24.1 Definite Integrals and the Fundamental Theorem 771 So  b a f(t)dt =s(b) − s(a), where s is an antiderivative of f .  net change in position on [a, b]  =  position at t = b  −  position at t = a  When we interpret the integrand as a rate function the Fundamental Theorem of Calculus becomes transparent. Because any two antiderivatives of f differ only by an additive constant and the con- stants will cancel in the course of subtraction, it follows that s(t) could be any antiderivative of f .  b a ds dt dt = s(t)    b a = s(b) − s(a) ◆ ◆ EXAMPLE 24.9 If f(t) is the rate at which water is flowing into or out of a reservoir and we let W(t) be the amount of water in the reservoir at time t, then dW dt = f(t).W(t) is an antiderivative of f(t).The net change in the amount of water in the reservoir over the time interval [a, b]is given by  b a f(t)dt =  b a dW dt dt = W(b)− W(a). Again, when we interpret the integrand as a rate function, the Fundamental Theorem of Calculus is transparent. It is when the integrand, f(t),isnot thought of as a rate function that the Fundamental Theorem is most surprising. ◆ PROBLEMS FOR SECTION 24.1 1. (a) Using a computer or programmable calculator, find upper and lower bounds for the area under one arc of cos x using Riemann sums. Explain how you can be sure your lower bound is indeed a lower bound and your upper bound is an upper bound. (Do not use the Fundamental Theorem of Calculus to do so.) Your upper and lower bounds should differ by no more than 0.01. (b) Use the Fundamental Theorem of Calculus to show that the area under one arc of the cosine curve is exactly 2. 2. An object’s velocity at time t, t in seconds, is given by v(t) = 10t + 3 meters per second. Find the net distance traveled from time t = 1tot=9. Do this in two ways. First, look at the appropriate signed area and solve geometrically, without the Fundamental Theorem. Then calculate the definite integral  9 1 (10t + 3)dt using the Fundamental Theorem of Calculus. 3. Use the Fundamental Theorem of Calculus to calculate  2 1 t 3 dt. 4. Evaluate  3 1 1 t dt. 772 CHAPTER 24 The Fundamental Theorem of Calculus 5. Find the area under the graph of y = e x between x = 0 and x = 1. 6. Find the area under the graph of y = e −x between (a) x =−1 and x = 0. (Why should the answer be the same as the answer to the previous problem?) (b) x = 0 and x = 1. 7. Aimee and Alexandra spent Friday afternoon eating hot cinnamon hearts. If they gobbled hearts at a rate of 1.5t + √ t hearts per minute, then how many hearts did they consume between time t = 0 and t = 9, t given in minutes? 8. Suppose the temperature of an object is changing at a rate of r(t) =−2e −t degrees Celsius per hour, where t is given in hours. (a) Is the object heating, or cooling? (b) Between time t = 0 and t = 1, how much has the temperature changed? (c) Between t = 1 and t = 2, how much has the temperature changed? (d) If the object was 100 degrees Celsius at time t = 0, how hot is it at time t = 1? 9. Let f(x)=  x 1 1 t dt, x>0. (a) Find f(1),f(5),f(10), and f(1/2). t y = 1 t (b) What is an alternative formula for f(x)? (c) Often mathematicians define the natural logarithm by ln x =  x 1 1 t dt for x>0. Suppose this was the definition you had been given. Use the Fundamental Theorem of Calculus to show that ln x is increasing and concave down for x>0. 10. Find the value of x>1such that the area under the graph of 1/t from 1 to x is 1. 11. The rate of change of water level in a tank is given by r(t) =2 sin  π 4 t  gallons per hour, where t is measured in hours. At time t = 0 there are 30 gallons of water in the tank. (a) Between time t = 0 and t =8, when will the water level in the tank be the highest? (b) What is the maximum amount of water that will ever be in the tank? (c) What is the minimum amount of water that will ever be in the tank? (d) Is the amount of water added to the tank between t = 0 and t = 1 less than, greater than, or equal to the amount lost between t = 4 and t = 5? (Try to answer without doing any computations.) 