Chapter 6 Section Antiderivatives and Indefinite Integrals

Transcription

1 Chapter 6 Section Antiderivatives and Indefinite Integrals Objectives: The student will be able to formulate problems involving antiderivatives. The student will be able to use the formulas and properties of antiderivatives and indefinite integrals. The student will be able to solve applications using antiderivatives and indefinite integrals. The Antiderivative Many operations in mathematics have inverses. For example, division is the inverse of multiplication. The inverse operation of finding a derivative, called the antiderivative, will now command our attention. A function F is an antiderivative of a function f if F (x) = f (x). A function F(x) is an antiderivative of f(x) if the derivative ( ) ( ) Ex: Let s find some antiderivatives of 2x: Uniqueness of Antiderivatives The following theorem says that antiderivatives are almost unique. Theorem 1: If a function has more than one antiderivative, then the antiderivatives differ by at most a constant. Indefinite Integrals Let f (x) be a function. The family of all functions that are antiderivatives of f (x) is called the indefinite integral and has the symbol ( ) The symbol is called an integral sign, and the function f (x) is called the integrand. The symbol dx indicates that anti-differentiation is performed with respect to the variable x. By the previous theorem, if F(x) is any antiderivative of f, then The arbitrary constant C is called the constant of integration. 1 P a g e

2 Notation: Let ( ) be the entire set of antidervatives of the function f(x). That is [ ( ) ] ( ) We write. ( ) ( ) Integration symbol Integrand Indicates the variable of integration Constant of integration Integrate f(x) Find the antiderivative of f(x) The and the dx indicate that antidifferentiation with respect to x on f(x) Lets use our knowledge of derivatives to find some antiderivatives. (Integrate) Indefinite Integral Formulas and Properties x e n x n 1 x dx C, n 1 dx e C 1 dx ln x C x x n 1 k f ( x) dx k f ( x) dx f ( x) g( x) dx f ( x) dx g( x) dx It is important to note that property 4 states that a constant factor can be moved across an integral sign. A variable factor cannot be moved across an integral sign. 2 P a g e

3 Examples: ( ) ( ) 3 P a g e

4 Application In spite of the prediction of a paperless computerized office, paper and paperboard production in the United States has steadily increased. In 1990 the production was 80.3 million short tons, and since 1970 production has been growing at a rate given by f (x) = 0.048x , where x is years after Find f (x), and the production levels in 1970 and P a g e

5 Section Integration by Substitution Objectives : The student will be able to integrate by reversing the chain rule and by using integration by substitution. The student will be able to use additional substitution techniques. The student will be able to solve applications. Reversing the Chain Rule Recall the chain rule: Reading it backwards, this implies that Special Cases Differentials If y = f (x) is a differentiable function, then 1. The differential dx of the independent variable x is any arbitrary real number. 2. The differential dy of the dependent variable y is defined as dy = f (x) dx A differential equation in x and y is an equation that involves x, and the derivatives of y A function f(x) is a solution of a differential equation if it satisfies the equation. So to solve a differential equation means to find all solutions. Sometimes in addition to the differential equation we may know certain values of f or f, called initial conditions. 5 P a g e

6 Examples: 1. If y = f (x) = x 5 2, then dy = f (x) dx = 5x 4 dx 2. If y = f (x) = e 5x, then dy = f (x) dx = 5e 5x dx 3. If y = f (x) = ln (3x 5), then dy = f (x) dx = dx Ex: Solve the differential equation ( ) subject to the initial condition f(0) = 2 Ex: Solve the differential equation ( ) subject to the initial condition f(1) = 5 6 P a g e

7 Ex: Solve the differential equation ( ) subject to the initial condition ( ) and f(1)=3 General Indefinite Integral Formulas u e n u n 1 u du C, n 1 du e C 1 du ln u C u u n 1 Integration by Substitution Step 1. Select a substitution that appears to simplify the integrand. In particular, try to select u so that du is a factor of the integrand. Step 2. Express the integrand entirely in terms of u and du, completely eliminating the original variable. Step 3. Evaluate the new integral, if possible. Step 4. Express the antiderivative found in step 3 in terms of the original variable. (Reverse the substitution.) Learn by Example: ****In many substitutions the substitution is the inside expression**** ( ) 7 P a g e

8 ( ) **Sometimes the substitution is correct but we are off by a constant: Luckily this type of adjustment is easy ( ) ( ) 8 P a g e

9 The natural logarithm function: Integration We know that call this the log rule There are many rational functions that can be integrated using u substitutions followed by the log rule Look for quantities in which the numerator is the derivative if the denominator Ex: * let u = denominator If a rational function has a numerator degree greater than or equal to that of the denominator, then long division may reveal a form to which you can apply the log rule Ex: 9 P a g e

10 = ( ) 10 P a g e

11 Application The marginal price of a supply level of x bottles of baby shampoo per week is given by ( ) ( ) Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle. 11 P a g e

12 Section The Definite Integral Objectives: 1. The student will be able to approximate areas by using left and right sums. 2. The student will be able to compute the definite integral as a limit of sums. 3. The student will be able to apply the properties of the definite integral. Introduction We have been studying the indefinite integral or antiderivative of a function. We now introduce the definite integral. This integral will be the area bounded by f (x), the x axis, and the vertical lines x = a and x = b, with notation ( ) Where b is the upper limit of integration and a is the lower limit of integration ( ) is called the definite integral of f from a to b The definition of the definite integral is closely related to the area under the graph of the function. Suppose we wish to find the area of the shaded region We will disuse two ways of doing this 1. inscribed rectangles 2. circumscribed rectangles Example: Under the graph of f(x) x and above the x-axis between 0 and 2 12 P a g e

