4.4 the Fundamental Theorem of Calculus 277

4.4 The Fundamental Theorem of Calculus 277

4.4 The Fundamental Theorem of Calculus

Evaluate a definite integral using the Fundamental Theorem of Calculus. Understand and use the Mean Value Theorem for Integrals. Find the average value of a function over a closed interval. Understand and use the Second Fundamental Theorem of Calculus. Understand and use the Net Change Theorem.

The Fundamental Theorem of Calculus You have now been introduced to the two major branches of calculus: differential calculus (introduced with the tangent line problem) and integral calculus (introduced with the area problem). So far, these two problems might seem unrelated—but there is a very close connection. The connection was discovered independently by Isaac Newton and Gottfried Leibniz and is stated in the Fundamental Theorem of Calculus. Informally, the theorem states that differentiation and (definite) integration are inverse operations, in the same sense that division and multiplication are inverse operations. To see how Newton and Leibniz might have anticipated this relationship, consider the approximations shown in Figure 4.27. The slope of the tangent line was defined using the quotient y͞x (the slope of the secant line). Similarly, the area of a region under a curve was defined using the product yx (the area of a rectangle). So, at least in the primitive approximation stage, the operations of differentiation and definite integration appear to have an inverse relationship in the same sense that division and multiplication are inverse operations. The Fundamental Theorem of Calculus states that the limit processes (used to define the derivative and definite integral) preserve this inverse relationship.

Δx Δx

Area of rectangle Tangent Δy Δy Area of Secant line region line under curve

Δy Δy Slope = Slope ≈ Δx Δx Area = ΔyΔx Area ≈ ΔyΔx (a) Differentiation (b) Definite integration Differentiation and definite integration have an “inverse” relationship. Figure 4.27

ANTIDIFFERENTIATION AND DEFINITE INTEGRATION Throughout this chapter, you have been using the integral sign to denote an antiderivative (a family of functions) and a definite integral (a number).

b Antidifferentiation:͵ f͑x͒ dx Definite integration: ͵ f͑x͒ dx a The use of the same symbol for both operations makes it appear that they are related. In the early work with calculus, however, it was not known that the two operations were related.The symbol ͐ was first applied to the definite integral by Leibniz and was derived from the letter S. (Leibniz calculated area as an infinite sum, thus, the letter S.͒

THEOREM 4.9 The Fundamental Theorem of Calculus If a function f is continuous on the closed interval ͓a, b͔ and F is an antiderivative of f on the interval ͓a, b͔, then b ͵ f ͑x͒ dx F͑b͒ F͑a͒. a

Proof The key to the proof is writing the difference F͑b͒ F͑a͒ in a convenient form. Let be any partition of ͓a, b͔. . . . a x0 < x1 < x2 < < xn1 < xn b By pairwise subtraction and addition of like terms, you can write ͑ ͒ ͑ ͒ ͑ ͒ ͑ ͒ ͑ ͒ . . . ͑ ͒ ͑ ͒ ͑ ͒ F b F a F xn F xn1 F xn1 F x1 F x1 F x0 n ͓ ͑ ͒ ͑ ͔͒ ͚ F xi F xi1 . i1

By the Mean Value Theorem, you know that there exists a number ci in the i th subinterval such that F͑x ͒ F͑x ͒ ͑ ͒ i i1 F ci . xi xi1 ͑ ͒ ͑ ͒ Because F ci f ci , you can let xi xi xi1 and obtain n ͑ ͒ ͑ ͒ ͑ ͒ F b F a ͚ f ci xi . i1 This important equation tells you that by repeatedly applying the Mean Value Theorem, ͑ ͒ ͑ ͒ you can always find a collection of ci ’s such that the constant F b F a is a Riemann sum of f on ͓a, b͔ for any partition. Theorem 4.4 guarantees that the limit of Riemann sums over the partition with ʈʈ → 0 exists. So, taking the limit ͑as ʈʈ → 0͒ produces b F͑b͒ F͑a͒ ͵ f ͑x͒ dx. a See LarsonCalculus.com for Bruce Edwards’s video of this proof.

GUIDELINES FOR USING THE FUNDAMENTAL THEOREM OF CALCULUS 1. Provided you can find an antiderivative of f, you now have a way to evaluate a definite integral without having to use the limit of a sum. 2. When applying the Fundamental Theorem of Calculus, the notation shown below is convenient.

b b ͵ f͑x͒ dx F͑x͒΅ F͑b͒ F͑a͒ a a ͐3 3 For instance, to evaluate 1 x dx, you can write 3 x4 3 34 14 81 1 ͵ x3 dx ΅ 20. 1 4 1 4 4 4 4 3. It is not necessary to include a constant of integration C in the antiderivative.

