SECTION 5.4 ]| The Fundamental Theorem of Calculus, Part II 603

5.4 The Fundamental Theorem of Calculus,PartIl

Preliminary Questionsx

1. Let 6) = | V3 +1dt.4

(a) Is the FTC needed to calculate G(4)?(b) Is the FTC needed to calculate G (4)?SOLUTION(a) No. G(4) = fi +1dt = 0.(b) Yes. By the FTC Il, G'(x) = Vx31, so G (4) = 65.2. Whichofthe following is an antiderivative F(x) of f(x) = x? satisfying F(2) = 0?

x 2 xd 2 2@ [ 21 dt of, 1 dt of 1? dt

xSOLUTION Thecorrect answeris (c): / 1? dt.23. Does every continuous function have an antiderivative? Explain.

x3 a

4. Let G(x) = / sint dt. Which of the following statements are correct?(a) G(x)is the composite function sin(x3).(b) G(x)is the composite function A(x3), where

A(x) = [ sin(t) dt

(c) G(x)is too complicated to differentiate.(d) The Product Rule is used to differentiate G(x).(e) The Chain Ruleis usedto differentiate G(x).(f) G(x) = 3x? sin(x3).SOLUTION Statements (b), (e), and (f) are correct. Exercises1. Write the area function of f(x) = 2x + 4 with lower limit a = 2 as an integral and find a formulaforit.SOLUTION Let f(x) = 2x + 4. The area function with lower limit a = 2 is

A(x) = [ fü) dt = Le +4)dt.Carrying out the integration, we find

x x/ (21+ 4) dt = (17 +41)| = (X? +4x)- ((-2)? +4-2)) = x? +4x +42 _ 2or (x + 2) . Therefore, A(x) = (x + 2)?.2. Find a formulafor the area function of f(x) = 2x + 4 with lowerlimit a = 0.SOLUTION Thearea function for f(x) = 2x + 4 with lower limit a = is given by

A(x) = fo +4 di =( +41) =x3244x.

x

03. Let G(x) = /(1? 2) dt. Calculate G(1), G (1) and G (2). Then find a formula for G(x).SOLUTION Let G(x) = f(t? 2) dt. Then G(1) = fi (1? 2) dt = 0. Moreover, G (x) = x? 2, so that G'(1) = 1 andG (2) = 2. Finally,

G(x) = fo 2)dt= Gr = 2) =E = 2) =o = 20) = e = 2x4 >

604 CHAPTER 5 || THEINTEGRALx

4. Find F(0), F (0), and F (3), where F(x) = / vr? +1dt.0SOLUTION By definition, F(0) = 5 1? +tdt = 0. By FTC, F (x) = Vx? +x, so that F (0) = v0? +0 = 0 andF'G) = V32 +3 = V12 = 243.

x5. Find G(1), G (0), and G (2/4), where G(x) = / tant dt.

1SOLUTION By definition, G(1) = E tant dí = 0. By FTC, G'(x) = tan x, so that G (0) = tan0 = 0 and G'(4) =tanF =1.

x du6. Find H( 2) and H ( 2), wh Hx) = |ind H( 2) an ( 2), where H(x) ,2+1SOLUTION By definition, H(-2) /2 di o myrre H! (x) ~, s0 H!(-2) 2niti 2) = =0. , H'(x) = , 2 ==.$ ms la wl y x2+1 5In Exercises 7-16, findformulasfor thefunctions represented bythe integrals.

x7. / uf du

2x 1SOLUTION F(x) =] u du = ge2 "E 5

"4 32LE , x8. I (121? 81) dt2

xSOLUTION F(x) = / (121? 81) d = (41? -41?)| = 4x? 4x? - 16.2x

9. / sinu du0

x

2

x x

SOLUTION F(x) = Î sinu du = ( cosu)| = 1 cosx.0 0x

10. / sec? 0 dO1/4

x xSOLUTION F(x) = / sec? 0 d0 = tan =tanx tan( 1/4) = tanx +1.

