).
(let INT = integral of f’(x) from -3 to 0)</p>
<p>INT = f(0) - f(-3)
f(-3) = f(0) - INT = answer</p>
<p>Oh! Thanks.</p>
<p>For 2d, I’m totally confused about CollegeBoard’s explanation. Could you elaborate?</p>
<p>Here’s what I said we had to do:
</p>
<p>Actually, I think I sort of messed up above (since you can’t find where the abs. max/min is without knowing the abs. max/min), but the general idea is right. Also, I do not have the College Board’s explanation, so I’ll just show you how I would work the problem.</p>
<p>1) Locate possible values of t for absolute extrema for x(t)
Because it’s a closed interval, we must find the absolute extrema. The first step to doing this is locating the critical values for x(t), and this is done by setting v(t)=0, since v(t) is the derivative of x(t). I got t = 0, 2.507.
Now we plug in the endpoints and the critical values into x(t), but since one endpoint is a critical value, we have three values of t to work with: 0, 2.507, 3.</p>
<p>2) Calculate x(t) for each of our three values.
But what is our x(t) equation? We don’t know. But what we can do is set up definite integrals (as we did for 4d) in such a way that we can solve for our unknowns. We’ll only need two integrals, since we already know that for t=0, x(0) = 1.</p>
<p>INT(from 0 to 2.507) v(t)dt = x(2.507) - x(0)
INT(from 0 to 3) v(t)dt = x(3) - x(0)</p>
<p>Rearrange and solve for x(2.507) and x(3). For our three values of t, we now have:</p>
<p>x(0) = 1
x(2.507) = -2.265
x(3) = -1.197</p>
<p>Since they’re asking for the farthest distance from the origin, we can disregard the negative signs. So the answer is (hopefully!) 2.265.</p>
<p>That is the answer. But I can’t visualize what is that. Why does “greatest distance from the origin” imply extrema? Or why does the greatest distance HAVE to be an extrema?</p>
<p>Also, I am given the value of function f at a point, let’s say 5, and f '(5) and f ‘’(5). I am told to fine f(4.5). How do I do that?</p>
<p>How do I find horizontal asymptotes using limits?</p>
<p>horizontal asymptotes are found by dividing the co-efficients. for example in (2x^2+3x)/(3x^2+6x), the horizontal asymptote is at x=2/3</p>
<p>y - y1 = m(x - x1)
y - y1 = f '(5) (x - x1)
Solve for the Y = mx + b equation, and then plug in 4.5 for x to get your answer.</p>
<p>Can anyone help me with number 15 in the AB Part B multiple choice questions found here: <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Now, I’m pretty sure that I have to plug this into my calculator, but when I do I keep getting answer B, but the answer key says that D is the correct answer.</p>
<p>What I’m doing is plugging in the equation into my TI-84 using fnInt like so: fnInt(f(x), x, 0, 2) where f(x) is v(t), x is what I’m taking the integral with respect to, and 0 to 2 are the numbers I’m taking the integral from.</p>
<p>I can’t figure out what I’m doing wrong, so any help would be appreciated.</p>
<p>
</p>
<p>Maybe I can help you visualize this better (although I think the problem might be your thinking of the origin as (0, 0) on an x vs t graph, not a y vs x graph).</p>
<p>Let’s start out with a simple function. How about x(t) = (x-1)^2 - 1. And let’s only look at the range [0, 3]. </p>
<p>Now, let’s graph this as a parametric function, so we can see where the particle is on a “y versus x” graph, rather than a “x versus t” graph. Put your calculator into parametric mode. For x(t) we insert our function above, and for y(t) we type in zero, because for this problem it’s not moving in that direction.</p>
<p>Now, staying at the “Y=” part of your calculator, move the cursor over to the FAR left. You should have it highlighted on a diagonal line. Hit enter twice; the diagonal line should now have changed into a “0” with a line next to it - looks something like “-0”.</p>
<p>Next, let’s set our window. Set Tmin=0, Tmax=3, Tstep=0.05, Xmin= -2, Xmax= 6, Ymin= -3, Ymax= 3.</p>
<p>Finally, let’s graph the function. Hit “Graph” and watch carefully (if you need to watch it again, go back to “Y=”, highlight the equals signs, and press enter twice). What we’re watching is the actual motion of the particle (the ‘O’ represents our particle) on a real y vs x graph.</p>
<p>Now can you see where the particle is “furthest from the origin”? It’s furthest from the origin when the “x” values is largest. So how do we find when the x-value is the largest? Find absolute extrema of x(t).</p>
<p>Hopefully that helps you understand the question much more easily than before.</p>