Solucao impares Calculo 6ed CAP.5

5

INTEGRALS

5.1

Areas and Distances

1. (a) Sincei is increasing, we can obtain a lower estimate by using

left endpoints. We are instructed to use ve rectangles, soq = 5.

O5= 5 S l=1i({l1) { [{ = ed q = 10  05 = 2]

= i({0) · 2 + i({1) · 2 + i({2) · 2 + i({3) · 2 + i({4) · 2

= 2 [i(0) + i(2) + i(4) + i(6) + i(8)]  2(1 + 3 + 4=3 + 5=4 + 6=3) = 2(20) = 40

Sincei is increasing, we can obtain an upper estimate by using

right endpoints.

U5= 5

S

l=1i({l) {

= 2 [i({1) + i({2) + i({3) + i({4) + i({5)]

= 2 [i(2) + i(4) + i(6) + i(8) + i(10)]  2(3 + 4=3 + 5=4 + 6=3 + 7) = 2(26) = 52

ComparingU₅toO₅, we see that we have added the area of the rightmost upper rectangle,i(10) · 2, to the sum and subtracted the area of the leftmost lower rectangle,i(0) · 2, from the sum.

(b)O10= 10 S l=1i({l1) { [{ = 10  0 10 = 1]

= 1 [i({0) + i({1) + · · · + i({9)]

= i(0) + i(1) + · · · + i (9)  1 + 2=1 + 3 + 3=7 + 4=3 + 4=9 + 5=4 + 5=8 + 6=3 + 6=7 = 43=2 U10= 10 S

l=1i({l) { = i(1) + i(2) + · · · + i(10)

= O10+ 1 · i(10)  1 · i(0)



add rightmost upper rectangle, subtract leftmost lower rectangle

 = 43=2 + 7  1 = 49=2

233

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3. (a)U₄ =S4 l=1i({l) {  { = @2  0₄ = ₈  = _S4 l=1i({l)  { = [i({1) + i({2) + i({3) + i({4)] {

=cos

8 + cos28 + cos38 + cos48



8

 (0=9239 + 0=7071 + 0=3827 + 0)

8  0=7908

Sincei is decreasing on [0> @2], an underestimate is obtained by using the right endpoint approximation, U4. (b)O4= 4 S l=1i({l1) { = _S4 l=1i ({l1)  { = [i({0) + i({1) + i({2) + i({3)] {

=cos 0 + cos 8 + cos28 + cos38  8  (1 + 0=9239 + 0=7071 + 0=3827) 8  1=1835

O4is an overestimate. Alternatively, we could just add the area of the leftmost upper rectangle and subtract the area of the rightmost lower rectangle; that is,O4 = U4+ i(0) ·₈  i₂·₈.

5. (a)i({) = 1 + {2and{ = 2  (1)

3 = 1 

U3= 1 · i(0) + 1 · i(1) + 1 · i(2) = 1 · 1 + 1 · 2 + 1 · 5 = 8.

{ = 2  (1)₆ = 0=5 

U6 = 0=5[i(0=5) + i(0) + i(0=5) + i(1) + i(1=5) + i(2)]

= 0=5(1=25 + 1 + 1=25 + 2 + 3=25 + 5) = 0=5(13=75) = 6=875

(b)O3 = 1 · i(1) + 1 · i(0) + 1 · i(1) = 1 · 2 + 1 · 1 + 1 · 2 = 5

O6 = 0=5[i(1) + i(0=5) + i(0) + i(0=5) + i(1) + i(1=5)]

= 0=5(2 + 1=25 + 1 + 1=25 + 2 + 3=25) = 0=5(10=75) = 5=375

(c)P3 = 1 · i(0=5) + 1 · i(0=5) + 1 · i(1=5)

= 1 · 1=25 + 1 · 1=25 + 1 · 3=25 = 5=75 P6 = 0=5[i(0=75) + i(0=25) + i(0=25)

+ i(0=75) + i(1=25) + i(1=75)]

= 0=5(1=5625 + 1=0625 + 1=0625 + 1=5625 + 2=5625 + 4=0625) = 0=5(11=875) = 5=9375

SECTION 5.1 AREAS AND DISTANCES ¤ 235 7.Here is one possible algorithm (ordered sequence of operations) for calculating the sums:

1 Let SUM = 0, X_MIN = 0, X_MAX = 1, N = 10 (depending on which sum we are calculating), DELTA_X = (X_MAX - X_MIN)/N, and RIGHT_ENDPOINT = X_MIN + DELTA_X. 2 Repeat steps 2a, 2b in sequence until RIGHT_ENDPOINTA X_MAX.

2a Add (RIGHT_ENDPOINT)^4 to SUM. 2b Add DELTA_X to RIGHT_ENDPOINT.

At the end of this procedure, (DELTA_X)·(SUM) is equal to the answer we are looking for. We nd that U10= 1₁₀ 10 S l=1  l 10 4  0=2533, U30= 1₃₀ 30 S l=1  l 30 4  0=2170, U50= 1₅₀ 50 S l=1  l 50 4  0=2101> and U100= 1 100 100_S l=1  l 100 4

 0=2050. It appears that the exact area is 0=2.

The following display shows the program SUMRIGHT and its output from a TI-83 Plus calculator. To generalize the program, we have input (rather than assign) values for Xmin, Xmax, and N. Also, the function,{4, is assigned toY₁, enabling us to evaluate any right sum merely by changingY1and running the program.

9.In Maple, we have to perform a number of steps before getting a numerical answer. After loading the student package [command: with(student);] we use the command

left_sum:=leftsum(1/(xˆ2+1),x=0..1,10 [or 30, or 50]); which gives us the expression in summation notation. To get a numerical approximation to the sum, we use evalf(left_sum);. Mathematica does not have a special command for these sums, so we must type them in manually. For example, the rst left sum is given by

(1/10)*Sum[1/(((i-1)/10)ˆ2+1)],{i,1,10}], and we use the N command on the resulting output to get a numerical approximation.