b b ͵ f͑x͒ dx ΄F͑x͒ C΅ ͓F͑b͒ C͔ ͓F͑a͒ C͔ F͑b͒ F͑a͒ a a

Evaluating a Definite Integral

See LarsonCalculus.com for an interactive version of this type of example. Evaluate each definite integral. 2 4 ͞4 ͵ ͑ 2 ͒ ͵ Ί ͵ 2 a. x 3 dx b. 3 x dx c. sec x dx 1 1 0 Solution 2 x3 2 8 1 2 a. ͵ ͑x2 3͒ dx ΄ 3x΅ ΂ 6΃ ΂ 3΃ 1 3 1 3 3 3 4 4 x3͞2 4 Ί 1͞2 ͑ ͒3͞2 ͑ ͒3͞2 b. ͵ 3 x dx 3͵ x dx 3΄ ͞ ΅ 2 4 2 1 14 1 1 3 2 1 ͞4 ͞4 c. ͵ sec2 x dx tan x΅ 1 0 1 0 0

y y = ⏐2x − 1⏐ A Definite Integral Involving Absolute Value

2 3 Evaluate ͵ Խ2x 1Խ dx. 0

2 Solution Using Figure 4.28 and the definition of absolute value, you can rewrite the integrand as shown. ͑2x 1͒, x < 1 1 Խ2x 1Խ Ά 2 1 2x 1, x 2

x From this, you can rewrite the integral in two parts. − 112 2 1͞2 2 y = −(2x − 1) y = 2x − 1 ͵ Խ2x 1Խ dx ͵ ͑2x 1͒ dx ͵ ͑2x 1͒ dx 0 0 1͞2 ͞ The definite integral of y on ͓0, 2͔ is 5. 1 2 2 2 ΄x2 x΅ ΄x2 x΅ Figure 4.28 0 1͞2 1 1 1 1 ΂ ΃ ͑0 0͒ ͑4 2͒ ΂ ΃ 4 2 4 2 5 2

y Using the Fundamental Theorem to Find Area y = 2x2 − 3x + 2 Find the area of the region bounded by the graph of 4 y 2x2 3x 2 3 the x -axis, and the vertical lines x 0 and x 2, as shown in Figure 4.29.

2 Solution Note that y > 0 on the interval ͓0, 2͔. 2 ͵ ͑ 2 ͒ Integrate between x 0 and x 2. 1 Area 2x 3x 2 dx 0 2x3 3x2 2 x ΄ 2x΅ Find antiderivative. 1234 3 2 0 16 The area of the region bounded by the ΂ 6 4΃ ͑0 0 0͒ Apply Fundamental Theorem. graph of y, the x -axis,x 0, and 3 x 2 is 10. 10 3 Simplify. Figure 4.29 3

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 280 Chapter 4 Integration The Mean Value Theorem for Integrals y In Section 4.2, you saw that the area of a region under a curve is greater than the area of an inscribed rectangle and less than the area of a circumscribed rectangle. The Mean Value Theorem for Integrals states that somewhere “between” the inscribed and circumscribed rectangles, there is a rectangle whose area is precisely equal to the area of the region under the curve, as shown in Figure 4.30. f f(c)

x THEOREM 4.10 Mean Value Theorem for Integrals acb If f is continuous on the closed interval ͓a, b͔, then there exists a number c in Mean value rectangle: the closed interval ͓a, b͔ such that b b f͑c͒͑b a͒ ͵ f͑x͒ dx ͵ ͑ ͒ ͑ ͒͑ ͒ a f x dx f c b a . Figure 4.30 a

Proof Case 1: If f is constant on the interval ͓a, b͔, then the theorem is clearly valid because c can be any point in ͓a, b͔. Case 2: If f is not constant on ͓a, b͔, then, by the Extreme Value Theorem, you can choose f ͑m͒ and f ͑M͒ to be the minimum and maximum values of f on ͓a, b͔. Because

f͑m͒ f ͑x͒ f ͑M͒ for all x in ͓a, b͔, you can apply Theorem 4.8 to write the following.

b b b ͵ f ͑m͒ dx ͵ f ͑x͒ dx ͵ f ͑M͒ dx See Figure 4.31. a a a b f ͑m͒͑b a͒ ͵ f ͑x͒ dx f ͑M͒͑b a͒ Apply Fundamental Theorem. a 1 b ͑ ͒ ͑ ͒ ͑ ͒ f m ͵ f x dx f M Divide by b a. b a a From the third inequality, you can apply the Intermediate Value Theorem to conclude that there exists some c in ͓a, b͔ such that b b ͑ ͒ 1 ͑ ͒ ͑ ͒͑ ͒ ͑ ͒ f c ͵ f x dx or f c b a ͵ f x dx. b a a a

f f(M) f f f(m)

ababab Inscribed rectangle Mean value rectangle Circumscribed rectangle (less than actual area) (equal to actual area) (greater than actual area) b b b ͵ f͑m͒ dx f͑m͒͑b a͒ ͵ f͑x͒ dx ͵ f͑M͒ dx f͑M͒͑b a͒ a a a Figure 4.31

See LarsonCalculus.com for Bruce Edwards’s video of this proof.

Notice that Theorem 4.10 does not specify how to determine c. It merely guarantees the existence of at least one number c in the interval.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.4 The Fundamental Theorem of Calculus 281 Average Value of a Function y The value of f ͑c͒ given in the Mean Value Theorem for Integrals is called the average value of f on the interval ͓a, b͔. Average value f Definition of the Average Value of a Function on an Interval If f is integrable on the closed interval ͓a, b͔, then the average value of f on the interval is 1 b ͵ f ͑x͒ dx. See Figure 4.32. x b a ab a 1 b Average value ͵ f͑x͒ dx b a a To see why the average value of f is defined in this way, partition ͓a, b͔ into n Figure 4.32 subintervals of equal width b a x . n

If ci is any point in the i th subinterval, then the arithmetic average (or mean) of the function values at the ci ’s is 1 a ͓ f ͑c ͒ f ͑c ͒ . . . f ͑c ͔͒. Average of f ͑c ͒, . . . , f ͑c ͒ n n 1 2 n 1 n By multiplying and dividing by ͑b a͒, you can write the average as n 1 ͑ ͒ b a an ͚ f ci ΂ ΃ n i1 b a n 1 ͑ ͒ b a ͚ f ci ΂ ΃ b a i1 n n 1 ͑ ͒ ͚ f ci x. b a i1 Finally, taking the limit as n → produces the average value of f on the interval ͓a, b͔, as given in the definition above. In Figure 4.32, notice that the area of the region under the graph of f is equal to the area of the rectangle whose height is the average value. This development of the average value of a function on an interval is only one of many practical uses of definite integrals to represent summation processes. In Chapter 7, you will study other applications, such as volume, arc length, centers of mass, and work.

