interval notation: the types of intervals

there are a few different types of intervals that commonly arise when studying math, called the open interval and the closed interval, notated respectively as (a, b) and [a, b].

interval notation for open intervals

the open interval uses parentheses, and they signify the fact that the interval contains all the real numbers x that are strictly between the numbers a and b, i.e. the interval does not actually contain the numbers a and b. another way of notating an open interval is the set of all x such that a < x < b.

interval notation for closed intervals

in the case of the closed interval, the square brackets are used to indicate that the endpoints are contained in the interval. therefore we can notate a closed interval as the set of x so that

interval notation for half-open intervals

there are slightly fancier intervals, called half-open intervals, notated as (a, b] and [a, b), which are the respective sets of all x so that , and .

an interval is called bounded when there is a real positive number m with the property that for any point x inside of the interval, we have that |x| < m.

observations on intervals

supposing as in the setup that a < b, then how many numbers are actually in the interval (a, b)? it turns out that there are uncountably infinite numbers in any interval (a, b) where a < b, no matter how close a and b are together.

it is a fact that actually, there are the same quantity of real numbers in the interval (0, 1) as there are in the entire real numbers, also represented by the interval . this seems counterintuitive, because one interval seems so much more vast than the other, but it is not a contradiction, but rather a beautiful subtly of set theory.

calculus and intervals

intervals arise regularly in calculus, and it will be important for you to know the difference between a closed interval and an open interval, since there are some theorems, like the intermediate value theorem, which requires that the interval upon which the function is defined is a closed and bounded interval.

closed and bounded intervals touch on one of the most important concepts in the broader study of calculus, that of compactness. many central theories in calculus revolve around compact sets, which in the setting of the real numbers are exactly the closed bounded intervals.

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free-response definite integrals:

you will not commonly be asked to evaluate common definite integrals on the free-response, but rather you will be asked to find an area or compute a volume, which will require computing a common definite integral. suppose we want to compute the volume of the solid obtained by revolving the function about the x-axis:

the cross sections when cutting perpendicular to the x-axis are circles with radius given by the function . the definite integral that needs to be evaluated is ,  since this is the area of a circle multiplied by the length of the interval from -6 to 6. we compute:

therefore to compute the integral we compute the sum of the integrals of the individual terms, since polynomials are sums of continuous functions:

recall the fundamental theorem of calculus (ftc):

theorem: if v(x) is a continuous function with an antiderivative v(x), then  where ,  are in the domain of v(x). 

the ftc says that we can pick any old antiderivative v(x) for v(x), so we need to compute a string of antiderivatives for the integrands of the terms in the sum. in the previous post we discussed but did not state:

the power rule: the derivative ‘=

we used this to find that the integral , and since we only need one antiderivative to evaluate definite integrals, we can take  for use in this case.

therefore we can evaluate (using the fact that , ,  and the ftc):

you can use your calculator to get 723.823 units cubed.

multiple-choice definite integrals:

here is a sample of a typical multiple-choice question asking for you to formulate a definite integral based on the same concept discussed above.

question: a solid is generated by revolving the region enclosed by the function , and the lines x=2, x=3, y=1 about the x-axis. which of the following definite integrals gives the volume of the solid? (hint: draw a picture)

the idea for this problem is to recognize that this solid is a difference of integrals. suppose that we had the volume of the function  when bounded by the lines  x = 2, x = 3, and rotated about the x-axis—then we would have the volume of the following solid:

given this volume, we would only need to subtract the volume of the following figure, derived by rotating y=1 bounded by x=2, x=3, about the x-axis:

from the upper volume, with radius :

therefore we need to subtract the two integrals, however using the integral laws we can express this in the form , which we follow up by substitution of our names for ,:

so the answer is a.
to compute the value of the integral we see that

this has the value .

