7 Using FTC to Find AreaExample 3 in the text: Find the area of the region bounded by the graph ofy= 2x2-3x+2, the x-axis and the vertical lines x=0 and x=2Note that all of the function is above the x-axis in the intervalStart by integrating the function over the closed interval [0,2]Then find the antiderivative and apply the FTCSimplify.
10 Absolute Value Integration Example: Example 2 in the text: keep in mind the definition of absolute value and where your function is positive and where it would be negative:From this, you can rewrite the integral in two parts:
14 Definition of the Average Value of a Function on an Interval and Figure 4.32
15 Let’s get started:You already know about functions and how to take the average of some finite set.Today we’re going to take the average over infinitely many values (those that the function takes on over some interval)…which means CALCULUS!Before I show how to do this…let’s talk about WHY we might want to do this?
16 Consider the following picture: How high would the water level be if the waves all settled?
17 Okay! So, now that you have seen that this an interesting question... Let’s forget about real life, and...Do Some Math
18 Suppose we have a “nice”function and we need to find its average value over the interval [a,b].
19 Let’s apply our knowledge of how to find the average over a finite set of values to this problem:First, we partition the interval [a,b] into n subintervals ofequal length to get back to the finite situation:In the above graph, we have n=8
20 Let us set up our notation: comes from the i-th intervalNow we can get an estimate for theaverage value:
31 Finding the x values where we GET the average value of the function.NOW take the equationand it’s average valueover the interval [0,2], average f = 5and set that equal to the function…And then solve for x…NO PROBLEM!!!!the answer is…
32 Mean Value Theorem (find the value of x that gives you the average value of your function)
33 Average Value of a Function Definition:If f is integrable on the closed interval [a,b], then the average value of f on the interval is:
35 Mean Value TheoremIf f is continuous on the closed interval [a,b], then there exists a number c in the closed interval [a,b] such that:
36 Mean Value Theorem Example: Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval:Now we need to find the x coordinate where we get this average y value:
37 MVT-IThe mean value theorem does not specify how to find c, just that there exists at least one number c in the interval that will give you the average value of the function.
40 Second Fundamental Theorem of Calculus Earlier you saw that the definite integral of f on the interval [a,b] was defined using the constant b as the upper limit of integration and x as the variable of integration. However, a slightly different situation may arise in which the variable x is used as the upper limit of integration. To avoid the confusion of using x in two different ways, t is temporarily used as the variable of integration
41 Theorem 4.11 The Second Fundamental Theorem of Calculus
43 The Definite Integral as a Function You can think of the function F(x) as accumulating the area under the curve f(t)=cost from t=0 to t=x. For x=0, the area is 0 and F(x)=0. for x=pi/2, F(pi/2)=1 gives the accumulated area under the cosine curve on the entire interval [0,pi/2]. This interpretation of an integral as an accumulation function is used often in applications of integration. See p. 288 for a graphical example of this.
45 The Second Fundamental Theorem of Calculus If f is continuous on an open interval I containing a, then, for every x in the interval,The proof of this is on p. 289 in the text.
46 Using the SFTCNote that is continuous on the entire real line. So, using the SFTC you can writeThe differentiation shown in tthis example is a straightforward application of the SFTC. The next example shows an application of this combined with the chain rule to find the derivative of a function
47 Using the SFTC Chain Rule Definition of dF/du Substitute the integral for F(x)Substitute u for x3Apply the SFTCRewrite as a function of x.
48 Find 𝐹′(𝑥) if 𝐹 𝑥 = 0 𝑥 2 𝑠𝑖𝑛 𝜃 2 𝑑𝜃 What is u and du? 𝑢= 𝑥 2 , 𝑑𝑢=2𝑥 𝑑 𝑑𝑢 0 𝑢 𝑠𝑖𝑛 𝜃 2 𝑑𝜃 𝑑𝑢 𝑑𝑥 =𝑠𝑖𝑛 𝑢 2 𝑑𝑢 =sin( 𝑥 2 ) 2 (2𝑥) =𝑠𝑖𝑛 𝑥 4 (2𝑥)
50 Antidifferentiation of a Composite Function Let g be a function whose range is an interval I and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, thenIf u = g(x), then du = g’(x)dx and
51 Pattern RecognitionIn this section you will study techniques for integrating composite functions.This is split into two parts, pattern recognition and change of variables.u-substitution is similar to the techniques used for the chain rule in differentiation.
53 Pattern Recognition for Finding the Antiderivative Let g(x)=5x and we have g’(x)=5So we have f(g(x))=f(5x)=cos5xFrom this, you can recognize that the integrand follows the f(g(x))g’(x) pattern. Using the trig integration rule, we getYou can check this by differentiating the answer to obtain the original integrand.
54 Multiplying and Dividing by a Constant Example 3 p. 297
58 Guidelines for Making a Change of Variables Choose a substitution u=g(x). Usually it is best to choose the INNER part of a composite function, such as a quantity raised to a power.Compute du=g’(x)dxRewrite the integral in terms of the variable uFind the resulting integral in terms of uReplace u by g(x) to obtain an antiderivative in terms of x.Check your answer by differentiating.
59 General Power Rule for Integration Theorem: The General Power Rule for IntegrationIf g is a differentiable function of x, thenEquivalently, if u=g(x), then
60 Theorem 4.14 Change of Variables for Definite Integrals
61 Change of Variables for Definite Integrals Theorem 4.14 p 301
62 Definite Integrals and Change of Variables Example 8 p. 301
63 Definite Integrals and Change of Variables Example 9 p. 302
64 Theorem 4.15 Integraion of Even and Odd Functions and Figure 4.39