Finding the Average Value of a Function

y Find the average value of f ͑x͒ 3x2 2x on the interval ͓1, 4͔.

(4, 40) Solution The average value is 40 1 b 1 4 ͑ ͒ ͑ 2 ͒ f(x) = 3x2 − 2x ͵ f x dx ͵ 3x 2x dx 30 b a a 4 1 1 1 4 3 2 20 ΄x x ΅ 3 1 1 10 Average ͓64 16 ͑1 1͔͒ value = 16 3 (1, 1) 48 x 1234 3 Figure 4.33 16. See Figure 4.33.

The Speed of Sound

At different altitudes in Earth’s atmosphere, sound travels at different speeds. The speed of sound s͑x͒ (in meters per second) can be modeled by 4x 341, 0 x < 11.5 295, 11.5 x < 22 ͑ ͒ 3 s x 4x 278.5, 22 x < 32 Ά3 2x 254.5, 32 x < 50 3 2x 404.5, 50 x 80 where x is the altitude in kilometers (see Figure 4.34). What is the average speed of sound over the interval ͓0, 80͔?

s The first person to fly at a speed greater than the speed of sound 350 was Charles Yeager. On October 340 14, 1947, Yeager was clocked at 295.9 meters per second at an 330 altitude of 12.2 kilometers. If 320 Yeager had been flying at an altitude below 11.275 kilometers, 310 this speed would not have 300 “broken the sound barrier.” The 290 photo shows an F/A-18F Super Speed of sound (in m/sec) Hornet, a supersonic twin-engine 280 strike fighter. A “green Hornet” x using a 50/50 mixture of biofuel 10 20 30 40 50 60 70 80 90 made from camelina oil became Altitude (in km) the first U.S. naval tactical Speed of sound depends on altitude. aircraft to exceed 1 mach. Figure 4.34

Solution Begin by integrating s͑x͒ over the interval ͓0, 80͔. To do this, you can break the integral into five parts. 11.5 11.5 11.5 ͵ s͑x͒ dx ͵ ͑4x 341͒ dx ΄2x2 341x΅ 3657 0 0 0 22 22 22 ͵ s͑x͒ dx ͵ 295 dx ΄295x΅ 3097.5 11.5 11.5 11.5 32 32 32 ͵ ͑ ͒ ͵ ͑3 ͒ 3 2 s x dx 4x 278.5 dx ΄8x 278.5x΅ 2987.5 22 22 22 50 50 50 ͵ ͑ ͒ ͵ ͑3 ͒ 3 2 s x dx 2x 254.5 dx ΄4x 254.5x΅ 5688 32 32 32 80 80 80 ͵ ͑ ͒ ͵ ͑3 ͒ 3 2 s x dx 2x 404.5 dx ΄ 4x 404.5x΅ 9210 50 50 50 By adding the values of the five integrals, you have 80 ͵ s͑x͒ dx 24,640. 0 So, the average speed of sound from an altitude of 0 kilometers to an altitude of 80 kilometers is 1 80 24,640 Average speed ͵ s͑x͒ dx 308 meters per second. 80 0 80

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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.4 The Fundamental Theorem of Calculus 283 The Second Fundamental Theorem of Calculus Earlier you saw that the definite integral of f on the interval ͓a, b͔ 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 in 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. (Remember that the definite integral is not a function of its variable of integration.) The Definite Integral as a Number The Definite Integral as a Function of x

Constant F is a function of x.

b x ͵ f ͑x͒ dx F͑x͒ ͵ f ͑t͒ dt a a

f is a f is a Constant function of x. Constant function of t.

Exploration The Definite Integral as a Function Use a graphing utility to graph Evaluate the function the function x x F͑x͒ ͵ cos t dt F͑x͒ ͵ cos t dt 0 0 at x 0, , , , and . for 0 x . Do you 6 4 3 2 recognize this graph? Explain. Solution You could evaluate five different definite integrals, one for each of the given upper limits. However, it is much simpler to fix x (as a constant) temporarily to obtain x x ͵ cos t dt sin t΅ 0 0 sin x sin 0 sin x. Now, using F͑x͒ sin x, you can obtain the results shown in Figure 4.35.

y y y y y

π π 2 π 3 π F(0) = 0 F = 1 F = F = F = 1 ()6 2 ()42 ()32 ()2

t π t π t π t π t x = 0 x = x = x = x = 6 4 3 2 x F͑x͒ ͵ cos t dt is the area under the curve f͑t͒ cos t from 0 to x. 0 Figure 4.35

You can think of the function F͑x͒ as accumulating the area under the curve f ͑t͒ cos t from t 0 to t x. For x 0, the area is 0 and F͑0͒ 0. For x ͞2, F͑͞2͒ 1 gives the accumulated area under the cosine curve on the entire interval ͓0, ͞2͔. This interpretation of an integral as an accumulation function is used often in applications of integration.

