Need help on Calculus AB

Test Preparation AP Test Preparation
1

<p>1) Do the following topics appear on the AB exam?
- Differential equations involving growth and decay
- Derivative and Integral of Inverse Trig functions
- Hyperbolic functions
- Area of a region between two curves/between intersection curves
- Arc Lengths
- Surfaces of Revolution
- Work done by a constant/variable force
- Fluid pressure and fluid force
- Moments, centers of mass, centroids
- Partial fractions</p>

<p>2) I need help on some integration problems. =O</p>

<p>*This is not homework; I'm self-studying calculus.</p>

<p>a) ∫ (√ x) / [(√ x) - 3] dx
b) ∫ [(1+ ln x)^2] / x dx; definite integral from e to 1
c) ∫ (x^2 - 2)/ (x+1) dx; definite integral from 2 to 0</p>

2

<p>Topics that appear from the list are:
-Differential Equations involving growth and decay
-Derivative and Integral of Inverse Trig Functions
-Area of a region between two curves/between intersection curves</p>

3

<p>For b, you do u-substitution with u being (1+ln x). du is (1/x) dx and you go from there.</p>

4

<p>For c, I think you divide using long or synthetic division, and what you get should be integratable (is that a word?)</p>

5

<p>thanks.</p>

<p>I’m also confused about some e^x questions.</p>

<p>1) derivative of 2/(e^x + e^-x)
2) indefinite integral of (e^-x)/(1+e^-x) dx</p>

6

<p>1) Factor out the 2 and raise the denominator to the -1 power. Then use chain rule. Sure, you can do quotient rule, but that would be a pain.</p>

<p>2) U-substitution. u is the denominator, and du is the numerator times -1.</p>

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<p>I may be wrong, but if partial fractions aren’t on AB then 2C in the OP is beyond the course.</p>

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<p>I am self studying too! Message me? I want to know how it is going!!!</p>

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<p>From what I know, partial fractions aren’t in the AB curriculum.</p>

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<p>byubound is right. Partial fraction isn’t part of the AB curriculum.</p>

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<p>Neither are arc length, improper integrals, integration by parts, and I don’t think there are parametric equations either. Not to mention there are no series.</p>

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<p>^all of those are BC materials, just to confirm.</p>

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<p>

</p>

<p>Not in this case. x^2 - 2 = (x+1)(x-1) - 1, so (x^2 - 2)/(x + 1) = x - 1 - (1 / (x+1)). All of these terms are integrable without partial fractions. :)</p>

<p>Partial fractions would require a denominator that can be factored (this denominator cannot be) and that cannot be factored using other techniques (i.e. u-substitution).</p>

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<p>Can someone explain Ce^kt and its relationship to differential equations? I read the textbook but don’t quite get it.</p>

<p>For anyone who uses T1-89, I am not able to graph any graph with an asymptote (1/x for example). Why is that happening? Also, how do I implicitly differentiate on the calculator?</p>

<p>Could someone explain the midpoint formula for Riemann Sums?</p>

15

<p>I can hit two of those three:</p>

<p>If you were to have a differential equation of the form dy/dt = ky, and you were to solve that equation for y that doesn’t involve dy/dt, you would end up with Ce^kt by separating the differential equation as follows:</p>

<p>dy/dt = ky
dy/y = kdt
ln |y| = kt + A (where A is a constant).
e^(ln |y|) = e^(kt + A)
|y| = e^(kt)*e^A
y = +/- e^A * e^(kt)
y = Ce^(kt), where C = e^A or C = -e^A, depending on the initial condition.</p>

<p>As far as the midpoint formula for Riemann sums, what you’re essentially doing is dividing the region up into the specified number of subintervals, and finding the area of rectangles to approximate the are under the curve. For the width, you’re using the width specified, and for the height, you’re using the midpoint (for the x values) on each subinterval to approximate the height of the curve over the entire interval. Conversely, the left Riemann sum uses the left endpoint of the subinterval and the right Riemann sum uses the right endpoint of the subinterval.</p>