24.1 Definite Integrals and the Fundamental Theorem 773 12. Let g(x) =  x 0 e −t 2 dt. (a) Where is g(x) zero? Positive? Negative? y t y = e –t 2 (b) Where is g(x) increasing? Decreasing? (c) Where is g(x) concave up? Concave down? (d) Is g(x) even, odd, or neither? (e) Although we cannot find an antiderivative for e −t 2 , we are able to get a lot of information about g(x). Sketch g(x). (f) Using a computer or programmable calculator, approximate g(1), g(−1), and g(2). Go back and look at all your answers to this question; make sure that they are consistent with one another. 13. Let h(x) =  x 0 sin(t 2 )dt. (a) Graph sin(t 2 ) on the domain [−3, 3]. (b) On (0, ∞), where is h(x) positive? Negative? (c) Is h(x) even, odd, or neither? Explain. (d) On [0, 3] where is h(x) increasing? Decreasing? Give exact answers. (e) What is the absolute maximum value of sin(t 2 )? What is the absolute minimum value of sin(t 2 )? (f) Where on [0, 3] does h attain its maximum and minimum values? Will your answers change if the domain is (0, ∞)? If the domain is (−∞, ∞)? (g) Numerically approximate the maximum value of h on [0, 3]. 14. (a) Find  1 −1 |x| dx using the area interpretation of the definite integral. (b) Show that |x 2 | 2 is not an antiderivative of |x| on [−1, 1] by showing that applying the Fundamental Theorem as if it were, gives the wrong answer. (c) Find an antiderivative of |x| on [−1, 0). Find an antiderivative of |x| on (0, 1]. 15. Find the following two definite integrals without using the Fundamental Theorem of Calculus. Instead, use the area interpretation of the definite integral. (a)  7 −3 (π + 1)dx (b)  7 −3 |−2x−4|dx 16. Evaluate the following. (If you haven’t done Problem 14, do that first.) (a)  3 −3 |x 2 − 4|dx (b)  5 0 |(x + 3)(x − 1)|dx 17. Calculate the following definite integrals by calculating the limit of Riemann sums. You’ll need to use the formulas provided. Check your answers using the Fundamental Theorem of Calculus. 774 CHAPTER 24 The Fundamental Theorem of Calculus (a)  5 0 xdx 1 + 2 + 3 +···+n= n(n + 1) 2 (b)  5 0 x 2 dx 1 2 + 2 2 + 3 2 +···n 2 = n(n + 1)(2n + 1) 6 (c)  2 0 x 3 dx 1 3 + 2 3 + 3 3 +···n 3 =  n(n + 1) 2  2 18. Read Appendix D: Proof by Induction. Then prove that the formulas provided in the previous problem are valid. 19. Given the graph below, compute  b a [f  (x)g(x) + g  (x)f (x)]dx. Assume that f and g are differentiable functions. a b x f(x) g(x) –5 3 4 6 (a, 4) (b, 5.5) (b, –5) y 20. (a) Write a Riemann sum with 10 equal subdivisions that gives an overestimate for the area under ln x on [1, 6]. Write your answer in two ways, once with summation notation and one without summation notation (using + ···+). State clearly and precisely the meaning of any notation used in your sum. (b) Consider the Riemann sum 49  k=0 ln(3 + k · (5/50)) · 5 50 . This is an underestimate for what integral? (The answer to this question is not unique.) 21. Put the following in ascending order, using < or = as appropriate.  2 1 ln xdx  2 0.5 ln xdx  2.5 1 ln xdx, 22. Put the following in ascending order, using < or = as appropriate. 0  6 1 ln xdx L 50 R 50 L 10 R 10 Explain your reasoning briefly. 24.2 The Average Value of a Function: An Application of the Definite Integral 775 23. (a) Which of the following is an antiderivative of ln x? i. 1 x ii. x ln x iii. x ln x − x iv. arctan(ln x) (b) Evaluate  6 1 ln xdx. 24.2 THE AVERAGE VALUE OF A FUNCTION: AN APPLICATION OF THE DEFINITE INTEGRAL In the modern world, a data set is often presented in a way to succinctly convey basic in- formation about the central tendency of the data. The average is one measure of central tendency of a data set. We’re familiar with computing averages when given a discrete set of data points. For example, to determine the average score on a quiz, sum all quiz scores and divide by the number of quizzes. In this section we will look at the average value of a continuous function and find that the definite integral enables us to compute this. For exam- ple, we will look at how to determine the average velocity of an object given a continuous velocity function and the average temperature of an object given a continuous temperature function. ◆ EXAMPLE 24.10 A dove flies at a velocity of w(t) = 3t 2 + 4 meters per second on the interval [1, 3]. A hawk flies at a velocity of v(t) = 51 2 (−t 2 + 4t − 3) meters per second over the same interval. (a) How far does each travel during this time? (b) What is the average velocity of each bird on the interval [1, 3]? SOLUTION (a) The net change in position (or distance traveled, as velocity is nonnegative here) is