In Example 6, note that the derivative of F is the original integrand (with only the variable changed). That is, d d d x ͓F͑x͔͒ ͓sin x͔ ΄͵ cos t dt΅ cos x. dx dx dx 0 This result is generalized in the next theorem, called the Second Fundamental Theorem of Calculus.

THEOREM 4.11 The Second Fundamental Theorem of Calculus If f is continuous on an open interval I containing a, then, for every x in the interval, d x ΄͵ f ͑t͒ dt΅ f ͑x͒. dx a

Proof Begin by defining F as x F͑x͒ ͵ f ͑t͒ dt. a Then, by the definition of the derivative, you can write F͑x x͒ F͑x͒ F͑x͒ lim x→0 x 1 x x x lim ͵ f ͑t͒ dt ͵ f ͑t͒ dt → ΄ ΅ x 0 x a a 1 x x a lim ͵ f ͑t͒ dt ͵ f ͑t͒ dt → ΄ ΅ x 0 x a x 1 x x lim ͵ f ͑t͒ dt . → ΄ ΅ x 0 x x From the Mean Value Theorem for Integrals ͑assuming x > 0͒, you know there exists a number c in the interval ͓x, x x͔ such that the integral in the expression above is equal to f ͑c͒ x. Moreover, because x c x x, it follows that c → x as x → 0. So, you obtain 1 F͑x͒ lim ΄ f ͑c͒ x΅ lim f ͑c͒ f͑x͒. x→0 x x→0 A similar argument can be made for x < 0. See LarsonCalculus.com for Bruce Edwards’s video of this proof.

Using the area model for definite integrals, f(t) the approximation Δx xx f ͑x͒ x Ϸ ͵ f ͑t͒ dt x can be viewed as saying that the area of the rectangle of height f ͑x͒ and width x is f(x) approximately equal to the area of the region lying between the graph of f and the x -axis t on the interval xx + Δx xx ͓x, x x͔ ͑ ͒ Ϸ ͵ ͑ ͒ f x x f t dt x as shown in the figure at the right.

Note that the Second Fundamental Theorem of Calculus tells you that when a function is continuous, you can be sure that it has an antiderivative. This antiderivative need not, however, be an elementary function. (Recall the discussion of elementary functions in Section P.3.)

The Second Fundamental Theorem of Calculus

d x Evaluate ΄͵ Ίt2 1 dt΅. dx 0 Solution Note that f ͑t͒ Ίt2 1 is continuous on the entire real number line. So, using the Second Fundamental Theorem of Calculus, you can write d x ΄͵ Ίt2 1 dt΅ Ίx2 1. dx 0

The differentiation shown in Example 7 is a straightforward application of the Second Fundamental Theorem of Calculus. The next example shows how this theorem can be combined with the Chain Rule to find the derivative of a function.

The Second Fundamental Theorem of Calculus

x3 Find the derivative of F͑x͒ ͵ cos t dt. ͞2 Solution Using u x3, you can apply the Second Fundamental Theorem of Calculus with the Chain Rule as shown. dF du F͑x͒ Chain Rule du dx d du dF ͓F͑x͔͒ Definition of du dx du x3 d du x3 ͵ Substitute͵ cos t dt for F͑x͒. ΄ cos t dt΅ du ͞2 dx ͞2 d u du ΄͵ cos t dt΅ Substituteu for x3. du ͞2 dx ͑cos u͒͑3x2͒ Apply Second Fundamental Theorem of Calculus. ͑cos x3͒͑3x2͒ Rewrite as function of x.

Because the integrand in Example 8 is easily integrated, you can verify the derivative as follows. x3 F͑x͒ ͵ cos t dt ͞2 x3 sin t΅ ͞2 sin x3 sin 2 sin x3 1 In this form, you can apply the Power Rule to verify that the derivative of F is the same as that obtained in Example 8. d ͓sin x3 1͔ ͑cos x3͒͑3x2͒ Derivative of F dx

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 286 Chapter 4 Integration Net Change Theorem The Fundamental Theorem of Calculus (Theorem 4.9) states that if f is continuous on the closed interval ͓a, b͔ and F is an antiderivative of f on ͓a, b͔, then b ͵ f͑x͒ dx F͑b͒ F͑a͒. a But because F͑x) f ͑x͒, this statement can be rewritten as b ͵ F͑x͒ dx F͑b͒ F͑a͒ a where the quantity F͑b͒ F͑a) represents the net change of F on the interval ͓a, b͔.

THEOREM 4.12 The Net Change Theorem The definite integral of the rate of change of quantity F͑x͒ gives the total change, or net change, in that quantity on the interval ͓a, b͔. b ͵ F͑x͒ dx F͑b͒ F͑a͒ Net change of F a

Using the Net Change Theorem

A chemical flows into a storage tank at a rate of ͑180 3t͒ liters per minute, where t is the time in minutes and 0 t 60. Find the amount of the chemical that flows into the tank during the first 20 minutes. Solution Let c͑t͒ be the amount of the chemical in the tank at time t. Then c͑t͒ represents the rate at which the chemical flows into the tank at time t. During the first 20 minutes, the amount that flows into the tank is 20 20 ͵ c͑t͒ dt ͵ ͑180 3t͒ dt 0 0 3 20 ΄180t t2΅ 2 0 3600 600 4200. So, the amount that flows into the tank during the first 20 minutes is 4200 liters.

