Tutorial 6 The Definite Integral MT129 – Calculus and Probability

Outline Anti-differentiation Areas Definite Integrals and the Fundamental Theorem Areas in the xy-Plane Applications of the Definite Integral MT129 – Calculus and Probability

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5 Anti-differentiation Definition Example Anti-differentiation: The process of determining f (x) given f ΄(x) If f ΄(x) = 2x, then f (x) = x2 EXAMPLE Find all antiderivatives of the function f (x) = 9x8 SOLUTION The derivative of x9 is exactly 9x8. Therefore, x9 is an antiderivative of 9x8. So is x9 + 5 and x9  17.2. It turns out that all antiderivatives of f (x) are of the form x9 + C (where C is any constant) as we will see next. MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7 Anti-differentiation and The Indefinite Integrals MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9 Rules of Integration MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9

Finding Anti-derivatives EXAMPLE Determine the following SOLUTION Using the rules of indefinite integrals, we have MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #10

Finding Anti-derivatives EXAMPLE Find the function f (x) for which and f (1) = 3. SOLUTION The unknown function f (x) is an anti-derivative of . One Anti-derivative is . Therefore, by Theorem I, Now, we want the function f (x) for which f (1) = 3. So, 3 = 1 + C and therefore, C = 2. Therefore, our function is MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #11

Anti-derivatives in Application EXAMPLE 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? SOLUTION (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 anti-derivative of v(t) for which s(0) = 400. The anti-derivative of v(t) is Therefore, C = 400. So, our antiderivative is s(t) = − 16t2 + 400. MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #13

Anti-derivatives in Application 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. s(t) = −16t2 + 400 This is the function s(t). 0 = −16t2 + 400 Replace s(t) with 0. −400 = −16t2 Subtract. 25 = t2 Divide. 5 = t Take the positive square root since t ≥ 0. So, it will take 5 seconds for the rock to reach the ground. MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #14

Anti-derivatives in Application CONTINUED (c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5). v(t) = −32t This is the function v(t). v(5) = −32(5) = −160 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). MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #15

Areas Under a Graph Definition Example Area Under the Graph of f (x) from a to b: An example of this is shown to the right MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #18

Calculating Definite Integrals EXAMPLE Calculate the following integral. SOLUTION The figure shows the graph of the function f (x) = x + 0.5. 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. The region consists of a rectangle and a triangle. By geometry, 1 Thus the area under the graph is 0.5 + 0.5 = 1, and hence 1 0.5 Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #32 MT129 – Calculus and Probability

The Definite Integral MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #34

Calculating Definite Integrals EXAMPLE Calculate the following integral SOLUTION 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 MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #35

The Fundamental Theorem of Calculus MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #37

The Fundamental Theorem of Calculus EXAMPLE Use the Fundamental Theorem of Calculus to calculate the following integral. SOLUTION An antiderivative of 3x1/3 – 1 – e0.5x is Therefore, by the fundamental theorem, MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #38

The Fundamental Theorem of Calculus EXAMPLE 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? SOLUTION (a) To compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2, we evaluate the following. (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 24.533 degrees Fahrenheit. MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #39

Area Under a Curve as an Anti-derivative MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #41

Properties of Definite Integrals MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #44

Area Between Two Curves MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #45

Finding the Area Between Two Curves EXAMPLE Let R be the region enclosed by the graphs of f (x) = x2 + 5x  7 and g (x) = x2 + 5x + 1. Find the area of the region R. SOLUTION MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #46

Finding the Area Between Two Curves EXAMPLE Find the area of the region between y = x2 – 3x and the x-axis (y = 0) from x = 0 to x = 4. SOLUTION Upon sketching the graphs we can see that the two graphs cross; and by setting x2 – 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 = x2 – 3x is on top. Consequently, MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #46

Finding the Area Between Two Curves CONTINUED Thus the total area is 4.5 + 1.833 = 6.333. MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #47

Finding the Area Between Two Curves CONTINUED y = x2 – 3x y = 0 MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #48

Finding the Area Between Two Curves EXAMPLE Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure. SOLUTION Since the two shaded regions are (1) disjoint and (2) have different functions on top, we will need a separate integral for each. Therefore MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #49

Finding the Area Between Two Curves CONTINUED Therefore, to represent all the shaded regions, we have MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #50

Finding the Area Between Two Curves EXAMPLE Two rockets are fired simultaneously straight up into the air. Their velocities (in meters per second) are v1(t) and v2(t), respectively, and v1(t) ≥ v2(t) for t ≥ 0. Let A denote the area of the region between the graphs of y = v1(t) and y = v2(t) for 0 ≤ t ≤ 10. What physical interpretation may be given to the value of A? SOLUTION Since v1(t) ≥ v2(t) for t ≥ 0, this suggests that the first rocket is always traveling at least as fast as the second rocket. Therefore, we have MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #51

Finding the Area Between Two Curves CONTINUED But again, since v1(t) ≥ v2(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). MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #52

Average Value of a Function Over an Interval MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #55

Average Value of a Function Over an Interval EXAMPLE Determine the average value of f (x) = 1 – x over the interval –1 ≤ x ≤ 1. SOLUTION 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. MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #56

Average Value of a Function Over an Interval EXAMPLE (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? SOLUTION The average temperature during the 12-hour period from t = 0 to t = 12 is MT129 – Calculus and Probability Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #57