AB Calculus Cram Sheet

[Courtesy of Will Felder. Edited/extended by EMH 5/10/00, 1/25/01, 8/3/02, 5/8/03, 4/22/04, 2/7/05.]




Trapezoid rule: A Dx (f(x0) + 2f(x1) + 2f(x2) + . . . + 2f(xn 1) + f(xn))


Left endpoint rule: A Dx (f(x0) + f(x1) + f(x2) + . . . + f(xn 1))


Right endpoint rule: A Dx (f(x1) + f(x2) + f(x3) + . . . + f(xn))


Chain rule: Dx(g(u(x))) = g(u(x)) u(x)


Derivative of an inverse: Dx(f 1(x)) = 1 / f (f 1(x))


Parametric chain rule: If y = y(t) and x = x(t), then dy/dx = (dy/dt) / (dx/dt).


Product rule: (uv) = uv + vu


Quotient rule: (u/v) = (vu uv)/v2


Change of base: logb x = (ln x) / (ln b)


Rewriting an exponential: bx = ex ln b


Exponential growth: Diffeq. y' = ky has solution y = cekx


Volume by disks: pr dx if axis of rotation is parallel to x-axis (use dy if parallel to y-axis)


Volume by washers: p(R2 r) dx if axis of rotation is parallel to x-axis (use dy if parallel to y-axis)


Volume by plane slicing A(x) dx if A(x) is an area function for cross sections perpendicular to the x-axis (use dy if perpendicular to y-axis)


Average value of f on [a, b] is (ab f (x) dx) / (b a).


f (x) f (a) + f '(a)(x a)
(This is the so-called linear approximator. Knowing the formula is not nearly as important as being able to apply it, however. In words, we would say, The value of a function at x can be approximated by the value of the function at some nearby known point, a, plus the slope there times the step size that we have taken. A positive step means we are going to the right of our starting point, a, and a negative step means we are going to the left.)

Derivatives and Antiderivatives



Should know derivatives and antiderivatives of all of these function families:


power functions (incl. negative and non-integer exponents)
trigonometric (dont forget sec x dx)


Need to know derivatives (not antiderivatives) for these families:


inverse trigonometric (esp. arctan); arcsin and arccos are good to know

u Substitution



If a function (or some altered form of it) and the functions derivative are both in the integrand, you can often do a u substitution.




x1(3 ln x + 7)11 dx

Let u = 3 ln x + 7, du = (3/x) dx.

Problem becomes
x1(3 ln x + 7)11 dx = 1/3 3x1(3 ln x + 7)11 dx
                                   = 1/3 u11 du
                                   = 1/3 (u12/12) + C
                                   = u12/36 + C
                                   = (3 ln x + 7)12/36 + C




If f is continuous on [a, b],


then " y (f(a), f(b)) [or, wlog, " y (f(b), f(a)) if f(a) > f(b)]
$ c (a, b) ' f(c) = y.


In words: For any intermediate value of a continuous function on a closed interval, there is at least one place in the interior of the open interval (a, b) where that intermediate value is actually attained. (Sometimes known as the Cape of Good Hope Theorem.) Interesting corollary: If f and g are both continuous on [a, b] and their difference is negative at one endpoint and positive at the other, then there is at least one place in (a, b) where f(x) = g(x).




If f is continuous on [a, b],


then $ x1, x2 [a, b] '
f(x1) is the maximum value of f on [a, b] and
f(x2) is the minimum value of f on [a, b].


Alternate (more cryptic) version of the theorem: If f is continuous on [a, b],
then $ x1, x2 [a, b] '
" x [a, b], f(x) f(x1) and f(x) f(x2).


In words: A continuous function on a closed interval (the conditions are crucial) attains its maximum and minimum values somewhere on that closed interval.




If f is differentiable on (a, b) and continuous on [a, b],


then $ c (a, b) ' f '(c) = (f(b) f(a)) / (b a).


In words: There is at least one place where (slope of tangent line) equals (average slope between a and b). Conditions are crucial to know: f differentiable on (a, b) and continuous on [a, b].




If f is integrable on [a, b] and g is any antiderivative of f,


then ab f(x) dx = g(b) g(a).


Equivalent form (sometimes called FTC2):
If h(x) = ax f(t) dt, then h(x) = f(x).

Diff. Cont. Integ.