x/4 7/4x11. / e du4

3 1 zul _1 3x 1,12SOLUTION F(x) =/ eYdu= e "| =2e*=--e4 3 4 3 3012. Î edt

x

0 0SOLUTION F(x) = / etd=| =-14e*x x

x213. | tat1

x? iodx àSOLUTION F(x) = | tdt=

SECTION 5.4 | The Fundamental Theorem of Calculus, Part Il 605

VX 4116. /2 1VE qi vx 1SOLUTION I =In|r| = In./x In2 = = Inx In2.1 4 2

In Exercises 17-20, express the antiderivative F(x) off(x) satisfying the given initial condition as an integral.17. f(x) = Vx3 +1, F(5)=0

xSOLUTION The antiderivative F(x) of Vx3 + satisfying F(5) = 0 is F(x) = / vi3+1dt.5

x+l18. y= ==» FM=0f=Zp FOSOLUTION The antiderivative F(x) of f(x) + tisfying F(7) 0 is F(x) [ +1 dte antiderivative F(x) of f(x) = satisfyin = 0is F(x) = - dt.| x?+9 6 7 1274919. f(x) =secx, F(0)=0

xSOLUTION The antiderivative F(x) of f(x) = sec satisfying F(0) = 0 is F(x) = / sect dt.020. f(x)=e"*, F(-4) =0SOLUTION Theantiderivative F(x) of f(x) = ex satisfying F( 4) = 0 is

F(x) = F O dr.4In Exercises 21-24, calculate the derivative.

d Rz21. (5-9%d= | 6-9)d xSOLUTION pyPren, | (1° 913) dt = x° 9x3.x Jo

ai

d 0SOLUTION ByFTCH, 28 cotu du = cot 6.1

d ft23. sec(5x 9) dxdt J100

=f10024 t d= uds A) an 1+ u2

SOLUTION By FTC II, 2E tan (=>) du = tan ( ).

25. Let A(x) = [ F(t) dt for f(x) in Figure 1.0(a) Calculate A(2), A(3), A (2), and A (3).(b) Find formulas for A(x) on [0, 2] and [2, 4] and sketch the graph of A(x).

Nu

BR

1.2.3 4FIGURE1

606 CHAPTER 5 | THE INTEGRAL

SOLUTION(a) A(2) = 2-2 = 4, the area under f(x) from x = 0 to x = 2, while A(3) = 2-3 + 4 = 6.5, the area under f(x) from x = 0to x = 3. By the FTC, A (x) = f(x) so A (2) = f(2) = 2 and A (3) = f(3) = 3.(b) For each x [0, 2], the region under the graph of y = f(x) is a rectangle of length x and height 2; for each x [2, 4], theregion is comprised of a square ofside length 2 and a trapezoid of height x 2 and bases 2 and x. Hence,

3x? +2, 2

SECTION 5.4 || The Fundamental Theorem of Calculus, Part Il 607

In Exercises 29-34, calculate the derivative.2d f*2. | iat

ax Jo 1+12d f* tdi A 2SOLUTION By the Chain Rule and the FTC, eo,edx Jo 141 x2+1 x? +1

d 1/x30. = cos? t dtdx 1

d fx, 3(1 1 1 3/1SOLUTION By the Chain Rule and the FTC, cos { dt = cos" |: [ = cos |.dx J x x2 x2 xd cos §

31. =| u* duds 6d cos Ss

SOLUTION By the Chain Rule and the FTC, 75 / u* du = cos* s( sins) = cos s sins.S J 6

a x32. / Vt dtdx Jy2Hintfor Exercise 32: F(x) = A(x*) A(x?).SOLUTION Let

x4 x4 x2=| vra=| vrar- | Vt dt.

x2 0 0Applying the Chain Rule combined with FTC, we have

Fx) = Vx4-4x3 Vx2-2x = 4x9 2x |x|.2

=|dx fxSOLUTION Let

x? x? JXG(x) =| tant dt = | tant ar f tant dt.

Vx 0 0Applying the Chain Rule combined with FTC twice, we have

1 tG' (x) = tan(x?) - 2x tan(./x) - =x7!/? = 2x tan(x?) u,3 2xd 3u34. Vx2 +14xdu Jy

SOLUTION Let

u3u 3uG(x) = Vx2+1dx= Vx2+14x- x2 +1 4x.u 0 0

Applying the Chain Rule combined with FTC twice, we haveG' (x) = 3V9u2 +14 Vu2 +1.