In Derive, we use the LEFT_RIEMANN command to get the left sums, but must dene the right sums ourselves. (We can dene a new function using LEFT_RIEMANN withn ranging from 1 to q instead of from 0 to q  1.) (a) Withi({) = 1

{2_{+ 1},0  {  1, the left sums are of the form Oq= 1_q q S l=1 1 _l1 q 2_{+ 1}. Specically,O10 0=8100,

O30 0=7937, and O50 0=7904. The right sums are of the form Uq= 1_q q S l=1 1 _l q 2_{+ 1}. Specically,U10 0=7600, U30 0=7770, and U50 0=7804.

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(b) In Maple, we use the leftbox (with the same arguments as left_sum) and rightbox commands to generate the graphs.

left endpoints,q = 10 left endpoints,q = 30 left endpoints,q = 50

right endpoints,q = 10 right endpoints,q = 30 right endpoints,q = 50

(c) We know that since| = 1@({2+ 1) is a decreasing function on (0> 1), all of the left sums are larger than the actual area, and all of the right sums are smaller than the actual area. Since the left sum withq = 50 is about 0=7904 ? 0=791 and the right sum withq = 50 is about 0=7804 A 0=780, we conclude that 0=780 ? U₅₀? exact area ? O₅₀? 0=791, so the exact area is between0=780 and 0=791.

11. Sincey is an increasing function, O₆will give us a lower estimate andU₆will give us an upper estimate.

O6= (0 ft@s)(0=5 s) + (6=2)(0=5) + (10=8)(0=5) + (14=9)(0=5) + (18=1)(0=5) + (19=4)(0=5) = 0=5(69=4) = 34=7 ft

U6= 0=5(6=2 + 10=8 + 14=9 + 18=1 + 19=4 + 20=2) = 0=5(89=6) = 44=8 ft

13. Lower estimate for oil leakage:U5= (7=6 + 6=8 + 6=2 + 5=7 + 5=3)(2) = (31=6)(2) = 63=2 L.

Upper estimate for oil leakage:O₅= (8=7 + 7=6 + 6=8 + 6=2 + 5=7)(2) = (35)(2) = 70 L.

15. For a decreasing function, using left endpoints gives us an overestimate and using right endpoints results in an underestimate. We will useP₆to get an estimate.w = 1, so

P6= 1[y(0=5) + y(1=5) + y(2=5) + y(3=5) + y(4=5) + y(5=5)]  55 + 40 + 28 + 18 + 10 + 4 = 155 ft

For a very rough check on the above calculation, we can draw a line from(0> 70) to (6> 0) and calculate the area of the triangle:1₂(70)(6) = 210. This is clearly an overestimate, so our midpoint estimate of 155 is reasonable.

17. i({) =4_{{, 1  {  16. { = (16  1)@q = 15@q and {l= 1 + l { = 1 + 15l@q.} D = lim qUq= limq q S l=1i({l) { = limq q S l=1 4 u 1 + 15l_q · 15_q.

19. i({) = { cos {, 0  {  ₂. { = (₂  0)@q =₂@q and {_l= 0 + l { =₂l@q. D = lim_qUq= lim_q q S l=1i({l) { = limq q S l=1 l 2qcos  l 2q  · _2q.

SECTION 5.2 THE DEFINITE INTEGRAL ¤ 237 21. lim q q S l=1 

4qtan l4qcan be interpreted as the area of the region lying under the graph of| = tan { on the interval 

0> 4

 ,

since for| = tan { on0>₄with{ = @4  0

q = 4q,{l= 0 + l { = l4q, and{l = {l, the expression for the area is D = lim q q S l=1i ({  l) { = lim_q q S l=1tan  l 4q  

4q. Note that this answer is not unique, since the expression for the area is the same for the function| = tan({  n) on the intervaln> n +₄, wheren is any integer.

23. (a)| = i({) = {5. { = 2  0 q = 2q and{l = 0 + l { = 2lq. D = lim_qUq= lim_q q S l=1i({l) { = limq q S l=1  2l q 5 · 2_q= lim_qSq l=1 32l5 q5 · 2_q = limq 64 q6 q S l=1l 5_. (b) Sq l=1l 5 CAS₌ q2(q + 1)2  2q2_{+ 2q  1} 12 (c) lim q 64 q6 · q2_{(q + 1)}2_2q2_{+ 2q  1} 12 = 6412qlim  q2_{+ 2q + 1}_2q2_{+ 2q  1} q2_{· q}2 = 16₃ lim q  1 + 2_q+ 1_q2  2 + 2_q 1_q2  =16 3 · 1 · 2 = 323 25.| = i({) = cos {. { = e  0 q = eqand{l= 0 + l { = elq. D = lim_qUq= lim q q S l=1i({l) { = limq q S l=1cos  el q  · e_q CAS = lim_q 5 9 9 7 e sin  e  1 2q+ 1  2q sin  e 2q   e_2q 6 : : 8CAS= sin e Ife =₂, thenD = sin₂ = 1.

5.2

The Definite Integral

1.i({) = 3 1₂{, 2  {  14. { = e  d

q = 14  26 = 2. Since we are using left endpoints,{_l = {l1.

O6= 6

S

l=1i({l1) {

= ({) [i({0) + i({1) + i({2) + i({3) + i({4) + i({5)]

= 2[i(2) + i(4) + i(6) + i(8) + i(10) + i(12)] = 2[2 + 1 + 0 + (1) + (2) + (3)] = 2(3) = 6

The Riemann sum represents the sum of the areas of the two rectangles above the{-axis minus the sum of the areas of the three rectangles below the{-axis; that is, the net area of the rectangles with respect to the {-axis.

3. i({) = h{ 2, 0  {  2. { = e  d

q = 2  04 = 12. Since we are using midpoints,{_l = {l=1₂({l1+ {l).

P4 = 4

S

l=1i({l) { = ({) [i({1) + i({2) + i({3) + i({4)]

=1 2  i1 4  + i3 4  + i5 4  + i7 4  =1 2 k (h1@4_{ 2) + (h}3@4_{ 2) + (h}5@4_{ 2) + (h}7@4_{ 2)}l  2=322986

The Riemann sum represents the sum of the areas of the three rectangles above the{-axis minus the area of the rectangle below the{-axis; that is, the net area of the rectangles with respect to the {-axis.