Another way to illustrate the Net Change Theorem is to examine the velocity of a particle moving along a straight line, where s͑t͒ is the position at time t. Then its velocity is v͑t͒ s͑t͒ and b ͵ v͑t͒ dt s͑b͒ s͑a͒. a This definite integral represents the net change in position, or displacement, of the particle. Christian Lagerek/Shutterstock.com

v When calculating the total distance traveled by the particle, you must consider the intervals where v͑t͒ 0 and the intervals where v͑t͒ 0. When v͑t͒ 0, the particle moves to the left, and when v͑t͒ 0, the particle moves to the right. To v(t) calculate the total distance traveled, integrate the absolute value of velocity Խv͑t͒Խ. So, the displacement of the particle on the interval ͓a, b͔ is A 1 A b 3 ͓ ͔ ͵ ͑ ͒ t Displacement on a, b v t dt A1 A2 A3 a b a A2 and the total distance traveled by the particle on ͓a, b͔ is b ͓a b͔ ͵ v͑t͒ dt A A A A1, A2, and A3 are the areas of the Total distance traveled on , Խ Խ 1 2 3. shaded regions. a Figure 4.36 (See Figure 4.36.)

Solving a Particle Motion Problem

The velocity (in feet per second) of a particle moving along a line is v͑t͒ t3 10t2 29t 20 where t is the time in seconds. a. What is the displacement of the particle on the time interval 1 t 5? b. What is the total distance traveled by the particle on the time interval 1 t 5? Solution a. By definition, you know that the displacement is 5 5 ͵ v͑t͒ dt ͵ ͑t3 10t2 29t 20͒ dt 1 1 t4 10 29 5 ΄ t3 t2 20t΅ 4 3 2 1 25 103 ΂ ΃ 12 12 128 12 32 . 3 32 So, the particle moves 3 feet to the right. ͐5 ͑ ͒ b. To find the total distance traveled, calculate 1 Խv t Խ dt. Using Figure 4.37 v and the fact that v͑t͒ can be factored as ͑t 1͒͑t 4͒͑t 5͒, you can determine that v͑t͒ 0 on ͓1, 4͔ and v͑t͒ 0 on ͓4, 5͔. So, the total distance traveled is 8 5 4 5 v(t) ͵ ͑ ͒ ͵ ͑ ͒ ͵ ͑ ͒ 6 Խv t Խ dt v t dt v t dt 1 1 4 4 5 4 ͵ ͑t3 10t2 29t 20͒ dt ͵ ͑t3 10t2 29t 20͒ dt 1 4 2 t4 10 29 4 t4 10 29 5 ΄ t3 t2 20t΅ ΄ t3 t2 20t΅ t 4 3 2 1 4 3 2 4 12345 45 7 ΂ ΃ −2 4 12 71 Figure 4.37 feet. 6

4.4 Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.

Graphical Reasoning In Exercises 1–4, use a graphing Finding the Area of a Region In Exercises 35–38, utility to graph the integrand. Use the graph to determine determine the area of the given region. whether the definite integral is positive, negative, or zero. 1 35.y x x2 36. y 4 x2 1.͵ dx 2. ͵ cos x dx 2 0 x 1 0 y y 2 2 ͵ Ί 2 ͵ Ί 3. x x 1 dx 4. x 2 x dx 1 2 2 4 1 Evaluating a Definite Integral In Exercises 5–34, evaluate the definite integral. Use a graphing utility to verify your x result. 1 x 12 2 1 5.͵ 6x dx 6. ͵ 8 dt 37.y cos x 38. y x sin x 0 3 y y 0 2 7.͵ ͑2x 1͒ dx 8. ͵ ͑7 3t͒ dt 4 1 1 1 3 1 2 9.͵ ͑t2 2͒ dt 10. ͵ ͑6x2 3x͒ dx 2 1 1 1 1 3 x x 11.͵ ͑2t 1͒2 dt 12. ͵ ͑4x3 3x2͒ dx π π π π 0 1 4 2 2 2 3 1 1 13.͵ ΂ 1΃ dx 14. ͵ ΂u ΃ du Finding the Area of a Region In Exercises 39–44, find the x2 u2 1 2 area of the region bounded by the graphs of the equations. 4 8 u 2 ͞ 15.͵ du 16. ͵ x1 3 dx 39. y 5x2 2, x 0, x 2, y 0 Ί 1 u 8 40. y x3 x, x 2, y 0 1 8 2 17.͵ ͑Ί3 t 2͒ dt 18. ͵Ί dx 41. y 1 Ί3 x, x 0, x 8, y 0 1 1 x 42. y 2Ίx x, y 0 1 x Ίx 2 19.͵ dx 20. ͵ ͑2 t͒Ίt dt 43. y x2 4x, y 0 0 3 0 44. y 1 x4, y 0 0 1 x x2 21.͵ ͑t 1͞3 t 2͞3͒ dt 22. ͵ dx Ί3 1 8 2 x Using the Mean Value Theorem for Integrals In 5 4 Exercises 45–50, find the value(s) of c guaranteed by the Mean ͵ ͵ ͑ ͒ 23. Խ2x 5Խ dx 24. 3 Խx 3Խ dx Value Theorem for Integrals for the function over the given 0 1 interval. 4 4 ͵ 2 ͵ 2 3 25. Խx 9Խ dx 26. Խx 4x 3Խ dx 45.f ͑x͒ x , ͓0, 3͔ 46. f ͑x͒ Ίx, ͓4, 9͔ 0 0 x2 9 47.y , ͓0, 6͔ 48. f ͑x͒ , ͓1, 3͔ 27.͵ ͑1 sin x͒ dx 28. ͵ ͑2 cos x͒ dx 4 x3 0 0 ͞4 ͞4 ͑ ͒ 2 ͑ ͒ 1 sin2 sec2 49.f x 2 sec x, ΄ , ΅ 50. f x cos x, ΄ , ΅ 29.͵ d 30. ͵ d 4 4 3 3 2 2 0 cos 0 tan 1 ͞6 Finding the Average Value of a Function In Exercises 31. ͵ sec2 x dx 51–56, find the average value of the function over the given ͞6 interval and all values of x in the interval for which the ͞2 function equals its average value. 32. ͵ ͑2 csc2 x͒ dx ͞4 4͑x2 1͒ 51.f ͑x͒ 9 x2, ͓3, 3͔ 52. f ͑x͒ , ͓1, 3͔ ͞3 x2 33. ͵ 4 sec tan d 3 3 2 ͞3 53.f ͑x͒ x , ͓0, 1͔ 54. f ͑x͒ 4x 3x , ͓0, 1͔ ͞2 34. ͵ ͑2t cos t͒ dt 55.f ͑x͒ sin x, ͓0, ͔ 56. f ͑x͒ cos x, ΄0, ΅ ͞2 2