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it also turns out that the structure of the ab and bc exams are the same, the only difference is question type.

ab free response: half of your ap score.

yes my friends, you heard it right, the 90-minute ap calculus ab free response is worth 50% of the test, according to the college board. there are six total questions broken into two parts: part a and part b.  you may use either a pencil or pen with black or dark blue ink.

the questions on the free response are weighted equally, therefore each question is 16.66% of your total score on the ab free response, and 8.33% of your total score on the exam. however, most questions have multiple parts, and they might be weighted differently, you don’t know. to get a feel for the length of questions you might want to try some practice problems.

do i need a graphing calculator for the ap calculus ab free response?

part a is 30 minutes long and consists of two questions. you need a graphing calculator for this part of the exam. part b is 60 minutes long and you may not use a graphing calculator; however, you can work on the problems from both part a and part b.

you might need to use your calculator to graph a function that you could never plot by hand, or integrate difficult integrands, so do your best to make sure that you have it when you take the test, and it has fully charged batteries.

in fact you can bring up to two calculators to the ap exam, but no more than that.

ab free response time management

the calculator on part a combined with no calculator on part b but the ability to go back and also work on part a problems is a curveball. it would probably be wise to try and complete the new questions in part b before returning to the questions in part a, since you are now not allowed to use the graphing calculator that the questions require.

make sure to read all the questions in the part you are working on before actually starting to solve anything, since you can do the easiest questions first (remember that each question has equal weight, so you should do the easy ones first.)

showing work for the ab free response.

distinct from the multiple choice questions on in section i of the ap exam, you need to show all of your work! you need to label all objects in a graph, such as the function you are plotting, specific points (such as inflection points or maximums and minimums,) axes, tables of values, etc.

the ap course description says that “[a]nswers without supporting work will usually not receive credit.” in addition, when you justify a solution, it means that you state “mathematical (noncalculator) reasons” for why a particular solution is correct.

you also have to use the standard mathematical notation, and not any type of programming, computational, or calculator syntax. the silver lining, however, is that you do not need to simplify your answers, and if you supply a decimal answer, it needs to be correct to 3 places after the decimal point

summary of the ap calculus ab free response.

the ap calculus ab free response can be a bit intimidating since you need to do something more than just correctly bubble in the correct answer, but if you know the structure of the ab free response and the requirements, you are one step closer to successfully navigating this portion of the exam.

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it might seem like there is clearly a difference, but sometimes it’s hard to articulate into words.

we have spoken about the definition of a function. simply put, it’s a rule that transforms one real number into another real number. a graph is a geometric representation of that rule.

a graph is a set.

if this is true, then according to the definition of a set, a graph is an unordered collection of objects. for this lesson, you need to know a little more about sets: the cartesian product of two sets a and b is again a set, denoted  a x b and read “a cross b.” it is the set of all elements of the form (a, b) with 

the cartesian product is named after the famous french philosopher rené descartes:

the points in a graph

the objects in the graph of a function are points  ordered pairs of real numbers in the cartesian product of the set of real numbers with itself.  we call these points cartesian coordinates.

we represent these points geometrically in what is known as the cartesian plane, or simply the plane:

the center of the plane, the point (0, 0) is called the origin, the horizontal axis is called the x-axis while the vertical axis is known as the y-axis. the two axes naturally divide the plane into quadrants. if a point (x, y) has both x, y positive, we say that (x, y) lies in the first quadrant; if (x, y) has x negative and y positive, then (x, y) lies in the second quadrant, etc.

the graph of a function

given a function f whose domain is the set of real numbers and whose codomain is the set of real numbers, we say that the graph is the set of all points in the set  of the form (x, f(x)) where x is a point in the domain of f.

therefore the graph is a set that is unique for a given function, which geometrically represents the function. we can often use the graph of a function in order to deduce properties of said function.

conclusion

the graph of a function and a function are closely related but not the same. therefore when you are explaining a solution to a problem, make sure that you use “the function” and “the graph of the function” in the right places, depending on which you really mean.