givenbyadefinite integral. Dove:  3 1 (3t 2 + 4)dt =(t 3 + 4t)    3 1 = (27 + 12) − (1 + 4) = 34 meters Hawk:  3 1 51 2 (−t 2 + 4t − 3)dt= 51 2  −t 3 3  + 2t 2 − 3t      3 1 = 51 2  (−9 + 18 − 9) −  −1 3 + 2 − 3  = 51 2  0 − −4 3  = 34 meters 776 CHAPTER 24 The Fundamental Theorem of Calculus 31 7 Dove's velocity t 13 Hawk's velocity t 13 Figure 24.8 Both birds travel the same distance; even though their velocity functions look quite different, the areas under each on the interval [1, 3] are the same. (b) To compute average velocity, we divide the net change in position by the change in time. This will be the same for each bird. average velocity = net change in position time = 34 meters 2 seconds = 17 meters per second Notice that the units (meters per second) are the units we expect for velocity. ◆ The key notion to grasp here is that in order to find the average value of the velocity function, we divide the net change in position by the change in time. We integrate the velocity function rather than use any specific values of it. average velocity on [a, b] = net change in position time =  b a v(t) dt b − a = 1 b − a  b a v(t) dt Notice that if we let s(t) denote position, where s  (t) = v(t), then  b a v(t) dt b − a = s(b) − s(a) b − a = s t . We can generalize the discussion. For instance, to find the average rate at which water enters a tank, we integrate the rate function over the time interval (finding the net change in the amount of water) and divide by the change in time. But not only can we generalize to other rate functions, we can generalize to find the average value of any integrable function. We define the average value of a function on a closed interval [a, b] as follows. 24.2 The Average Value of a Function: An Application of the Definite Integral 777 Definition The average value of f(x)on [a, b]is 1 b − a  b a f(x)dx. We can interpret the average value of f graphically. Denote by f ave the average value of f(x)on [a, b]. Then  b a f(x)dx=f ave (b − a). The latter expression can be interpreted as the area of a rectangle with base of length b − a and height of f ave . Thus the average value is the constant function with the same signed area under it on [a, b]asf.For instance, an eagle traveling at a constant velocity of 17 meters per second would travel the same distance as the dove and the hawk over the interval [1, 3]. 17 velocity velocity t 13 t 13 v dove v hawk v eagle v eagle Figure 24.9 SOME COMMON MISCONCEPTIONS: There are several conceptual potholes to be avoided when thinking about average value. We point them out in the example below so you can bypass the potholes instead of falling into them. The situation: A dove travels at 3t 2 + 4 meters per second, a hawk at 51 2 (−t 2 + 4t − 3) meters per second, and an eagle at 17 meters per second. We’ve established that on the interval [1, 3] all three birds travel the same total distance and hence have the same average velocity. Mistake I. The average velocity on [1, 3] is not the average of the velocities at t = 1 and t = 3. We are interested in the average over an interval; the values of v(t) between t = 1 and t = 3 are vitally important. Notice that the average of the final and initial velocities gives different results for each of the three birds. Especially alarming is the case of the hawk; because v(3) and v(1) are both zero, the average of the two is 0+0 2 = 0. But the hawk’s velocity is positive for all t ∈ (1, 3), so it is nonsensical to conclude that its average velocity on [1, 3] is zero. Mistake II. The average velocity on [1, 3] is not half the difference between the velocities at t = 1 and t = 3. This computation also gives different results for the dove and the hawk and assigns both the hawk and the eagle the nonsensical value of zero as an average velocity. The eagle is flying at 17 meters per second on [1, 3]; certainly its average velocity is not zero. 778 CHAPTER 24 The Fundamental Theorem of Calculus ◆ EXAMPLE 24.11 Ahawk’svelocity is given by v(t) = 51 2 (−t 2 + 4t − 3) meters per second over the interval [0, 3]. v(t) positive indicates the hawk is flying east. v(t) negative indicates the hawk is flying west. (a) What is the hawk’s average velocity on [0, 3]? (b) What is the hawk’s average speed on [0, 3]? SOLUTIONS (a) The hawk’s velocity on [0, 3] is graphed in Figure 