57. Velocity The graph shows the velocity, in feet per second, 63. Respiratory Cycle The volume V, in liters, of air in the of a car accelerating from rest. Use the graph to estimate the lungs during a five-second respiratory cycle is approximated distance the car travels in 8 seconds. by the model V 0.1729t 0.1522t2 0.0374t3, where t is v v the time in seconds. Approximate the average volume of air in 150 100 the lungs during one cycle.

120 80 64. Average Sales A company fits a model to the monthly sales data for a seasonal product. The model is 90 60 t t 60 40 S͑t͒ 1.8 0.5 sin΂ ΃, 0 t 24 4 6 30 20 where S is sales (in thousands) and t is time in months. t t Velocity (in feet per second) Velocity (in feet per second) 481216 20 12345 (a) Use a graphing utility to graph f ͑t͒ 0.5 sin͑t͞6͒ for Time (in seconds) Time (in seconds) 0 t 24. Use the graph to explain why the average ͑ ͒ Figure for 57 Figure for 58 value of f t is 0 over the interval. (b) Use a graphing utility to graph S͑t͒ and the line 58. Velocity The graph shows the velocity, in feet per second, g͑t͒ t͞4 1.8 in the same viewing window. Use the of a decelerating car after the driver applies the brakes. Use the graph and the result of part (a) to explain why g is called graph to estimate how far the car travels before it comes to the trend line. a stop. 65. Modeling Data An experimental vehicle is tested on a straight track. It starts from rest, and its velocity v (in meters WRITING ABOUT CONCEPTS per second) is recorded every 10 seconds for 1 minute (see 59. Using a Graph The graph of f is shown in the figure. table). y t 0102030405060 4

3 v 0 5 21 40 62 78 83

2 f (a) Use a graphing utility to find a model of the form 1 v at3 bt2 ct d for the data. x (b) Use a graphing utility to plot the data and graph the model. 1234567 (c) Use the Fundamental Theorem of Calculus to approximate ͐7 ͑ ͒ (a) Evaluate 1 f x dx. the distance traveled by the vehicle during the test. (b) Determine the average value of f on the interval ͓1, 7͔. (c) Determine the answers to parts (a) and (b) when the graph is translated two units upward. 66. HOW DO YOU SEE IT? The graph of f is shown in the figure. The shaded region A has an ͑ ͒ 60. Rate of Growth Let r t represent the rate of growth area of 1.5, and ͐6 f ͑x͒ dx 3.5. Use this ͑ ͒ 0 of a dog, in pounds per year. What does r t represent? information to fill in the blanks. ͐6 ͑ ͒ What does 2 r t dt represent about the dog? y

61. Force The force F (in newtons) of a hydraulic cylinder in a press is proportional to the square of sec x, where x is the distance (in meters) that the cylinder is extended in its cycle. A f ͓ ͞ ͔ ͑ ͒ B The domain of F is 0, 3 , and F 0 500. x (a) Find F as a function of x. 23456 (b) Find the average force exerted by the press over the interval ͓0, ͞3͔. 2 6 62. Blood Flow The velocity v of the flow of blood at a (a)͵ f ͑x͒ dx ᭿ (b) ͵ f ͑x͒ dx ᭿ distance r from the central axis of an artery of radius R is 0 2 6 2 v k͑R2 r 2͒ (c)͵ Խf ͑x͒Խ dx ᭿ (d) ͵ 2 f ͑x͒ dx ᭿ 0 0 where k is the constant of proportionality. Find the average rate 6 (e) ͵ ͓2 f ͑x͔͒ dx ᭿ of flow of blood along a radius of the artery. (Use 0 and R as 0 the limits of integration.) (f) The average value of f over the interval ͓0, 6͔ is ᭿.

