Chapter 6 The Definite Integral

 Antidifferentiation  Areas and Riemann Sums  Definite Integrals and the Fundamental Theorem  Areas in the xy-Plane  Applications of the Definite Integral Chapter Outline

§ 6.1 Antidifferentiation

 Antidifferentiation  Finding Antiderivatives  Theorems of Antidifferentiation  The Indefinite Integral  Rules of Integration  Antiderivatives in Application Section Outline

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5 Antidifferentiation DefinitionExample Antidifferentiation: The process of determining f (x) given f ΄(x) If, then

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #6 Finding AntiderivativesEXAMPLE SOLUTION Find all antiderivatives of the given function. The derivative of x 9 is exactly 9x 8. Therefore, x 9 is an antiderivative of 9x 8. So is x and x It turns out that all antiderivatives of f (x) are of the form x 9 + C (where C is any constant) as we will see next.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7 Theorems of Antidifferentiation

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #8 The Indefinite Integral

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9 Rules of Integration

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #10 Finding AntiderivativesEXAMPLE SOLUTION Determine the following. Using the rules of indefinite integrals, we have

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #11 Finding AntiderivativesEXAMPLE SOLUTION Find the function f (x) for which and f (1) = 3. The unknown function f (x) is an antiderivative of. One antiderivative is. Therefore, by Theorem I, Now, we want the function f (x) for which f (1) = 3. So, we must use that information in our antiderivative to determine C. This is done below.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #12 Finding Antiderivatives So, 3 = 1 + C and therefore, C = 2. Therefore, our function is CONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #13 Antiderivatives in ApplicationEXAMPLE SOLUTION A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v(t) = -32t feet per second. (a) Find s(t), the height of the rock above the ground at time t. (b) How long will the rock take to reach the ground? (c) What will be its velocity when it hits the ground? (a) We know that s΄(t) = v(t) = -32t and we also know that s(0) = 400. We can now use this information to find an antiderivative of v(t) for which s(0) = 400. The antiderivative of v(t) is To determine C, Therefore, C = 400. So, our antiderivative is s(t) = -16t

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #14 Antiderivatives in Application s(t) = -16t CONTINUED (b) To determine how long it will take for the rock to reach the ground, we simply need to find the value of t for which the position of the rock is at height 0. In other words, we will find t for when s(t) = 0. This is the function s(t). 0 = -16t Replace s(t) with = -16t 2 Subtract. 25 = t 2 Divide. 5 = t Take the positive square root since t ≥ 0. So, it will take 5 seconds for the rock to reach the ground.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #15 Antiderivatives in Application v(t) = -32t CONTINUED This is the function v(t). Replace t with 5 and solve. So, the velocity of the rock, as it hits the ground, is 160 feet per second in the downward direction (because of the minus sign). (c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5). v(5) = -32(5) = -160

§ 6.2 Areas and Riemann Sums

 Area Under a Graph  Riemann Sums to Approximate Areas (Midpoints)  Riemann Sums to Approximate Areas (Left Endpoints)  Applications of Approximating Areas Section Outline

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #18 Area Under a Graph DefinitionExample Area Under the Graph of f (x) from a to b: An example of this is shown to the right

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #19 Area Under a Graph In this section we will learn to estimate the area under the graph of f (x) from x = a to x = b by dividing up the interval into partitions (or subintervals), each one having width where n = the number of partitions that will be constructed. In the example below, n = 4. A Riemann Sum is the sum of the areas of the rectangles generated above.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #20 Riemann Sums to Approximate AreasEXAMPLE SOLUTION Use a Riemann sum to approximate the area under the graph f (x) on the given interval using midpoints of the subintervals The partition of -2 ≤ x ≤ 2 with n = 4 is shown below. The length of each subinterval is -22 x1x1 x2x2 x3x3 x4x4

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #21 Riemann Sums to Approximate Areas Observe the first midpoint is units from the left endpoint, and the midpoints themselves are units apart. The first midpoint is x 1 = -2 + = = Subsequent midpoints are found by successively adding CONTINUED midpoints: -1.5, -0.5, 0.5, 1.5 The corresponding estimate for the area under the graph of f (x) is So, we estimate the area to be 5 (square units).