Differentiability implies continuity, which implies integrability.
However, C does not imply D (think of a cusp).
Differentiability means the left- and right-hand limits of the difference quotient exist and are equal. That does not happen at a cusp.

What does ~C imply? You learned in geometry that D C is equivalent to its contrapositive, namely, ~C ~D. Therefore, if a function f is not continuous at a point, then f is not differentiable there. Reason: A function that is not continuous will not have a difference quotient limit and hence cannot be differentiable.

While continuity is sufficient to imply integrability, the converse is false. That is to say, there are examples of Riemann integrable functions that are not continuous. For example, they could be piecewise continuous with step discontinuities, or they could have asymptotic discontinuities leading to a so-called convergent improper integral. Luckily, you do not have to deal with any of these situations in AB Calculus. Just remember that C I. Of course, just as in the previous paragraph, we can form a valid contrapositive, namely ~I ~C. In English, a function that is not Riemann integrable is not continuous.




Derivative = limit of difference quotient = slope of tangent line = instantaneous rate of change.


A function f is differentiable iff f has local linearity with finite slope. In other words, f is differentiable iff f looks like a straight but nonvertical line when you zoom in closely enough.


Derivative at a point: f ' (c) = limxc [ (f(x) f(c)) / (x c) ]


Derivative function: f ' (x) = limh0 [ (f(x + h) f(x)) / h ]


Definite integral: limDx0 of Riemann sums = limDx0 S f(xi) Dx = ab f(x) dx. Notice how, in the limit, the summation (S) is replaced by the stretchy S and the step size (Dx) is replaced by dx. In order for the definite integral to exist, the sufficient condition you are expected to know for the AP exam is that f is continuous on [a, b]. Although this condition can be weakened somewhat while still preserving Riemann integrability, do not worry about that for the AP exam.


Critical point: f = 0 or DNE.


Stationary point (a.k.a. plateau point): f = 0 but does not change sign.


Local min.: f = 0 and is changing from negative to positive.


Local max.: f = 0 and is changing from positive to negative.


Although it is sometimes useful to check f when checking for max. or min., the second deriv. test is not 100% reliable.
If you have a critical point where f = 0 and f < 0 (negative concavity), then you have a local max.
If you have a critical point where f = 0 and f > 0 (positive concavity), then you have a local min.
However, a local max. or local min. is not necessarily associated with a helpful sign of f . There are many examples of mins. and maxes where f = 0 or DNE, and in those cases, the second deriv. test gives no information.


Inflection point: A point of continuity for f where f changes sign. (Frequently, this is a point where f has a strict local max. or min. However, that is not necessarily the case; for example, think of the origin, which is an inflection point for the function y = f(x) = x1/3 even though f (0) and f (0) are undefined.)


Some students think (wrongly) that point of inflection means f = 0.
True enough, there are many examples where f does equal 0 at a point of inflection. However, there are examples where f = 0 but no inflection point occurs (look at f(x) = x4 at the origin). There are also examples where an inflection point occurs but f 0 (look at f(x) = x1/3 at the origin, or for a more interesting example, look at f(x) = x2 sgn x at the origin).

Bottom line: If you want to know what a point of inflection is, you need to read the correct definition above.


Concavity: A synonym for f .





Techniques for Multiple Choice

Techniques for Free Response

1. Pace yourself. Keep brainpower in reserve for free response.
2. Get the answer any way you can. Work is not graded for multiple choice.
3. Circle the hard ones and come back to them later.
4. If you can positively rule out one or more choices, choose randomly from those that remain. Do not make an educated guess, since you will probably fall into a trap.
5. In an integral problem where a common mistake would be to be off by a factor of 2, look closely at the two choices that differ by a factor of 2. The correct answer is probably one of these.

1. If you cant get part (a), skip it and do the others. Part (a) may be worth only a point.
2. A few lines of accurate work are usually enough. Long, tedious problems are rare.
3. Keep intermediate results in full precision (can use STO to save to a variable). Write ". . ." on paper if you are omitting some digits.
4. Round final answers to 3 decimal places.
5. Show all steps. Dont make leaps of logic. You may use and \ symbols as transitions (e.g., "f diff. at x (given) f cont. at x"), but its easier just to put one thought on each line.
6. Dont waste time erasing large areas. Just mark them out with a quick X.
7. Avoid using the word it.