5. { = (e  d)@q = (8  0)@4 = 8@4 = 2.

(a) Using the right endpoints to approximateU₀8i({) g{, we have

4

S

l=1i({l) { = 2[i(2) + i(4) + i(6) + i(8)]  2[1 + 2 + (2) + 1] = 4.

(b) Using the left endpoints to approximateU₀8i({) g{, we have

4

S

l=1i({l1) { = 2[i(0) + i(2) + i(4) + i(6)]  2[2 + 1 + 2 + (2)] = 6.

(c) Using the midpoint of each subinterval to approximateU₀8i({) g{, we have

4

S

l=1i({l) { = 2[i(1) + i(3) + i(5) + i(7)]  2[3 + 2 + 1 + (1)] = 10.

7. Sincei is increasing, O5U₀25i({) g{  U5. Lower estimate= O₅=S5

l=1i({l1) { = 5[i(0) + i(5) + i(10) + i(15) + i(20)]

= 5(42  37  25  6 + 15) = 5(95) = 475 Upper estimate= U5=

5

S

l=1i({l) { = 5[i(5) + i(10) + i(15) + i(20) + i(25)]

= 5(37  25  6 + 15 + 36) = 5(17) = 85

9. { = (10  2)@4 = 2, so the endpoints are 2, 4, 6, 8, and 10, and the midpoints are 3, 5, 7, and 9. The Midpoint Rule givesU₂10{3+ 1 g{ S4

l=1i({l) { = 2



33_{+ 1 +}₅3_{+ 1 +}₇3_{+ 1 +}₉3_{+ 1}_{ 124=1644.}

11. { = (1  0)@5 = 0=2, so the endpoints are 0, 0.2, 0.4, 0.6, 0.8, and 1, and the midpoints are 0.1, 0.3, 0.5, 0.7, and 0.9. The Midpoint Rule gives

U1 0 sin({2) g{  5 S l=1i({l) { = 0=2 

SECTION 5.2 THE DEFINITE INTEGRAL ¤ 239 13.In Maple, we use the command with(student); to load the sum and box commands, then

m:=middlesum(sin(xˆ2),x=0..1,5);which gives us the sum in summation notation, then M:=evalf(m); which givesP₅ 0=30843908, conrming the result of Exercise 11. The command middlebox(sin(xˆ2),x=0..1,5) generates the graph. Repeating forq = 10 and q = 20 gives P₁₀ 0=30981629 and P₂₀ 0=31015563.

15.We’ll create the table of values to approximateU₀sin { g{ by using the program in the solution to Exercise 5.1.7 withY1= sin {, Xmin = 0,

Xmax = , and q = 5, 10, 50, and 100. The values ofUqappear to be approaching2.

q Uq 5 1=933766 10 1=983524 50 1=999342 100 1=999836 17.On[2> 6], lim q q S l=1{lln(1 + { 2 l) { =U6₂{ ln(1 + {2) g{= 19.On[1> 8], lim q q S l=1 s 2{ l + ({l)2{ = U8 1  2{ + {2_g{. 21.Note that{ = 5  (1) q = 6q and{l= 1 + l { = 1 + 6lq. ] ₅ 1(1 + 3{) g{ = limq q S l=1i({l) { = limq q S l=1  1 + 3  1 + 6l_q  6 q = limq 6 q q S l=1  2 + 18l_q  = lim_qq6 _Sq l=1(2) + q S l=1 18l q  = lim_q6_q  2q + 18_q Sq l=1l  = lim_qq6  2q + 18_q · q(q + 1)₂  = lim_q  12 + 108_q2 · q(q + 1)₂  = lim q  12 + 54q + 1 q  = lim q  12 + 541 + 1 q  = 12 + 54 · 1 = 42 23.Note that{ = 2  0 q = 2qand{l= 0 + l { = 2lq. ] 2 0  2  {2_{g{ = lim} q q S l=1i({l) { = limq q S l=1  2  4l_q22  2 q  = lim q 2 q _Sq l=12  4q2 q S l=1l 2 = lim_q2_q  2q  4_q2 Sq l=1l 2_{= lim} q  4  8_q3 · q(q + 1)(2q + 1)₆  = lim q  4  4₃· q + 1_q · 2q + 1_q  = lim q  4  4₃  1 + 1_q  2 + 1_q  = 4 4 3 · 1 · 2 = 43

TX.10

25. Note that{ = 2  1 q = 1q and{l= 1 + l { = 1 + l(1@q) = 1 + l@q. ] 2 1 { 3_{g{ = lim} q q S l=1i({l) { = limq q S l=1  1 + l_q 3₁ q  = lim_qq1 Sq l=1  q + l q 3 = lim q 1 q4 q S l=1  q3_{+ 3q}2_{l + 3ql}2_{+ l}3_{= lim} q 1 q4 _Sq l=1q 3₊Sq l=13q 2_{l +}Sq l=13ql 2₊Sq l=1l 3 = lim_qq1₄  q · q3_{+ 3q}2Sq l=1l + 3q q S l=1l 2₊Sq l=1l 3 = lim q  1 + 3_q2 · q(q + 1)₂ + 3_q3 · q(q + 1)(2q + 1)₆ + 1_q4 · q2(q + 1)₄ 2  = lim_q  1 + 3₂· q + 1_q + 1₂· q + 1_q · 2q + 1_q + 1₄· (q + 1)_q2 2  = lim_q % 1 + 3₂  1 + 1_q  + 1₂  1 + 1_q  2 + 1_q  + 1₄  1 + 1_q 2& = 1 + 3₂+1 2· 2 +14 = 3=75 27. ] e d { g{ = limq e  d q q S l=1  d + e  d_q l  = lim_q  d(e  d) q q S l=11 + (e  d) 2 q2 q S l=1l  = lim_q  d(e  d) q q + (e  d) 2 q2 · q(q + 1)₂  = d (e  d) + lim_q(e  d)₂ 2  1 + 1_q  = d(e  d) +1 2(e  d)2= (e  d)  d +1 2e 12d  = (e  d)1 2(e + d) =12  e2_{ d}2 29. i({) = {