In Exercises 35-38, with f(x) as in Figure 3 let

A(x) = E FO dt and B(x) = [ FOdt.

FIGURE 3

608 CHAPTER 5 | THE INTEGRAL

35. Find the min and max of A(x) on [0, 6].SOLUTION The minimum values of A(x) on [0, 6] occur where A (x) = f(x) goes from negative to positive. This occurs at oneplace, where x = 1.5. The minimum value of A(x) is therefore A(1.5) = 1.25. The maximum values of A(x) on [0, 6] occurwhere A (x) = f(x) goes from positive to negative. This occurs at one place, where x = 4.5. The maximum value of A(x) istherefore A(4.5) = 1.25.36. Find the min and max of B(x) on [0,6].SOLUTION The minimum values of B(x) on [0, 6] occur where B (x) = f(x) goes from negative to positive. This occurs at oneplace, where x = 1.5. The minimum value of A(x) is therefore B(1.5) = 0.25. The maximum values of B(x) on [0, 6] occurwhere B (x) = f(x) goes from positive to negative. This occurs at one place, where x = 4.5. The maximum value of B(x)istherefore B(4.5) = 2.25.37. Find formulas for A(x) and B(x) valid on [2, 4].

2 2SOLUTION Ontheinterval [2, 4], A (x) = B (x) = f(x) =1.AQ) = / f(0) dt = 1 and B(2) = / f(t) dt = 0. HenceA(x) = (x 2) 1 and B(x) = (x 2). ° 238. Find formulas for A(x) and B(x) valid on [4, 5].

4SOLUTION Onthe interval [4,5], A (x) = B (x) = f(x) = 2(x 4.5) = 9 2x. A(4) = / f(t) dt = 1 and B(4) =04/ f(t) dt = 2. Hence A(x) = 9x x? 19 and B(x) = 9x x? 18.239. Let A(x) = [ F(t) dt, with f(x) as in Figure 4.0(a) Does A(x) have a local maximum at P?(b) Where does A(x) have a local minimum?(c) Where does A(x) have a local maximum?(d) True or false? A(x) < 0 for all x in the interval shown.

yRAS

FIGURE 4 Graph of f(x).

SOLUTION(a) In order for A(x) to have a local maximum, 4/(x) = f(x) must transition from positive to negative. As this does not happenat P, A(x) does not have a local maximum at P.(b) A(x) will have a local minimum when A (x) = f(x) transitions from negative to positive. This happens at R, so A(x) has alocal minimum at À.(e) A(x) will have a local maximum when A (x) = f(x) transitions from positive to negative. This happensat S, so A(x) has alocal maximum at S.(d) It is true that A(x) < 0 on J sincethe signed area from 0 to x is clearly always negative fromthe figure.

x40. Determine f(x), assuming that [ f(t) dt =x? +x.0

xSOLUTION Let F(x) = | ft) dt = x? +x. Then F'x)= f(x) = 2x + 1.

041. Determine the function g(x) and all values of c such that

x/ g(t)dt =x? +x-6c

SOLUTION By the FTC II we haved 2g(x) = (*4+x-6) = 2x41dx

and therefore,x

/ gOdt=x2?+x-(c?+c)Cc

We must choose c so that c2 + c = 6. We can take c = 2 orc = 3.

SECTION 5.4 | The Fundamental Theorem of Calculus, Part Il 609

b42. Find a < b such that / (x? 9) dx has minimalvalue.

a

SOLUTION Let a be given, and let F,(x) = ie (t? 9) dt. Then F(x) = x? 9, and the critical points are x = +3. BecauseF/(=3) = 6 and F//(3) = 6, we see that F,(x) has a minimum at x = 3. Now, we find a minimizing [2 (x? 9) dx. LetG(x) = E (1? 9) dx. Then G'(x) = (x? 9), yielding critical points x = 3 or x = 3. With x = 3,

3 1 3G(-3) = / (x? 9) dx = -x? - 9x = -36._3 3 LaWith x = 3,

3G(3) = / (x? 9 dx =0.3

bHence a = 3 and b = 3 are the values minimizing / (x? 9) dx.

ax

In Exercises 43 and 44, let A(x) = / FO dt.a

43. ÉS Area Functions and Concavity Explain why the following statements are true. Assume f(x) is differentiable.(a) Ifc is an inflection point of A(x), then f (c) = 0.(b) A(x) is concave up if f(x) is increasing.(c) A(x) is concave downif f(x) is decreasing.