<p>So, for instance, if I wanted to use a midpoint Riemann sum to approximate f(x) using 4 subintervals of equal size over [1, 13]…</p>

<p>The entire interval is 12 units wide, so each subinterval is 3 units wide.</p>

<p>The first subinterval spans [1, 4], so that rectangle formed has a midpoint approximation of 3<em>f(2.5), (since 2.5 is halfway between 1 and 4).
Similarly, the three other subintervals have areas of 3</em>f(5.5), 3<em>f(8.5), and 3</em>f(11.5). Sum these four areas to get an approximation of the definite integral.</p>

<p>Unfortunately, I’m not really familiar with the TI-89, so I can’t help you with the middle question.</p>

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<p>how about product of sum formula?</p>

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<p>Probably not…</p>

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<p>@TheMathProf: Thanks. Now that makes some sense to me. Could you do an example with the Right and Left approximation (preferable the hardest type of question involving it on the AB exam)? Could you also do an example with the 2nd Fundamental Theorem? I’m really confused on how to show work using that theorem.</p>

<p>Anyone knows why my T1-89 can’t graph functions with asymptotes?</p>

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<p>Here goes with the Riemann sum piece:</p>

<p>Consider a function f(x) with the following points: (1, 3), (2, 6), (3, 10), (5, 15), (8, 21), (13, 28).</p>

<p>All Riemann sums would consider the five intervals [1, 2], [2, 3], [3, 5], [5, 8], and [8, 13].</p>

<p>When performing a left Riemann sum, you’ll use the left endpoints:</p>

<p>[1, 2]: width = 1, height = f(1) = 3, area = 3
[2, 3]: width = 1, height = f(2) = 6, area = 6
[3, 5]: width = 2, height = f(3) = 10, area = 20
[5, 8]: width = 3, height = f(5) = 15, area = 45
[8, 13]: width = 5, height = f(8) = 21, area = 105
Left Riemann Sum = 3 + 6 + 20 + 45 + 105 = 179</p>

<p>Conversely, a right Riemann sum uses the right endpoints:</p>

<p>[1, 2]: width = 1, height = f(2) = 6, area = 6
[2, 3]: width = 1, height = f(3) = 10, area = 10
[3, 5]: width = 2, height = f(5) = 15, area = 30
[5, 8]: width = 3, height = f(8) = 21, area = 63
[8, 13]: width = 5, height = f(13) = 28, area = 140
Right Riemann sum = 6 + 10 + 30 + 63 + 140 = 249</p>

<p>They’ll rarely ask you to compute both in the same question.</p>

<p>Sometimes, they’ll ask you whether that Riemann sum is an overestimate or an underestimate of the integral. In this particular case, since f(x) is an increasing function, the Left Riemann sum is an underestimate (since there are parts of f(x) under the curve that are not being considered for the area), and the Right Riemann sum is an overestimate (since there are parts that aren’t under f(x) that are being considered for the area).</p>

<p>As far as the 2nd Fundamental Theorem, I’m presuming you’re referring to the d/dx[Integral(a, x) of f(t) dt] = f(x) piece (I’ve seen this referred to as the Fundamental Theorem or the Second Fundamental Theorem, depending on the text). You actually rarely need to show work with these kinds of problems. The ones that I’ve found that most students struggle with has to do with when they give you a graph of a function f(x), and they tell you that g(x) = integral(0, x) of f(t) dt, and then they ask you questions about what g is doing.</p>

<p>The key to these questions is to consistently connect what you’re looking for in g to what they give you in f.</p>

<p>For instance, if they ask you when the graph of g has a relative maximum, well, we know g has a relative maximum when g’ changes from positive to negative. Since g’ = f, we look for where the graph of f changes from positive to negative, and that tells us where g has a relative maximum.</p>

<p>Similarly, if they ask for where g has a point of inflection, we know that g has a point of inflection when g" changes signs. We know that g" = f ', and f ’ changes signs when the graph of f changes either from increasing to decreasing or from decreasing to increasing. So we identify those points.</p>

<p>Hope that helps.</p>

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<p>So MathProf, what would the midpoint Riemann sum be?</p>