1 + {5,d = 2, e = 6, and { = 6  2_q = 4_q. Using Theorem 4, we get{l = {l= 2 + l { = 2 + 4l_q,

so ] 6 2 { 1 + {5g{ = lim_qUq= lim_q q S l=1 2 + 4l_q 1 +  2 + 4l_q 5 · 4_q. 31. { = (  0)@q = @q and {_l = {l = l@q. ]  0 sin 5{ g{ = limq q S l=1(sin 5{l)  q  = lim q q S l=1  sin 5l q   q CAS =  lim q 1 qcot  5 2q  CAS =  2 5  = 2 5

33. (a) Think ofU₀2i({) g{ as the area of a trapezoid with bases 1 and 3 and height 2. The area of a trapezoid is D =1₂(e + E)k, soU₀2i({) g{ =1₂(1 + 3)2 = 4. (b)U₀5i({) g{ =U₀2i({) g{ trapezoid +U₂3i({) g{ rectangle +U₃5i({) g{ triangle = 1 2(1 + 3)2 + 3 · 1 + 12· 2 · 3 = 4 + 3 + 3 = 10

(c)U₅7i({) g{ is the negative of the area of the triangle with base 2 and height 3.U₅7i({) g{ = 1₂· 2 · 3 = 3. (d)U₇9i({) g{ is the negative of the area of a trapezoid with bases 3 and 2 and height 2, so it equals

1 2(E + e)k = 12(3 + 2)2 = 5. Thus, U9 0 i({) g{ = U5 0 i({) g{ + U7 5 i({) g{ + U9 7 i({) g{ = 10 + (3) + (5) = 2.

SECTION 5.2 THE DEFINITE INTEGRAL ¤ 241 35.U₀31₂{  1g{ can be interpreted as the area of the triangle above the x-axis

minus the area of the triangle below the x-axis; that is,

1 2(1) ₁ 2  1 2(2)(1) =14  1 = 34. 0

37.U_30 1 +9  {2g{ can be interpreted as the area under the graph of i({) = 1 +9  {2_{between{ = 3 and { = 0. This is equal to one-quarter}

the area of the circle with radius3, plus the area of the rectangle, so U0

3



1 +9  {2_{g{ =}1

4 · 32+ 1 · 3 = 3 +94.

39.U_12 |{| g{ can be interpreted as the sum of the areas of the two shaded triangles; that is, 1₂(1)(1) +1₂(2)(2) = 1₂+4₂ =5₂.

0

41.Usin2{ cos4{ g{ = 0 since the limits of intergration are equal.

43.U₀1(5  6{2) g{ =U₀15 g{  6U₀1{2g{ = 5(1  0)  61₃= 5  2 = 3 45.U₁3h{ + 2g{ =U₁3h{ · h2g{ = h2U₁3h{g{ = h2(h3 h) = h5 h3

47.U_22 i({) g{ +U₂5i({) g{ U_21i({) g{ =U_25 i({) g{ +U_12i({) g{ [by Property 5 and reversing limits]

=U_15 i({) g{ [Property 5]

49.U₀9[2i({) + 3j({)] g{ = 2U₀9i({) g{ + 3U₀9j({) g{ = 2(37) + 3(16) = 122

51.Using Integral Comparison Property 8,p  i({)  P  p(2  0) U₀2i({) g{  P(2  0)  2p U2

0 i({) g{  2P.

53.If1  {  1, then 0  {2  1 and 1  1 + {2 2, so 1 1 + {2 2 and 1[1  (1)] U_11 1 + {2_{g{ }_{2 [1  (1)] [Property 8]; that is, 2 }U1

1



1 + {2_{g{  2}_2.

55.If1  {  4, then 1 {  2> so 1(4  1) U4₁{ g{  2(4  1); that is, 3 U4₁{ g{  6.

57.If ₄  {  ₃, then1  tan { 3, so 1₃ ₄U@3_@4tan { g{ 3₃ ₄or ₁₂ U@3_@4tan { g{  ₁₂3. 59.The only critical number ofi({) = {h{on[0> 2] is { = 1. Since i(0) = 0, i(1) = h1 0=368, and

i(2) = 2h2_{ 0=271, we know that the absolute minimum value of i on [0> 2] is 0, and the absolute maximum is h}1_{. By}

Property 8> 0  {h{ h1for0  {  2  0(2  0) U₀2{h{g{  h1(2  0)  0 U₀2{h{g{  2@h. 61.{4+ 1 {4= {2, soU₁3{4+ 1 g{ U₁3{2g{ = 1₃33 13= 26₃.

63. Using right endpoints as in the proof of Property 2, we calculate Ue dfi({) g{ = lim_q q S l=1fi({l) { = limqf q S l=1i({l) { = f limq q S l=1i({l) { = f Ue di({) g{.

65. Since |i({)|  i({)  |i({)|, it follows from Property 7 that

Ue_d|i({)| g{ Ue_di({) g{ Ue_d|i({)| g{  Uedi({) g{ 

Ue

d|i({)| g{

Note that the denite integral is a real number, and so the following property applies:d  e  d  |e|  d for all real numberse and nonnegative numbers d.

67. To show thati is integrable on [0> 1] > we must show that lim

q q

S

l=1i({ 

l) { exists. Let n denote a positive integer and divide

the interval[0> 1] into n equal subintervals  0> 1 q  ,  1 q> 2q  ,=== ,  q  1 q > 1 

. If we choose{_l to be a rational number in the ith

subinterval, then we obtain the Riemann sumSq

l=1i({  l) · 1_q = 0, so lim_q q S l=1i({ 

l) · 1_q = lim_q0 = 0. Now suppose we

choose{_l to be an irrational number. Then we get Sq

l=1i({  l) · 1_q = q S l=11 · 1q = q · 1q = 1 for each q, so lim q q S l=1i({ 

l) · 1_q = lim_q1 = 1. Since the value of lim_q q

S

l=1i({ 

l) { depends on the choice of the sample points {l, the

limit does not exist, andi is not integrable on [0> 1].