SOLUTION(a) Ifx = c is an inflection point of A(x), then A (c) = f'(c) = 0.(b) If A(x) is concave up, then A (x) > 0. Since A(x) is the area function associated with f(x), A (x) = f(x) by FTC Il, soA" (x) = f'(x). Therefore f (x) > 0, so f(x) is increasing.(c) If A(x) is concave down, then A (x) < 0. Since A(x) is the area function associated with f(x), A (x) = f(x) by FTC II, soAl(x) = f (x). Therefore, f'(x) < 0 and so f(x) is decreasing.44, Match the property of A(x) with the corresponding property of the graph of f(x). Assume f(x)is differentiable.Area function A(x)(a) A(x)is decreasing.(b) A(x) has a local maximum.(c) A(x) is concave up.(d) A(x) goes from concave up to concave down.Graph of f(x)(i) Lies below the x-axis.(ii) Crosses the x-axis from positive to negative.(iii) Has a local maximum.(iv) f(x) is increasing.SOLUTION Let A(x) = [5 f(t) dt be an area function of f(x). Then A (x) = f(x) and A (x) = f (x).(a) A(x) is decreasing when A (x) = f(x) 0, i.e., when f(x) is increasing. This correspondsto choice (iv).(d) A(x) goes from concave up to concave down (at xp) when A (x) = f'(x) changes sign from + to 0 to as x increasesthrough xo, i.e., when f(x) has a local maximum at xo. This is choice(iii).

x45. Let A(x) = / FO dt, with f(x) as in Figure 5. Determine:

0(a) The intervals on which A(x) is increasing and decreasing(b) The values x where A(x) has a local min or max(c) The inflection points of A(x)(d) The intervals where A(x) is concave up or concave down FIGURE 5

610 CHAPTER 5 || THEINTEGRAL

SOLUTION(a) A(x) is increasing when A (x) = f(x) > 0, which correspondsto the intervals (0, 4) and (8, 12). A(x) is decreasing whenA'(x) = f(x) < 0, whichcorrespondsto the intervals (4, 8) and (12, 00).(b) A(x) has a local minimum when A (x) = f(x) changes from to +, corresponding to x = 8. A(x) has a local maximumwhen A (x) = f(x) changes from + to , corresponding to x = 4 and x = 12.(c) Inflection points of A(x) occur where A (x) = f'(x) changessign, or where f changes from increasing to decreasing or viceversa. Consequently, A(x) hasinflection points at x = 2, x = 6, and x = 10.(d) A(x) is concave up when A (x) = f (x) is positive or f(x) is increasing, which correspondsto the intervals (0, 2) and (6, 10).Similarly, A(x) is concave down when f(x) is decreasing, which corresponds to the intervals (2, 6) and (10, 00).46. Let f(x) = x* 5x 6 and F(x) = [ FO dt.

0(a) Findthecritical points of F(x) and determine whether they are local minimaor local maxima.(b) Find the points ofinflection of F(x) and determine whether the concavity changes from up to downor from down to up.(c) [GU] Plot f(x) and F(x) on the same set of axes and confirm your answers to (a) and (b).SOLUTION(a) If F(x) = fp (t? 5t 6) dt, then F'(x) = x? 5x 6 and F (x) = 2x 5. Solving F (x) = x? 5x 6 = 0 yieldscritical points x = 1 and x = 6. Since F ( 1) = 7 < O, there is a local maximum value of F at x = 1. Moreover, sinceF"(6) = 7 > 0,there is a local minimum value of F at x = 6.(b) As notedin part (a),

F'(x)=x?- 5x-6 and F"(x)=2x 5.A candidate pointofinflection occurs where F (x) = 2x 5 = 0. Thus x = 3. F (x) changes from negativeto positive at thispoint, so there is a point of inflection at x = 8 and concavity changes from downto up.(c) From the graph below, weclearly note that F(x) has a local maximum at x = 1, a local minimum at x = 6 and a point ofinflection at x = 3.