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question: a solid is generated by revolving the region enclosed by the function , and the lines x =2, x =3, y =1 about the x-axis. which of the following definite integrals gives the volume of the solid? (hint: draw a picture)

calculating the antiderivatives of an integrand

when we look at an integral we call the values a, b the bounds of integration, while we call the function f(x) the integrand. the goal when computing an integral is to first compute an antiderivative f(x) for the integrand, for then the fundamental theorem of calculus will allow us to evaluate the right side of  , which is much easier to evaluate than taking a limit of riemann sums as discussed in the previous post.

therefore we need to learn how to compute antiderivatives in order to learn how to compute integrals.

the method for computing antiderivatives of a function f(x) usually consists recognizing it as the derivative of some function that you know f(x).

we need a little background now, because we are going to develop a toolkit for computing antiderivatives. a set is an unordered collection of objects. we can describe the elements of set inside of curly brackets: the integers , , the real numbers are everything in between. when there are no bounds of integration on an integral, we call an indefinite integral. the indefinite integral of a continuous function is the set of antiderivatives  for the integrand  so  (the set notation is read as “the set of f such that f is an antiderivative for f.”

so there is more than one antiderivative for a continuous function f? actually, the number of antiderivatives for a continuous function is actually uncountable, and certainly infinite. to see this we need two laws for derivatives: the derivative of a sum of differentiable functions is the sum of the derivatives, and the derivative of a constant function is 0. recall that a function f(x) is differentiable if the limit  exists for every point x in the domain of f, and we write this value as f'(x).

let f(x), g(x) be differentiable functions, then we can see using the definition of the derivative that

next, we want to suppose that our function for some i.e. for some real number c,) then the derivative is zero at every point x:

now we see why there are so many antiderivatives of an arbitrary continuous function f(x): let f(x) be any antiderivative of f(x). this means that f'(x) = f(x) for every x in the domain of f. but then f(x) + c  where c is a real number is also an antiderivative for f , since we have that

where the first equal sign uses the sum rule for derivatives and the second equal sign uses the fact that the derivative of a constant function is zero.

therefore we write that the indefinite integral of a continuous function is a set of antiderivatives that differ by a real constant:

solution to warm-up: a

calculating the antiderivatives of a polynomial

warm-up:

(find the solution at the conclusion)

today we want to discuss the computation of the antiderivatives of a polynomial function. polynomials, as you might recall, are sums and differences of different powers of the independent variable. some functions that are polynomials:

some functions that are not polynomials:

notice that h, g, are very similar functions to h, g, so it is important to verify if you are dealing with non-polynomial functions (like ) or simply constants (like ). we now want to understand how to compute the indefinite integral of one of these polynomial functions—which means finding antiderivatives of polynomial functions.

in order to understand antiderivatives, we must first understand derivatives—so the first question we ask is:

what is the derivative of the function f(x) = ?

we can answer this question by computing the limit from the definition:

now what is ? if we recall the binomial formula (and pascal’s triangle) we see that 

substituting this back into the limit we see that:

since all the terms in the numerator have a power of “h”, then we can cancel and are left with:

now since the only term without an “h” is the first term, when  we see that all that we are left with is . therefore we have that

we now have figured out the derivatives of all sorts of functions! we have that ‘ = ‘ = etc. the last pieces of information that we need to know, which i will not prove because it requires advanced structures, is that the integral of a sum of continuous functions is the sum of the integrals. therefore if f(x), g(x) are continuous:

and given a real number  and a continuous function f(x), the integral of (c.f)(x) is c times the integral of f(x):

this allows us to compute the integral of a sum of polynomial functions, since computing the integral will consist of finding antiderivatives for each term in the sum and then adding them together. this means we only need to know the antiderivative of . since a polynomial takes the form , where the  is a sequence of real constants, we can compute the integral of one as follows:

so what is the function f that is an antiderivative of  ?  it will have the form , and we can solve for  by differentiating, so that . therefore we have that

therefore the antiderivatives for are as follows:

where c  is an arbitrary real constant.

warm-up solution: c

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