Evaluating a Definite Integral In Exercises 67–72, find F Using the Second Fundamental Theorem of Calculus In Exercises 81–86, use the Second Fundamental Theorem of .8 ؍ and x ,5 ؍ x ,2 ؍ as a function of x and evaluate it at x ͧ ͨ x Calculus to find F x . ͑ ͒ ͵ ͑ ͒ 67. F x 4t 7 dt x x t2 0 81.F͑x͒ ͵ ͑t2 2t͒ dt 82. F͑x͒ ͵ dt 2 x 2 1 t 1 ͑ ͒ ͵ ͑ 3 ͒ 68. F x t 2t 2 dt x x 2 ͑ ͒ ͵ Ί 4 ͑ ͒ ͵ Ί4 83.F x t 1 dt 84. F x t dt x x 1 1 ͑ ͒ ͵ 20 ͑ ͒ ͵ 2 69.F x dv 70. F x dt x x v2 t3 1 2 85.F͑x͒ ͵ t cos t dt 86. F͑x͒ ͵ sec3 t dt x x 0 0 ͑ ͒ ͵ ͑ ͒ ͵ 71.F x cos d 72. F x sin d 1 0 Finding a Derivative In Exercises 87–92, find Fͧxͨ. 73. Analyzing a Function Let x2 x 87.F͑x͒ ͵ ͑4t 1͒ dt 88. F͑x͒ ͵ t3 dt x x x ͑ ͒ ͵ ͑ ͒ g x f t dt sin x x2 0 1 89.F͑x͒ ͵ Ίt dt 90. F͑x͒ ͵ dt t 3 where f is the function whose graph is shown in the figure. 0 2 3 2 ͑ ͒ ͑ ͒ ͑ ͒ ͑ ͒ ͑ ͒ x x (a) Estimate g 0 , g 2 , g 4 , g 6 , and g 8 . ͑ ͒ ͵ 2 ͑ ͒ ͵ 2 91.F x sin t dt 92. F x sin d (b) Find the largest open interval on which g is increasing. 0 0 Find the largest open interval on which g is decreasing. 93. Graphical Analysis Sketch an approximate graph of g on (c) Identify any extrema of g. the interval 0 x 4, where (d) Sketch a rough graph of g. x ͑ ͒ ͵ ͑ ͒ g x f t dt. y y 0 4 6 f Identify the x -coordinate of an extremum of g. To print an 5 3 4 2 enlarged copy of the graph, go to MathGraphs.com 3 1 y 2 t f 1 −1 132 456 78 2 t −2 f −1 132 4 78 −3 1 −2 −4 t 2 4 Figure for 73 Figure for 74 −1 74. Analyzing a Function Let −2

x ͑ ͒ ͵ ͑ ͒ 94. Area The area A between the graph of the function g x f t dt 0 4 g͑t͒ 4 where f is the function whose graph is shown in the figure. t2 (a) Estimate g͑0͒, g͑2͒, g͑4͒, g͑6͒, and g͑8͒. and the t- axis over the interval ͓1, x͔ is (b) Find the largest open interval on which g is increasing. x 4 Find the largest open interval on which g is decreasing. A͑x͒ ͵ ΂4 ΃ dt. t2 (c) Identify any extrema of g. 1 (d) Sketch a rough graph of g. (a) Find the horizontal asymptote of the graph of g. (b) Integrate to find A as a function of x. Does the graph of A Finding and Checking an Integral In Exercises 75–80, have a horizontal asymptote? Explain. (a) integrate to find F as a function of x , and (b) demonstrate the Second Fundamental Theorem of Calculus by differentiating Particle Motion In Exercises 95–100, the velocity function, the result in part (a). in feet per second, is given for a particle moving along a x x straight line. Find (a) the displacement and (b) the total 75.F͑x͒ ͵ ͑t 2͒ dt 76. F͑x͒ ͵ t͑t2 1͒ dt distance that the particle travels over the given interval. 0 0 x x 95. v͑t͒ 5t 7, 0 t 3 ͑ ͒ ͵ Ί3 ͑ ͒ ͵ Ί 77.F x t dt 78. F x t dt 96. v͑t͒ t2 t 12, 1 t 5 8 4 x x 97. v͑t͒ t3 10t2 27t 18, 1 t 7 79.F͑x͒ ͵ 2 t dt 80. F͑x͒ ͵ t t dt sec sec tan 98. v͑t͒ t3 t2 t t ͞4 ͞3 8 15 , 0 5

1 True or False? In Exercises 111 and 112, determine whether 99. v͑t͒ , 1 t 4 Ίt the statement is true or false. If it is false, explain why or give 100. v͑t͒ cos t, 0 t 3 an example that shows it is false. 111. If F͑x͒ G͑x͒ on the interval ͓a, b͔, then 101. Particle Motion A particle is moving along the x -axis. The position of the particle at time t is given by F ͑b͒ F ͑a͒ G͑b͒ G͑a͒. x͑t͒ t3 6t2 9t 2, 0 t 5. 112. If f is continuous on ͓a, b͔, then f is integrable on ͓a, b͔.

Find the total distance the particle travels in 5 units of time. 113. Analyzing a Function Show that the function 102. Particle Motion Repeat Exercise 101 for the position ͞ 1 x 1 x 1 function given by f͑x͒ ͵ dt ͵ dt 2 2 0 t 1 0 t 1 x͑t͒ ͑t 1͒͑t 3͒2, 0 t 5. x 103. Water Flow Water flows from a storage tank at a rate of is constant for > 0. ͑500 5t͒ liters per minute. Find the amount of water that 114. Finding a Function Find the function f͑x) and all values flows out of the tank during the first 18 minutes. of c such that

104. Oil Leak At 1:00 P.M., oil begins leaking from a tank at a x 2 rate of ͑4 0.75t͒ gallons per hour. ͵ f͑t͒ dt x x 2. c (a) How much oil is lost from 1:00 P.M. to 4:00 P.M.? 115. Finding Values Let (b) How much oil is lost from 4:00 P.M. to 7:00 P.M.? x s (c) Compare your answers to parts (a) and (b). What do you G͑x͒ ͵ ΄s͵ f ͑t͒ dt΅ ds notice? 0 0 where f is continuous for all real t. Find (a)G͑0͒, (b) G͑0͒, Error Analysis In Exercises 105–108, describe why the state- (c)G͑x͒, and (d) G͑0͒. ment is incorrect.