F(x)

47. Sketch the graph of an increasing function f(x) such that both f (x) and A(x) = F F(t) dt are decreasing.x

SOLUTION If f (x) is decreasing, then f (x) mustbe negative. Furthermore, if A(x) = | f(t) dt is decreasing, then A (x) =0f(x) must also be negative. Thus, we need a function which is negative but increasing and concave down. The graph of one such

function is shown below.

x48. ES Figure 6 showsthe graph of f(x) = x sin x. Let F(x) = / tsint dt.0 (a) Locate the local max and absolute max of F(x) on [0, 3z].(b) Justify graphically: F(x) hasprecisely one zeroin [x, 2x].(c) How many zeros does F(x) have in [0, 315]?(d) Find the inflection points of F(x) on [0, 37]. For each one, state whether the concavity changes from up to down or from downto up. FIGURE6 Graph of f(x) = x sinx.

SECTION 5.4 || The Fundamental Theorem of Calculus, Part Il 611

SOLUTION Let F(x) = fo !sint dt. A graph of f(x) = xsin x is depicted in Figure 6. Note that F (x) = f(x) and F"(x) =Fi).(a) For F to have a local maximum at xg (0, 377) we must have F'(xp) = f(xo) = O and F = f must changesign from + to0 to as x increases through xo. This occurs at x = a. The absolute maximum of F(x) on [0, 3x] occurs at x = 3z since (fromthe figure) the signed area between x = 0 and x = is greatest for x = c = 3x1.(b) Atx = x, the value ofF is positive since f(x) > 0 on (0, x). As x increases along the interval [zr, 271], we see that F decreasesas the negatively signed area accumulates. Eventually the additional negatively signed area outweighs the prior positively signedarea and F attains the value 0, say at b (x, 2m). Thereafter, on (b, 27), we see that f is negative and thus F becomes andcontinues to be negative as the negatively signed area accumulates. Therefore, F(x) takes the value 0 exactly oncein the interval[7, 2).(c) F(x) has two zeroesin [0, 377]. Oneis described in part (b) and the other must occur in the interval [27, 377] because F(x) < 0at x = 2x but clearly the positively signed area over [27311] is greater than the previous negatively signedarea.(d) Since f is differentiable, we have that F is twice differentiable on 7. Thus F(x) has an inflection point at xo providedF" (xo) = f (x0) = 0 and F" (x) = f (x) changes sign at xo. If F = f changes sign from + to 0 to at xg, then f has alocal maximum at x9. There is clearly such a value x9 in the figure in the interval [7/2, x] and another around 52/2. Accordingly,F has twoinflection points where F(x) changes from concave up to concave down. If F = f changes sign from to 0 to + atXo, then f has a local minimum at xo. Fromthe figure, there is such an x9 around 32/2; so F hasoneinflection point where F(x)changes from concave downto concave up.

Find the smallest positive critical point of

F(x) = [ cos(t3/?) dt

and determine whether it is a local min or max. Then find the smallest positive inflection point of F(x) and use a graph ofy= cos(x3/2) to determine whether the concavity changes from up to downor from downto up.SOLUTION critical point of F(x) occurs where F/(x) = cos(x3/2) = 0. The smallest positive critical points occurs wherex3/2 = x/2, so that x = (xc/2)2/3, F'(x) goes from positive to negative at this point, so x = (sc/2)2/3 correspondsto localmaximum..

Candidate inflection points of F(x) occur where F(x) = 0. By FTC, F (x) = cos(x3/?), so F(x) = (3/2)x!/? sin(x3/2),Finding the smallest positive solution of F (x) = 0, weget:

(3/2)x1/2 sin(x3/2) = 0sin(x3/?) =0 (since x > 0)

x3/2 = x1213 x= 2.14503.llx

From the plot below, wesee that F/(x) = cos (x3/2) changes from decreasing to increasing at n2/3 so F(x) changes from concavedownto concave up atthat point.