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #22 Approximating Area With Midpoints of IntervalsCONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #23 Riemann Sums to Approximate AreasEXAMPLE SOLUTION Use a Riemann sum to approximate the area under the graph f (x) on the given interval using left endpoints of the subintervals The partition of 1 ≤ x ≤ 3 with n = 5 is shown below. The length of each subinterval is 3 x1x1 x2x2 x3x3 x4x4 x5x

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #24 Riemann Sums to Approximate Areas The corresponding Riemann sum is CONTINUED So, we estimate the area to be (square units).

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #25 Approximating Area Using Left EndpointsCONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #26 Applications of Approximating AreasEXAMPLE SOLUTION The velocity of a car (in feet per second) is recorded from the speedometer every 10 seconds, beginning 5 seconds after the car starts to move. See Table 2. Use a Riemann sum to estimate the distance the car travels during the first 60 seconds. (Note: Each velocity is given at the middle of a 10-second interval. The first interval extends from 0 to 10, and so on.) Since measurements of the car’s velocity were taken every ten seconds, we will use. Now, upon seeing the graph of the car’s velocity, we can construct a Riemann sum to estimate how far the car traveled.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #27 Applications of Approximating AreasCONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #28 Applications of Approximating Areas Therefore, we estimate that the distance the car traveled is 2800 feet. CONTINUED

§ 6.3 Definite Integrals and the Fundamental Theorem

 The Definite Integral  Calculating Definite Integrals  The Fundamental Theorem of Calculus  Area Under a Curve as an Antiderivative Section Outline

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #31 The Definite Integral Δx = (b – a)/n, x 1, x 2, …., x n are selected points from a partition [a, b].

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #32 Calculating Definite IntegralsEXAMPLE SOLUTION Calculate the following integral. The figure shows the graph of the function f (x) = x Since f (x) is nonnegative for 0 ≤ x ≤ 1, the definite integral of f (x) equals the area of the shaded region in the figure below

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #33 Calculating Definite Integrals The region consists of a rectangle and a triangle. By geometry, CONTINUED Thus the area under the graph is = 1, and hence

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #34 The Definite Integral

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #35 Calculating Definite IntegralsEXAMPLE SOLUTION Calculate the following integral. The figure shows the graph of the function f (x) = x on the interval -1 ≤ x ≤ 1. The area of the triangle above the x-axis is 0.5 and the area of the triangle below the x-axis is 0.5. Therefore, from geometry we find that

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #36 Calculating Definite IntegralsCONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #37 The Fundamental Theorem of Calculus

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #38 The Fundamental Theorem of CalculusEXAMPLE SOLUTION Use the Fundamental Theorem of Calculus to calculate the following integral. An antiderivative of 3x 1/3 – 1 – e 0.5x is. Therefore, by the fundamental theorem,

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #39 The Fundamental Theorem of CalculusEXAMPLE SOLUTION (Heat Diffusion) Some food is placed in a freezer. After t hours the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where (a) Compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2. (b) What does the area in part (a) represent? (a) To compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2, we evaluate the following.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #40 The Fundamental Theorem of Calculus (b) Since the area under a graph can represent the amount of change in a quantity, the area in part (a) represents the amount of change in the temperature between hour t = 0 and hour t = 2. That change is degrees Fahrenheit. CONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #41 Area Under a Curve as an Antiderivative

§ 6.4 Areas in the xy-Plane

 Properties of Definite Integrals  Area Between Two Curves  Finding the Area Between Two Curves Section Outline

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #44 Properties of Definite Integrals

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #45 Area Between Two Curves

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #46 Finding the Area Between Two CurvesEXAMPLE SOLUTION Find the area of the region between y = x 2 – 3x and the x-axis (y = 0) from x = 0 to x = 4. Upon sketching the graphs we can see that the two graphs cross; and by setting x 2 – 3x = 0, we find that they cross when x = 0 and when x = 3. Thus one graph does not always lie above the other from x = 0 to x = 4, so that we cannot directly apply our rule for finding the area between two curves. However, the difficulty is easily surmounted if we break the region into two parts, namely the area from x = 0 to x = 3 and the area from x = 3 to x = 4. For from x = 0 to x = 3, y = 0 is on top; and from x = 3 to x = 4, y = x 2 – 3x is on top. Consequently,