69. lim q q S l=1 l4 q5 = limq q S l=1 l4 q4 · 1_q = limq q S l=1  l q 4 ₁

q. At this point, we need to recognize the limit as being of the form lim

q q

S

l=1i({l) {, where { = (1  0)@q = 1@q, {l= 0 + l { = l@q, and i({) = {

4_{. Thus, the denite integral}

isU1₀{4g{. 71. Choose{l= 1 + l_q and{_l = {l1{l= v 1 + l  1_q   1 + l_q  . Then U2 1 {2g{ = lim_qq1 q S l=1 1  1 +l  1 q  1 + l q  = lim qq q S l=1 1 (q + l  1)(q + l) = lim qq q S l=1  1 q + l  1 1q + l 

[by the hint] = lim

qq _q1S l=0 1 q + l q S l=1 1 q + l  = lim_qq  1 q+ 1q + 1+ · · · +2q  11    1 q + 1+ · · · +2q  11 + 12q  = lim_qq  1 q 12q  = lim_q1 1 2  =1 2

5.3

The Fundamental Theorem of Calculus

1. One process undoes what the other one does. The precise version of this statement is given by the Fundamental Theorem of Calculus. See the statement of this theorem and the paragraph that follows it on page 387.

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 243 3. (a)j({) =U₀{i(w) gw.

j(0) =U0

0 i(w) gw = 0

j(1) =U₀1i(w) gw = 1 · 2 = 2 [rectangle],

j(2) =U₀2i(w) gw =U₀1i(w) gw +U₁2i(w) gw = j(1) +U₁2i(w) gw = 2 + 1 · 2 +1

2· 1 · 2 = 5 [rectangle plus triangle],

j(3) =U₀3i(w) gw = j(2) +U₂3i(w) gw = 5 +1

2· 1 · 4 = 7,

j(6) = j(3) +U₃6i(w) gw [the integral is negative since i lies under the {-axis] = 7 +1

2· 2 · 2 + 1 · 2



= 7  4 = 3

(d)

(b)j is increasing on (0> 3) because as { increases from 0 to 3, we keep adding more area. (c)j has a maximum value when we start subtracting area; that is, at { = 3.

5. (a) By FTC1 withi(w) = w2andd = 1, j({) =U₁{w2gw 

j0_{({) = i({) = {}2_. (b) Using FTC2,j({) =U₁{w2gw =1₃w3{ 1= 13{313  j0({) = {2. 7.i(w) = 1 w3_{+ 1}andj({) = ] { 1 1 w3_{+ 1}gw, so by FTC1, j0({) = i({) = 1

{3_{+ 1}. Note that the lower limit,1, could be any

real number greater than1 and not affect this answer.

9.i(w) = w2sin w and j(|) =U₂|w2sin w gw, so by FTC1, j0(|) = i(|) = |2sin |.

11.I ({) = ]  {  1 + sec w gw = ] {   1 + sec w gw  I0_{({) =  g} g{ ] {   1 + sec w gw = 1 + sec { 13.Letx = 1 {. Thengxg{ =  1{2. Also, gk_g{ = gk_gxgx_g{, so k0_{({) = g} g{ ] 1@{ 2 arctan w gw = ggx ] x

2 arctan w gw · gxg{= arctan x gxg{ = arctan(1@{){2

.

15.Letx = tan {. Then gx

g{= sec2{. Also, g|g{ = g|gxgxg{, so |0_{= g} g{ ] tan { 0 t w +w gw = g_gx]x 0 t

w +w gw · gx_g{ =tx +x gx_g{ =ttan { +tan { sec2_{.

17.Letz = 1  3{. Then gz g{ = 3. Also, g|g{ = g|gzgzg{, so |0_{= g} g{ ] 1 13{ x3 1 + x2 gx = g_gz ] 1 z x3 1 + x2gx · gz_g{ =  g_gz ] z 1 x3 1 + x2gx · gz_g{ =  z 3 1 + z2(3) = 3(1  3{) 3 1 + (1  3{)2 19. ] 2 1  {3_{ 2{}_{g{ =}{4 4  {2 2 1=  24 4  22    (1)4 4  (1)2  = (4  4) 1 4 1  = 0 3 4  =3 4

TX.10

21. U4₁(5  2w + 3w2) gw =5w  w2+ w34₁= (20  16 + 64)  (5  1 + 1) = 68  5 = 63 23. U₀1{4@5g{ = k 5 9{9@5 l1 0= 5 9  0 = 59 25. ] 2 1 3 w4gw = 3 ] 2 1 w 4_{gw = 3}w3 3 2 1= 33  1 w3 2 1= 1  1 8 1  = 7₈ 27. U2₀{(2 + {5) g{ =U2₀(2{ + {6) g{ ={2+1₇{72₀=4 +128₇  (0 + 0) = 156₇ 29. ] 9 1 {  1_ { g{ = ] 9 1  {  { 1{  g{ =] 9 1 ({ 1@2_{ {}1@2_{) g{ =}k2 3{3@2 2{1@2 l9 1 =2 3· 27  2 · 3  2 3 2  = 12 4 3  =40 3

31. U₀@4sec2w gw =tan w@4₀ = tan₄  tan 0 = 1  0 = 1

33. U2₁(1 + 2|)2g| =U2₁(1 + 4| + 4|2) g| =| + 2|2+4₃|32₁=2 + 8 +32₃1 + 2 +₃4= 62₃ 13₃ =49₃ 35. ] 9 1 1 2{g{ = 12 ] 9 1 1 {g{ =12 k ln |{|l9 1= 1 2(ln 9  ln 1) =12ln 9  0 = ln 91@2= ln 3 37. ]  3@2 1@2 6  1  w2 gw = 6 ]  3@2 1@2 1  1  w2gw = 6  sin1_w3@2 1@2 = 6 k sin13 2   sin11 2 l = 6 3 6  = 6 6  =  39. U_11 hx+1gx =hx+11_1= h2 h0= h2 1 [or start with hx+1= hxh1]

41. Ifi({) = + sin { if 0  { ? @2 cos { if @2  {   then U 0i({) g{ = U@2 0 sin { g{ + U @2cos { g{ =   cos {@2₀ + [sin {]

@2=  cos2 + cos 0 + sin   sin2

= 0 + 1 + 0  1 = 0

Note thati is integrable by Theorem 3 in Section 5.2.