Further Insights and Challenges50. Proof of FTC If The proof in the text assumes that f(x) is increasing. To proveit for all continuous functions, let m(h)and M(h) denote the minimum and maximumof f(t) on [x,x + A] (Figure 7). The continuity of f(x) implies that lim m(h) =h=>0lim M(h) = f(x). Show that for h > 0,h 0

hm(h) < A(x + h) A(x) < hM(h)For h < 0, the inequalities are reversed. Prove that A (x) = f(x).

612 CHAPTER 5 | THE INTEGRAL

SHpo) M(h) m(h) + xa x x+h

FIGURE 7 Graphical interpretation of A(x + h) A(x).

SOLUTION Let f(x) be continuous on [a, b]. For À > 0, let m(h) and M(h) denote the minimum and maximum values of f on[x,x +h]. Since f is continuous, we have lim m(h)= lim M(h) = f(x). If h > 0, then since m(h) < f(x) < M(h) onh>0+ h 0+[x, x + h], we have

x+hx+hFO dt< / M(h) dt =hM(h).

x+h x+hhm(h) =| " m(h) dt < M(h). LettingIn other words, hm(h) < A(x + h) A(x) < hM(h). Since h > 0,it follows that m(h) <h => 0+ yields

A(x +h) A(x)f(x) < ;Jim > < f(x),whence

o AURA) AR) .,janAI)

by the Squeeze Theorem.If h < 0, then

[., [., E, [,

Since h < 0, we have h > 0 and thus- A(x) A(x +h)I < MUm(h) sh < M(h)

orA(x h) A(xm(h) < ere < M(h).h

Letting h > 0 givesA(x +h) A(xfe) < tim ÍEHDAD- sí,h 0- h

so thatA(x h) A(xim ARMANI _ pe)h 0- h

by the Squeeze Theorem. Since the one-sided limits agree, we therefore haveA h) AAG) = limA+AG) _ un,h 0 h

51. ProofofFTCI FTC lasserts that pe f(t) dt = F(b) F(a) if F (x) = f(x). Use FTC II to give a new proof ofFTC I asfollows. Set A(x) = f> f(t) dt.(a) Show that F(x) = A(x) + C for some constant.

b(b) Show that F(b) F(a) = A(b) A(a) = / FO dt.

SOLUTION Let F/(x) = f(x) and A(x) = f> f(t) dt.(a) Then by the FTC, Part IL, A (x) = f(x) and thus A(x) and F(x) are both antiderivatives of f(x). Hence F(x) = A(x) + Cfor some constant C.(b)

F(b) F(a) = (Alb) + C)- (Ala) +C)- ["oa - [roa

Abb) - Ala)b b/ FfO0d-0= / 10d

which proves the FTC,Part I.

SECTION 5.5 || Net Change asthe Integral ofa Rate 613

52. Can Every Antiderivative Be Expressed as an Integral? The area function /a F(0) dt is an antiderivative of f(x) for everyvalue of a. However,notall antiderivatives are obtained in this way. The general antiderivative of f(x) = x is F(x) = 4x? +C.Show that F(x) is an area function if C < 0 but not if C > 0.SOLUTION Let f(x) = x. The general antiderivative of f(x) is F(x) = 4x? + C. Let A(x) = Ie f@dt = tdt =

x41? = 3x? _ La? be an area function of f(x) = x. To express F(x) as an area function, we mustfind a value for a suchathat 4x? _ La? = 3x? + C, whence a = +/ 2C. If C < 0, then 2C > 0 and we may choose either a = Y-2C ora = y 2C. However, if C > 0, then there is no real solution for a and F(x) cannot be expressed as an area function.53. Prove the formula

|, fd = foe@v'@) feu)X Ju(x)SOLUTION Write

v(x) 0 v(x) v(x) u(x)Lo F(x) dx = Lo f(x) dx +/ f(x) dx = FO) dx- | F(x) dx.

Then, by the Chain Rule and the FTC,

Eley Oeaz, oa IlF(v))v'(&) ud).