24.10. The average velocity on [0, 3] is the average value of the velocity function v(t): v ave = 1 3  3 0 v(t) dt. This is the net displacement divided by the time elapsed. Hawk's velocity 123 t Figure 24.10 v ave = 1 3  3 0 v(t) dt = 1 3  3 0 51 2 (−t 2 + 4t − 3)dt = 1 3 · 51 2  −t 3 3 + 2t 2 − 3t      3 0 = 51 6 [ −9 + 18 − 9 ] = 17 2 [0] = 0 The average velocity of the hawk on [0, 3] is zero because its net displacement on the interval [0, 3] is zero. It flys the same distance west on the interval [0, 1] as it does east on the interval [1, 3]. (b) The average speed on [0, 3] is the average value of the speed function |v(t)|: |v| ave = 1 3  3 0 |v(t)| dt. This is the total distance traveled divided by the time elapsed. 24.2 The Average Value of a Function: An Application of the Definite Integral 779 |v| ave = 1 3  3 0 |v(t)| dt To integrate |v(t)| we must take off the absolute value signs. This means we must split the interval of integration into pieces such that on each subinterval the sign of v(t) does not change. |v| ave = 1 3   1 0 |v(t)| dt +  3 1 |v(t)| dt  = 1 3  1 0 −  v(t)  dt + 1 3  3 1 v(t) dt = 1 3  1 0 −  51 2 (−t 2 + 4t − 3)  dt + 1 3 [34] = 1 3  − 51 2   − t 3 3 + 2t 2 − 3t      1 0 + 34 3 = 1 3 (34) + 34 3 = 1 3 (34 + 34) = 68 3 = 22 2 3 v(t) 123 Area = t 34 3 Area = 34 3 Figure 24.11 The hawk’s average speed is 22 2 3 meters per second. ◆ EXERCISE 24.3 On the interval [1, 3] the hawk’s average speed and average velocity are equal, because v(t) ≥ 0 on [1, 3]. On [1, 3] the hawk’s average speed is 17 meters per second. On [0, 1] the hawk’s average speed is 1 1  1 0 |(v(t)| dt = =34 meters per second. If we average 17 meters per second and 34 meters per second we get 25 1 2 meters per second, which is not the hawk’s average speed. If, instead, we take a weighted average, giving the average speed on [1, 3] twice the weight of the average speed on [0, 1] we get 34 + 2 · 17 3 = 22 2 3 . Explain. 780 CHAPTER 24 The Fundamental Theorem of Calculus ◆ EXAMPLE 24.12 Suppose that the temperature in Toronto is given by T(t)=56 + √ t degrees on the interval [4, 9], where t is measured in hours. What is the average temperature over this time period? SOLUTION average temperature = 1 9 − 4  9 4  56 + √ t  dt = 1 5  56t + 2 3 t 3/2     9 4 = 1 5  (504 + 18) −  224 + 16 3  = 1 5 878 3 ≈ 58.53 degrees Notice that in this case the integral alone has no readily apparent meaning because the integrand is not a rate function. It is only after dividing by the time interval that we obtain a meaningful result. ◆ PROBLEMS FOR SECTION 24.2 1. Find the average value of 3 sin x +5 on the interval [0, 2π]. Do this in two ways, first geometrically and then using the Fundamental Theorem of Calculus. 2. (a) Suppose f is an odd function. Can you determine the average value of f on [−a, a]? If so, what is the average value? (b) Suppose f is an even function. Are the following equal? If not, can you determine which is largest? Explain your answer. i. the average value of f on [−a, a] ii. the average value of f on [0, a] iii. the average value of f on [−a,0] 3. Find the average value of sin x on [0, π ]. 4. The velocity of an object on [0, 6] is given by v(t) =−t 2 +4t. (a) Find the average velocity on [0, 6]. (b) Find the average speed on [0, 6]. 5. The function v(t) gives the velocity (in miles per hour) of a bicyclist on a cross-country trip. We let t =0 denote noon (t measured in hours) and designate v(t) to be positive when the bicyclist is going east. Interpret the following in terms of distance, position, velocity, etc. Your interpretation should be intelligible to a twelve year old with no calculus background. (a) v(5) (b)  5 0 v(t) dt (c)  5 0 |v(t)|dt (d) |v(5)| (e) v  (5) (f)  5 0 v(t) dt 5 . added to the tank between t = 0 and t = 1 less than, greater than, or equal to the amount lost between t = 4 and t = 5? (Try to answer without doing any computations.) 24.1 Definite Integrals and. integrand as a rate function the Fundamental Theorem of Calculus becomes transparent. Because any two antiderivatives of f differ only by an additive constant and the con- stants will cancel. area under the graph of y = e −x between (a) x =−1 and x = 0. (Why should the answer be the same as the answer to the previous problem?) (b) x = 0 and x = 1. 7. Aimee and Alexandra spent Friday