43. i({) = {4is not continuous on the interval[2> 1], so FTC2 cannot be applied. In fact, i has an innite discontinuity at { = 0, soU1_2{4_{g{ does not exist.}

45. i() = sec  tan  is not continuous on the interval [@3> ], so FTC2 cannot be applied. In fact, i has an innite discontinuity at{ = @2, soU_@3sec  tan gdoes not exist.

47. From the graph, it appears that the area is about60. The actual area is U27 0 {1@3g{ = k 3 4{4@3 l27 0 = 3 4· 81  0 = 2434 = 60=75. This is34 of the

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 245 49.It appears that the area under the graph is about 2₃of the area of the viewing

rectangle, or about2₃  2=1. The actual area is U

0 sin { g{ = [ cos {]0 = ( cos )  ( cos 0) =  (1) + 1 = 2.

51.U_12 {3g{ =1₄{42_1= 4 1₄ =15₄ = 3=75 53.j({) = ] 3{ 2{ x2_{ 1} x2_{+ 1}gx = ] 0 2{ x2_{ 1} x2_{+ 1}gx + ] 3{ 0 x2_{ 1} x2_{+ 1}gx =  ] 2{ 0 x2_{ 1} x2_{+ 1}gx + ] 3{ 0 x2_{ 1} x2_{+ 1}gx  j0_{({) = (2{)}2 1 (2{)2_{+ 1}· g_g{(2{) + (3{) 2_{ 1} (3{)2_{+ 1}· g_g{(3{) = 2 · 4{ 2_{ 1} 4{2_{+ 1}+ 3 · 9{ 2_{ 1} 9{2_{+ 1}

55.| =U{3_{w sin w gw =U1_{w sin w gw +U₁{3w sin w gw = U₁{w sin w gw +U₁{3w sin w gw 

|0_{= }4_{{ (sin}_{{ ) · g} g{( _{{ ) + {}3@2_sin({3_{) · g} g{  {3_{= }4{ sin{ 2{ + {3@2sin({3)(3{2) = 3{7@2_sin({3_{)  sin}{ 24_{ 57.I ({) = ] { 1 i(w) gw  I 0_{({) = i({) =}]{ 2 1  1 + x4 x gx % sincei(w) = ] w2 1  1 + x4 x gx &  I00_{({) = i}0_{({) =} s 1 + ({2₎4 {2 · g_g{  {2₌  1 + {8 {2 · 2{ = 2  1 + {8 { . SoI00(2) =  1 + 28₌_257.

59.By FTC2,U₁4i0({) g{ = i(4)  i(1), so 17 = i(4)  12  i(4) = 17 + 12 = 29.

61. (a) The Fresnel functionV({) =U₀{sin₂w2gw has local maximum values where 0 = V0({) = sin₂w2and V0_{changes from positive to negative. For{ A 0, this happens when}

2{2= (2q  1) [odd multiples of ] 

{2 _{= 2(2q  1)  { =}_{4q  2, q any positive integer. For { ? 0, V}0_{changes from positive to negative where} 

2{2= 2q [even multiples of ]  {2= 4q  { = 2



q. V0_{does not change sign at{ = 0.}

(b)V is concave upward on those intervals where V00({) A 0. Differentiating our expression for V0({), we get V00_{({) = cos} 2{2  2 2{  = { cos 2{2 

. For{ A 0, V00({) A 0 where cos(₂{2) A 0  0 ? ₂{2? ₂ or  2q 1 2   ?  2{2?  2q +1 2 

, q any integer  0 ? { ? 1 or4q  1 ? { ?4q + 1, q any positive integer. For{ ? 0, V00({) A 0 where cos(₂{2) ? 0  2q 3₂ ? ₂{2?2q 1₂, q any integer 

4q  3 ? {2_{? 4q  1 } _{4q  3 ? |{| ?}_{4q  1 } _{4q  3 ? { ?}_{4q  1 }

4q  3 A { A 4q  1, so the intervals of upward concavity for { ? 0 are4q  1> 4q  3,q any positive integer. To summarize:V is concave upward on the intervals (0> 1),3> 1,3>5,7> 5, 

7> 3,= = = .

(c) In Maple, we use plot({int(sin(Pi*tˆ2/2),t=0..x),0.2},x=0..2);. Note that Maple recognizes the Fresnel function, calling it FresnelS(x). In Mathematica, we use

Plot[{Integrate[Sin[Pi*tˆ2/2],{t,0,x}],0.2},{x,0,2}]. In Derive, we load the utility le FRESNELand plot FRESNEL_SIN(x). From the graphs, we see thatU₀{sin₂w2gw = 0=2 at {  0=74.

63. (a) By FTC1,j0({) = i({). So j0({) = i({) = 0 at { = 1> 3> 5> 7, and 9. j has local maxima at { = 1 and 5 (since i = j0 changes from positive to negative there) and local minima at{ = 3 and 7. There is no local maximum or minimum at { = 9, since i is not dened for { A 9.

(b) We can see from the graph thatU₀1i gw ?U₁3i gw ?U₃5i gw ? U₅7i gw ?U₇9i gw. So j(1) =U₀1i gw, j(5) =U5 0 i gw = j(1)  U3 1 i gw + U5 3 i gw, and j(9) = U9 0 i gw = j(5)  U7 5 i gw + U9 7 i gw. Thus,

j(1) ? j(5) ? j(9), and so the absolute maximum of j({) occurs at { = 9. (c)j is concave downward on those intervals where j00? 0. But j0({) = i({),

soj00({) = i0({), which is negative on (approximately)1₂> 2,(4> 6) and (8> 9). So j is concave downward on these intervals.