54. Use the result of Exercise 53 to calculatexd (® sin! di

dx Jin x

SOLUTION By Exercise 53,xdf. Es. Lt:sint dt = e* sine* sinlnx.ax Jinx x

5.5 Net Changeas the Integral of a Rate

Preliminary Questions1. A hot metal object is submerged in cold water. The rate at which the object cools (in degrees per minute) is a function f(r) of

time. Which quantity is represented by the integral 5 fa) d ?SOLUTION Thedefinite integral /,a F(t) at represents the total drop in temperature of the metal objectin the first Tminutesafterbeing submerged in the cold water.2. A plane travels 560 km from Los Angeles to San Francisco in 1 hour. If the plane's velocity at time 1 is v(t) km/h, whatis the

value ofJo v(t) dt?SOLUTION Thedefinite integral h v(t) dt represents the total distance traveled by the airplane during the one hourflight fromLos Angeles to San Francisco. Therefore the value of Jo v(t) dt is 560 km.3. Whichof the following quantities would be naturally represented as derivatives and which asintegrals?

(a) Velocity ofa train(b) Rainfall during a 6-month period(c) Mileage per gallon of an automobile(d) Increase in the U.S. population from 1990 to 2010SOLUTION Quantities (a) and (c) involve rates of change, so these would naturally be represented as derivatives. Quantities (b)and (d) involve an accumulation, so these would naturally be represented as integrals.

614 CHAPTER 5 | THE INTEGRAL

Exercises1. Water flows into an empty reservoir at a rate of 3000 + 201 liters per hour. Whatis the quantity of water in the reservoir after

5 hours?SOLUTION The quantity of water in the reservoir after five hoursis

55[ (3000 + 201) dt = (3000: + 1012) = 15,250 gallons.0 0

2. A population of insects increases at a rate of 200 + 107 + 0.251? insects per day. Find the insect population after 3 days,assuming that there are 35 insects at f = 0.SOLUTION Theincreasein the insect population over three days is

3 1 1 3 2589[ 200 + 101 + 212 dr = [2001 + 512 + 13]| = = 647.25.0 4 12 lo 4Accordingly, the population after 3 days is 35 + 647.25 = 682.25 or 682 insects.3. A survey shows that a mayoral candidate is gaining votes at a rate of 2000/ + 1000 votes per day, where 1 is the number of

days since she announced her candidacy. How many supporters will the candidate have after 60 days, assuming that she had nosupporters at 1 = 0?SOLUTION The number of supportersthe candidate has after 60 daysis

60 60/ (20007 + 1000) dí = (10001? + 10001)| = 3,660,000.0 0

4. A factory producesbicycles at a rate of 95 + 31? 1 bicycles per week. How many bicycles were produced from the beginningof week 2 to the end of week 3?SOLUTION Therate of production is r(t) = 95 + 312 bicycles per week and the period from the beginning of week 2 tothe end of week 3 corresponds to the second andthird weeks of production. Accordingly, the number ofbikes produced from thebeginning ofweek 2 to the end of week is

3 3 1 3/ r(t) dt = (95 + 31? -1) dt = (951+23 =1?)| =2121 1 2 JIbicycles.5. Find the displacementofa particle moving in a straight line with velocity v(t) = 41 3 m/s overthe timeinterval[2,5].SOLUTION The displacement is given by

5 5I (41 3) dt = (21? -31)| = (50 15) (8 6) = 33m.2 2

6. Find the displacementover the time interval[1, 6] of a helicopter whose(vertical) velocity at time f is v(t) = 0.0217 + t m/s.SOLUTION Given v(t) = 1? + 1 m/s, the change in height over [1, 6] is

6 6 61 1 1 2841) dt = r 41) dt = [ r? +-1?)| = ~ 18.93 m.i ve | 50 * (= +7), 5 m7. A cat falls from a tree (with zero initial velocity) at time t = 0. How far doesthe cat fall between + = 0.5 and = 1 s? Use

Galileo s formula v(t) = 9.8f m/s.SOLUTION Given v(t) = 9.81 m/s, the total distance the cat falls during the interval 14, 1] is

/ Î1/2 1/2 1/2

8. A projectile is released with an initial (vertical) velocity of 100 m/s. Use the formula v(t) = 100 9.8r for velocity todetermine the distancetraveled duringthe first 15 seconds.SOLUTION Thedistancetraveled is given by

15 100/9.8 15/ 1100 9.81] dt = / (100 9.81) dt + [ (9.812 100) dt0 0 100/9.8

100/9.8 15= (1007 - 4.917) + (4.977 - 1001) = 622.9 m.0 100/9.8