1 1 ͵ 2 ͓ 1͔ ͑ ͒ 105. x dx x 1 1 1 2 1 1 2 1 1 3 Demonstrating the Fundamental Theorem 106. ͵ dx ΄ ΅ x3 x2 4 2 2 Use a graphing utility to graph the function 3͞4 ͞ ͵ 2 ͓ ͔3 4 2 107. sec x dx tan x ͞4 2 y1 sin t ͞4 3͞2 on the interval 0 t . Let F͑x͒ be the following function of x. ͵ ͓ ͔3͞2 108. csc x cot x dx csc x ͞2 2 x ͞ 2 F͑x͒ ͵ sin2 tdt 0 109. Buffon’s Needle Experiment A horizontal plane is ruled with parallel lines 2 inches apart. A two-inch needle is (a) Complete the table. Explain why the values of F are increasing. tossed randomly onto the plane. The probability that the needle will touch a line is 2 5 x 0 ͞2 6 3 2 3 6 2 P ͵ sin d 0 F͑x͒ where is the acute angle between the needle and any one of the parallel lines. Find this probability. (b) Use the integration capabilities of a graphing utility to graph F. (c) Use the differentiation capabilities of a graphing utility to graph F͑x͒. How is this graph related to the graph in part (b)? (d) Verify that the derivative of

θ 1 1 y t sin 2t 2 4

is sin2 t. Graph y and write a short paragraph about how this graph is related to those in parts (b) and (c). 110. Proof Prove that ͑ ͒ d v x ΄͵ f ͑t͒ dt΅ f ͑v͑x͒͒v͑x͒ f ͑u͑x͒͒u͑x͒. dx u͑x͒

Section 4.4 (page 288)

5 5 y 3 1. 3. 93. 95. (a)2 ft to the right 113 2 (b) 10 ft −55 f 1 g −55 x 1234 −2 −5 −1 Positive Zero 10 1 1 2 −2 5. 12 7.2 9.3 11.3 13.2 15. 3 1 27 25 64 17.4 19.18 21.20 23.2 25. 3 An extremum of g occurs ͞ Ί ͞ 1 27. 2 29.4 31. 2 3 3 33. 0 35. 6 at x 2. 52 32 63 37. 1 39.3 41. 20 43. 3 97. (a) 0 ft (b) 2 ft 99. (a) 2 ft to the right (b) 2 ft 45. 3Ί 3 2͞2 Ϸ 1.8899 47. 2Ί3 Ϸ 3.4641 101. 28 units 103. 8190 L 49. ±arccos Ί͞2 Ϸ ±0.4817 105. f͑x͒ x2 has a nonremovable discontinuity at x 0. 1 ͑ ͒ 2 ͞ 51. Average value 6 53. Average value 4 107. f x sec x has a nonremovable discontinuity at x 2. x ±Ί3 Ϸ ±1.7321 x Ί3 2͞2 Ϸ 0.6300 109. 2͞ Ϸ 63.7% 111. True 55. Average value 2͞ 57. About 540 ft 1 1 1 113. f͑x͒ ΂ ΃ 0 x Ϸ 0.690, x Ϸ 2.451 ͑1͞x͒2 1 x2 x2 1 4 ͐7 ͑ 10 ͑ ͒ ͑ ͒ 59. (a) 8 (b)3 (c) 1 f x) dx 20; Average value 3 Because f x 0, f x is constant. ͑ ͒ 2 Ί ͞ Ϸ ͑ ͒ ͐x ͑ ͒ 61. (a) F x 500 sec x (b) 1500 3 827 N 115. (a) 0 (b) 0 (c) xf x 0 f t dt (d) 0 63. About 0.5318 L 65. (a) v 0.00086t3 0.0782t2 0.208t 0.10 (b) 90 (c) 2475.6 m

−10 70 −10 67. F͑x͒ 2x2 7x 69. F͑x͒ 20͞x 20 F͑2͒ 6 F͑2͒ 10 F͑5͒ 15 F͑5͒ 16 ͑ ͒ ͑ ͒ 35 F 8 72 F 8 2 71. F͑x͒ sin x sin 1 F͑2͒ sin 2 sin 1 Ϸ 0.0678 F͑5͒ sin 5 sin 1 Ϸ 1.8004 F͑8͒ sin 8 sin 1 Ϸ 0.1479 73. (a) g͑0͒ 0, g͑2͒ Ϸ 7, g͑4͒ Ϸ 9, g͑6͒ Ϸ 8, g͑8͒ Ϸ 5 (b) Increasing:͑0, 4͒; Decreasing: ͑4, 8͒ (c) A maximum occurs at x 4. (d) y

x 2468 1 2 3 4͞3 75. 2 x 2x 77. 4 x 12 79. tan x 1 81. x2 2x 83. Ίx4 1 85. x cos x 87. 8 89. cos xΊsin x 91. 3x2 sin x 6