(d) 65. lim q q S l=1 l3 q4 = lim_q1  0_q q S l=1  l q 3 = ] 1 0 { 3_{g{ =}{4 4 1 0= 14

67. Supposek ? 0. Since i is continuous on [{ + k> {], the Extreme Value Theorem says that there are numbers x and y in [{ + k> {] such that i(x) = p and i(y) = P, where p and P are the absolute minimum and maximum values of i on [{ + k> {]. By Property 8 of integrals, p(k) U_{+k{ i(w) gw  P(k); that is, i(x)(k)  U_{{+ki(w) gw  i(y)(k). Sincek A 0, we can divide this inequality by k: i(x)  1

k ] {+k { i (w) gw  i(y). By Equation 2, j({ + k)  j({) k = 1k ] _{+k

{ i(w) gw for k 6= 0, and hence i(x)  j({ + k)  j({)k  i(y), which is Equation 3 in the

SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ¤ 247 69. (a) Leti({) ={  i0({) = 1@(2{ ) A 0 for { A 0  i is increasing on (0> ). If {  0, then {3 0, so

1 + {3 _{ 1 and since i is increasing, this means that i}_{1 + {}3_{ i(1) } _{1 + {}3_{ 1 for {  0. Next let}

j(w) = w2_{ w  j}0_{(w) = 2w  1  j}0_{(w) A 0 when w  1. Thus, j is increasing on (1> ). And since j(1) = 0,}

j(w)  0 when w  1. Now let w =1 + {3_{, where{  0.} _{1 + {}3_{ 1 (from above)  w  1  j(w)  0 }



1 + {3__{1 + {}3_{ 0 for {  0. Therefore, 1 }_{1 + {}3_{ 1 + {}3_{for{  0.}

(b) From part (a) and Property 7: U₀11 g{ U1₀1 + {3g{ U₀1(1 + {3) g{   {1₀U1₀1 + {3_{g{ }_{{ +}1 4{4 1 0  1  U1 0  1 + {3_{g{  1 +}1 4 = 1=25. 71.0 ? { 2 {4_{+ {}2_{+ 1} ? { 2 {4 = 1_{2 on[5> 10], so 0  ] 10 5 {2 {4_{+ {}2_{+ 1}g{ ? ] 10 5 1 {2g{ =   1_{ 10 5 =  110  1₅  = 1₁₀= 0=1= 73.Using FTC1, we differentiate both sides of6 +

] { d i(w) w2 gw = 2  { to get i({) {2 = 2 1₂_{  i({) = {3@2.

To ndd, we substitute { = d in the original equation to obtain 6 + ] _d d i(w) w2 gw = 2  d  6 + 0 = 2d  3 =d  d = 9.

75. (a) LetI (w) =U₀wi(v) gv. Then, by FTC1, I0(w) = i(w) = rate of depreciation, so I (w) represents the loss in value over the interval[0> w]. (b)F(w) = 1 w  D +] w 0 i(v) gv 

= D + I (w)_w represents the average expenditure per unit ofw during the interval [0> w], assuming that there has been only one overhaul during that time period. The company wants to minimize average expenditure. (c)F(w) = 1 w  D +]w 0 i(v) gv  . Using FTC1, we haveF0(w) =  1 w2  D +] w 0 i(v) gv  + 1_wi(w). F0_{(w) = 0  w i(w) = D +}] w 0 i(v) gv  i(w) = 1w  D +]w 0 i(v) gv  = F(w).

5.4

Indefinite Integrals and the Net Change Theorem

1. g g{  {2_{+ 1 + F}_{= g} g{ k {2_{+ 1}1@2_{+ F}l₌1 2  {2_{+ 1}1@2_{· 2{ + 0 = } { {2_{+ 1} 3. g g{  sin { 1 3sin3{ + F  = g_g{sin { 1 3(sin {)3+ F  = cos { 1 3 · 3(sin {)2(cos {) + 0

= cos {(1  sin2_{{) = cos {(cos}2_{{) = cos}3_{

5. ] ({2_{+ {}2_{) g{ = {}3 3 + { 1 1 + F = 13{3 1{+ F

TX 10

7. ] {41₂{3+1₄{  2g{ = { 5 5  12{ 4 4 + 14{ 2 2  2{ + F =15{581{4+18{2 2{ + F 9. ] (1  w)(2 + w2_{) gw =}] _{(2  2w + w}2_{ w}3_{) gw = 2w  2 w}2 2 + w 3 3  w 4 4 + F = 2w  w2+13w314w4+ F 11. ] _{3_{ 2}_{ { g{ = ] _{3 {  2{ 1@2 {  g{ =] ({2_{ 2{}1@2_{) g{ = {}3 3  2 { 1@2 1@2 + F = 13{3 4  { + F 13. U(sin { + sinh {) g{ =  cos { + cosh { + F

15. U(  csc  cot ) g =1₂2+ csc  + F 17. U(1 + tan2) g =Usec2 g = tan  + F

19. U cos { +1₂{g{ = sin { +1₄{2+ F. The members of the family in the gure correspond toF = 5, 0, 5, and 10.

21. U₀2(6{2 4{ + 5) g{ =6 ·1₃{3 4 ·1₂{2+ 5{2₀=2{3 2{2+ 5{2₀= (16  8 + 10)  0 = 18 23. U_10 (2{  h{) g{ ={2 h{0_1= (0  1) 1  h1= 2 + 1@h 25. U_22 (3x + 1)2gx =U2_29x2+ 6x + 1gx =9 ·1₃x3+ 6 ·1₂x2+ x2_2=3x3+ 3x2+ x2_2 = (24 + 12 + 2)  (24 + 12  2) = 38  (14) = 52 27. U4₁w (1 + w) gw =U4₁(w1@2+ w3@2) gw = k 2 3w3@2+25w5@2 l4 1 = ₁₆ 3 +645  2 3+25  = 14 3 +625 =25615 29. ] 1 2  4|3_{+ 2} |3  g| =  4 ·1 4|4+ 2 · 1_2|2 1 2=  |4_{ 1} |2 1 2= (1  1)   16 1 4  = 63 4 31. U1₀{  3  { +4_{_{g{ =}U1 0({4@3+ {5@4) g{ = k 3 7{7@3+49{9@4 l1 0= ₃ 7+49   0 = 55 63 33. U4₁s5@{ g{ =5U₁4{1@2g{ =5 k 2{l4 1 =  5 (2 · 2  2 · 1) = 25 35. U₀(4 sin   3 cos ) g = 4 cos   3 sin ₀ = (4  0)  (4  0) = 8