13 Inscribed Rectangles: The height of each rectangle is the minimum value of f(x) in the subinterval. Every rectangle lies completely under the graph of f(x). The area of the inscribed rectangles will under estimate the area of the region. Let n = number of rectangles let n meaning that you let the width of the rectangles get very small Circumscribed Rectangles: Every rectangle lies above the graph of f(x). The area of the circumscribed rectangles will overestimate the area of the region. Let n = number of rectangles let n meaning that you let the width of the rectangles get very small 13 P a g e

14 Fact 1: If we let the n = we will be able to get the precise area. Fact 2: The area of the inscribed rectangles is less than the area of the region. The area of the circumscribed rectangles is more than the area of the region. But as n, these areas inscribed and circumscribed approach the same number. And that is the true area. Definition: Area of a region Let f be continuous and non-negative on the interval [a,b]. The area of the region bounded by the graph of f, the x-axis and the vertical lines x=a and x=b is given by: ( ) Height Width Add up the areas of the rectangles Where Our accuracy can be improved if we increase the number or rectangles, and let x get smaller. The error in our process can be calculated if the function is monotone. That is, if the function is only increasing or only decreasing. Let L n and R n be the approximate areas, using n rectangles of equal width, and the left or right endpoints, respectively. If the function is increasing, convince yourself by looking at the picture that L n < Area < R n If f is decreasing, the inequalities go the other way. If you use L n to estimate the area, then Error = Area L n < R n L n. If you use R n to estimate the area, then Error = Area R n < R n L n. Either way you get the same error bound. Theorem 1 It is not hard to show that R n L n = f (b) f (a) x, and that for n equal subintervals, Theorem 2 If f (x) is either increasing or decreasing on [a, b], then its left and right sums approach the same real number I as n. This number I is the area between the graph of f and the x axis from x = a to x = b. 14 P a g e

15 Definite Integral as Limit of Sums We now come to a general definition of the definite integral. Let f be a function on interval [a, b]. Partition [a, b] into n subintervals at points a = x 0 < x 1 < x 2 < < x n 1 < x n = b. The width of the k th subinterval is x k = (x k x k 1 ). In each subinterval, choose an arbitrary point c k Such that x k 1 < c k < x k. Then define S n is called a Riemann sum. Notice that L n and R n are both special cases of a Riemann sum. A Visual Presentation of a Riemann Sum 15 P a g e

16 Summation Notation = Sigma Notation This is a way of expressing sums of many numbers compactly. i is called the index of summation is the i th term of the summation Upper and lower bounds of the summation are 1 and n Properties of Summation ( ) Summation Formulas ( ) ( )( ) ( ) Ex: ( ) 16 P a g e

17 The Definite Integral Theorem 3. Let f be a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as n. This limit I of the Riemann sums for f on [a, b] is called the definite integral of f from a to b, denoted ( ) The integrand is f (x), the lower limit of integration is a, and the upper limit of integration is b. Negative Values If f (x) is positive for some values of x on [a, b] and negative for others, then the definite integral symbol ( ) represents the cumulative sum of the signed areas between the graph of f (x) and the x axis, where areas above are positive and areas below negative. Ex: Calculate the definite integrals by referring to the figure with the indicated areas. Find the following: b f(x)dx a c f(x)dx b = = c f(x)dx a = 17 P a g e

18 Definite Integral Properties Ex: Assume we know that Find the following: ( ) 18 P a g e

19 Sec The Fundamental Theorem of Calculus Objectives: The student will be able to evaluate definite integrals. The student will be able to calculate the average value of a function using the definite integral. Fundamental Theorem of Calculus If f is a continuous function on the closed interval [a, b], and F is any antiderivative of f, then b a f ( x) dx F( x) b a F( b) F( a) Evaluating Definite Integrals By the fundamental theorem we can evaluate We simply calculate F( b) F( a) b a f ( x) dx easily and exactly. Using the Fundamental Theorem of Calculus : 1. We don t have to use limit definition to evaluate a definite integral (as long as we can find an antiderivative of f) 2. The theorem says ( ) can be evaluated like this: ( ( ) ) ( ( ) ) 3. If is not necessary to include a constant of integration (+c) Definite Integral Properties 19 P a g e

20 Ex: Find the following using the fundamental theorem of calculus 20 P a g e

21 ( ) ( ) 21 P a g e

22 Change of variables (u- substitution) for definite integrals When using u-substitutions with a definite integral it is often convenient to determine the limits of integration for the variable u. Ex: From past records a management service determined that the rate of increase in maintenance cost for an apartment building (in dollars per year) is given by M (x) = 90x 2 + 5,000, where M(x) is the total accumulated cost of maintenance for x years. Write a definite integral that will give the total maintenance cost from the end of the second year to the end of the seventh year. Evaluate the integral. 22 P a g e

23 Definition: An application of definite integral. If f is integrable on a closed interval [a,b] then the average value of f on the interval is: ( ) Note this is the area under the curve divided by the width. Hence, the result is the average height or average value. Ex: The total cost (in dollars) of printing x dictionaries is C(x) = 20, x a) Find the average cost per unit if 1000 dictionaries are produced. b) Find the average value of the cost function over the interval [0, 1000]. 23 P a g e