Test 5 will be held in class on Friday Dec 3. It will cover sections 5.3, 5.4, 5.5, 6.1 and 6.2. A more detailed description of each section, with points to pay attention to, is provided below:5.3 The Fundamental Theorem of Calculus
- know that the fundamental theorem of calculus applies to integrals of continuous functions, and that the integral from (any constant) to x of a continuous function is, in fact, a differentiable function
- know the two different parts of the fundamental theorem and how they are useful
- Part 1: if a function g is defined as a definite integral from (any constant) to x, then the derivative of g is the integrand evaluated at x
- Part 2: the value of an integral from a to b is the value of an antiderivative of the integrand at b minus its corresponding value at a
- there is a variant of Part 1 that follows from the chain rule: if a function g is defined as a definite integral from (any constant) to h(x), then the derivative of g is the integrand evaluated at h(x) multiplied by h'(x)
5.4 Indefinite Integrals and the Total Change Theorem
- what is an indefinite integral? Answer: the general antiderivative of the integrand. In particular, it must contain an arbitrary additive constant.
- so the value of the definite integral from a to b is the value of the indefinite integral evaluated at b minus its value at a
- know the list of indefinite integrals given on page 402 as well as the two rules (at the top of the table) for finding indefinite integrals of complicated functions in terms of simpler functions
- know the total change theorem and how to use it: the value of the integral of f'(x) (the instantaneous change in f) from a to b is the total change in f, i.e. the value f(b) - f(a). Thus, for example, the total change in displacement is the integral of velocity.
5.5 The Substitution Rule
- know how to simplify an integral by making a substitution u = g(x). This involves two steps:
- write down du = g'(x)dx and use it as a way of eliminating dx in exchange for du
- solve u = g(x) for x, say x = h(y), and use this as a way of eliminating x in exchange for u
- this will leave an integral in u alone to evaluate
- know what to look for in deciding upon an appropriate substitution. This is mainly the internal part of a composite function in the integrand in question
- when you evaluate definite integrals by substitution, take the x integration from a to b and change it to a u integration from c = g(a) to d = g(b)
- know the special results on integrals of even and odd functions
6.1 Areas Between Curves
- this involves the "strip" method for computing areas. The area between two curves y = f1(x) and y = f2(x) is pictured as a collection of very thin strips of width dx parallel to the y-axis and height |f1(x) - f2(x)|. The height*dx is the area of the strip and we add these up by inserting an integral. The specifics of this method can involve:
- finding the intersection points of the two curves so that the limits of integration can be found
- viewing the area in terms of strips of width dy parallel to the x-axis
- dividing the integration interval into a number of sub-intervals depending on the sign of f1(x) - f2(x)
6.2 Volumes
- know the methods for using integrals to compute volumes. These invariably involve "slicing" the volume into disks that are either parallel to the x-axis or the y-axis. Thus the slice method assumes:
- given the area function A(x) of a volume sliced along the x-axis, or
- given the area function A(y) of a volume sliced along the y-axis, or
- given a curve y = f(x) and taking the volume to be a volume of revolution about the x-axis
- given a curve x = f(y) and taking the volume to be a volume of revolution about the y-axis