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Integration. Antidfferentiation The Fundamental Theorem of Calculus (FTC) and the Algebra of Integration Second FTC. Theorem 4.5 The Definite Integral as the Area of a Region. Fundamental Theorem of Calculus. 4.3. The Fundamental Theorem of Calculus.
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Integration Antidfferentiation The Fundamental Theorem of Calculus (FTC) and the Algebra of Integration Second FTC
The Fundamental Theorem of Calculus • If a function f is continuous on the closed [a,b] and F is an antiderivative of f on the interval [a,b], then:
FTC and the Constant of Integration It is not necessary to include a constant of integration C in the antiderivative because:
Using FTC to Find Area Example 3 in the text: Find the area of the region bounded by the graph of y= 2x2-3x+2, the x-axis and the vertical lines x=0 and x=2 Note that all of the function is above the x-axis in the interval Start by integrating the function over the closed interval [0,2] Then find the antiderivative and apply the FTC Simplify.
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:
Average Value of a Function Julia S. Lucas
Definition of the Average Value of a Function on an Interval and Figure 4.32
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?
Consider the following picture: • How high would the water level be if the waves all settled?
Okay! So, now that you have seen that this an interesting question... Let’s forget about real life, and... Do Some Math
Suppose we have a “nice”function and we need to find its average value over the interval [a,b].
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 of equal length to get back to the finite situation: In the above graph, we have n=8
Let us set up our notation: comes from the i-th interval Now we can get an estimate for the average value:
In a more condensed form, we now get: But we want to get out of the finite, and into the infinite! How do we do this? Take Limits!!!
In this way, we get the average value of f(x) over the interval [a,b]: So, if f is a “nice” function (i.e. we can compute its integral) then we have a precise solution to our problem.
So we’ve solved our problem! If I give you the equation and ask you to find it’s average value over the interval [0,2], you’ll all say NO PROBLEM!!!! the answer is…
Now we can answer our fish tank question!(That is, if the waves were described by an integrable function)
Theorem 4.10 Mean Value Theorem for Integrals and Figure 4.30
Finding the x values where we GET the average value of the function. First take the equation and find it’s average value over the interval [0,2], NO PROBLEM!!!! the answer is…
Finding the x values where we GET the average value of the function. NOW take the equation and it’s average value over the interval [0,2], average f = 5 and set that equal to the function… And then solve for x… NO PROBLEM!!!! the answer is…
Mean Value Theorem (find the value of x that gives you the average value of your function)
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:
Average Value Example • Find the average value of
Mean Value Theorem • If f is continuous on the closed interval [a,b], then there exists a number c in the closed interval [a,b] such that:
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:
MVT-I • The 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.
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
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.
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.
Using the SFTC • Note that is continuous on the entire real line. So, using the SFTC you can write • The differentiation shown in tthisexample 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
Using the SFTC • Chain Rule • Definition of dF/du • Substitute the integral for F(x) • Substitute u for x3 • Apply the SFTC • Rewrite as a function of x.
Find if What is u and du?
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, then • If u = g(x), then du = g’(x)dx and