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #47 Finding the Area Between Two CurvesCONTINUED Thus the total area is =

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #48 Finding the Area Between Two CurvesCONTINUED y = x 2 – 3x y = 0

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #49 Finding the Area Between Two CurvesEXAMPLE SOLUTION Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure. Since the two shaded regions are (1) disjoint and (2) have different functions on top, we will need a separate integral for each. Therefore

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #50 Finding the Area Between Two Curves Therefore, to represent all the shaded regions, we have CONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #51 Finding the Area Between Two CurvesEXAMPLE SOLUTION Two rockets are fired simultaneously straight up into the air. Their velocities (in meters per second) are v 1 (t) and v 2 (t), respectively, and v 1 (t) ≥ v 2 (t) for t ≥ 0. Let A denote the area of the region between the graphs of y = v 1 (t) and y = v 2 (t) for 0 ≤ t ≤ 10. What physical interpretation may be given to the value of A? Since v 1 (t) ≥ v 2 (t) for t ≥ 0, this suggests that the first rocket is always traveling at least as fast as the second rocket. Therefore, we have

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #52 Finding the Area Between Two CurvesCONTINUED But again, since v 1 (t) ≥ v 2 (t) for t ≥ 0, we know that. So, this implies that. This means that the position of the first rocket is always at least as high (up in the air) as that of the second rocket. That is, the first rocket is always higher up than the second rocket (or at the same height).

§ 6.5 Applications of the Definite Integral

 Average Value of a Function Over an Interval  Consumers’ Surplus  Future Value of an Income Stream  Volume of a Solid of Revolution Section Outline

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #55 Average Value of a Function Over an Interval

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #56 Average Value of a Function Over an IntervalEXAMPLE SOLUTION Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1. Using (2) with a = -1 and b = 1, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is equal to An antiderivative of 1 – x is. Therefore, So, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is 1.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #57 Average Value of a Function Over an IntervalEXAMPLE SOLUTION (Average Temperature) During a certain 12-hour period the temperature at time t (measured in hours from the start of the period) was degrees. What was the average temperature during that period? The average temperature during the 12-hour period from t = 0 to t = 12 is

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #58 Consumers’ Surplus

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #59 Consumers’ SurplusEXAMPLE SOLUTION Find the consumers’ surplus for the following demand curve at the given sales level x. Since 20 units are sold, the price must be Therefore, the consumers’ surplus is

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #60 Consumers’ Surplus That is, the consumers’ surplus is $20. CONTINUED

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #61 Future Value of an Income Stream

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #62 Future Value of an Income StreamEXAMPLE SOLUTION (Future Value) Suppose that money is deposited daily into a savings account at an annual rate of $2000. If the account pays 6% interest compounded continuously, approximately how much will be in the account at the end of 2 years? Divide the time interval from 0 to 2 years into daily subintervals. Each subinterval is then of duration years. Let t 1, t 2,..., t n be points chosen from these subintervals. Since we deposit money at an annual rate of $2000, the amount deposited during one of the subintervals is dollars. If this amount is deposited at time t i, the dollars will earn interest for the remaining 2 – t i years. The total amount resulting from this one deposit at time t i is then

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #63 Future Value of an Income Stream Add the effects of the deposits at times t 1, t 2,..., t n to arrive at the total balance in the account: CONTINUED This is a Riemann sum for the function on the interval 0 ≤ t ≤ 2. Since is very small when compared with the interval, the total amount in the account, A, is approximately That is, the approximate balance in the account at the end of 2 years is $4250.

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #64 Volume of a Solid of Revolution

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #65 Volume of a Solid of RevolutionEXAMPLE SOLUTION Find the volume of a solid of revolution generated by revolving about the x-axis the region under the following curve. Here g(x) = x 2, and