37. ] @4 0 1 + cos2_ cos2_ g = ] @4 0  1 cos2_+ cos 2_ cos2_  g =] @4 0 (sec 2_{ + 1) g} =tan  + @4₀ =tan 4 +4   (0 + 0) = 1 + 4

SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ¤ 249 39. ] 64 1 1 +3_{ _{ g{ =]64 1  1 {1@2 + { 1@3 {1@2  g{ =] 64 1  {1@2_{+ {}(1@3)  (1@2)_{g{ =}]64 1 ({ 1@2_{+ {}1@6_{) g{} =k2{1@2₊6 5{5@6 l64 1 =  16 +192 5  2 +6 5  = 14 +186 5 =2565 41. ] 1@3 0 w2_{ 1} w4_{ 1}gw = ] 1@3 0 w2_{ 1} (w2_{+ 1)(w}2_{ 1)}gw = ] 1@3 0 1 w2_{+ 1}gw = 

arctan w1@₀ 3= arctan1@3 arctan 0 = 6  0 = 6 43.U_12 ({  2 |{|) g{ =U_10 [{  2({)] g{ +U₀2[{  2({)] g{ =U_10 3{ g{ +U₀2({) g{ = 31₂{2_10 1₂{22₀ = 30 1 2   (2  0) = 7 2 = 3=5

45.The graph shows that| = { + {2 {4has{-intercepts at { = 0 and at { = d  1=32. So the area of the region that lies under the curve and above the {-axis is Ud 0({ + {2 {4) g{ = ₁ 2{2+13{315{5 d 0 =1 2d2+13d315d5   0  0=84 47.D =U₀22|  |2g| =|2₃1|32₀=4 8₃ 0 = 4₃

49.Ifz0(w) is the rate of change of weight in pounds per year, then z(w) represents the weight in pounds of the child at age w. We know from the Net Change Theorem thatU₅10z0(w) gw = z(10)  z(5), so the integral represents the increase in the child’s weight (in pounds) between the ages of5 and 10.

51.Sinceu(w) is the rate at which oil leaks, we can write u(w) = Y0(w), where Y (w) is the volume of oil at time w. [Note that the minus sign is needed becauseY is decreasing, so Y0(w) is negative, but u(w) is positive.] Thus, by the Net Change Theorem, U120

0 u(w) gw = 

U120

0 Y0(w) gw =  [Y (120)  Y (0)] = Y (0)  Y (120), which is the number of gallons of oil that leaked

from the tank in the rst two hours (120 minutes).

53.By the Net Change Theorem,U₁₀₀₀5000U0({) g{ = U(5000)  U(1000), so it represents the increase in revenue when production is increased from1000 units to 5000 units.

55.In general, the unit of measurement forU_dei({) g{ is the product of the unit for i({) and the unit for {. Since i({) is measured in newtons and{ is measured in meters, the units forU₀100i({) g{ are newton-meters. (A newton-meter is abbreviated N·m and is called a joule.)

57. (a) Displacement=U₀3(3w  5) gw =3₂w2 5w3 0= 272  15 = 32 m (b) Distance traveled=U₀3|3w  5| gw =U₀5@3(5  3w) gw +U_5@33 (3w  5) gw =5w 3 2w2 5@3 0 + ₃ 2w2 5w 3 5@3=253 32·259 +272  15  ₃ 2·259 253  =41 6 m

59. (a)y0(w) = d(w) = w + 4  y(w) =1₂w2+ 4w + F  y(0) = F = 5  y(w) =1₂w2+ 4w + 5 m@s (b) Distance traveled=U₀10|y(w)| gw =U₀101₂w2+ 4w + 5gw = U₀101₂w2+ 4w + 5gw =1₆w3+ 2w2+ 5w10

0 = 500 3 + 200 + 50 = 41623 m

TX 10

61. Sincep0({) = ({), p =U₀4({) g{ =U₀49 + 2{g{ =k9{ +4₃{3@2l4

0= 36 + 32

3  0 = 1403 = 4623 kg.

63. Letv be the position of the car. We know from Equation 2 that v(100)  v(0) =U₀100y(w) gw. We use the Midpoint Rule for 0  w  100 with q = 5. Note that the length of each of the ve time intervals is 20 seconds = 20

3600hour = 1801 hour.

So the distance traveled is U100

0 y(w) gw  1801 [y(10) + y(30) + y(50) + y(70) + y(90)] = 1801 (38 + 58 + 51 + 53 + 47) =247180  1=4 miles.

65. From the Net Change Theorem, the increase in cost if the production level is raised from 2000 yards to 4000 yards isF(4000)  F(2000) =U₂₀₀₀4000F0({) g{. U4000 2000 F0({) g{ = U4000 2000  3  0=01{ + 0=000006{2_{g{ =}_{3{  0=005{}2_{+ 0=000002{}34000 2000= 60,0002,000 = $58,000

67. (a) We can nd the area between the Lorenz curve and the line| = { by subtracting the area under | = O({) from the area under| = {. Thus,

coefcient of inequality= area between Lorenz curve and line| = { area under line| = { =

U1 0 [{  O({)] g{_U 1 0 { g{ = U1 0 [{  O({)] g{ [{2_@2]1 0 = U1 0 [{  O({)] g{ 1@2 = 2 U₁ 0 [{  O({)] g{

(b)O({) =₁₂5{2+₁₂7{  O(50%) = O₂1=₄₈5 +₂₄7 = 19₄₈= 0=39583, so the bottom 50% of the households receive at most about40% of the income. Using the result in part (a),

coefcient of inequality= 2U1₀[{  O({)] g{ = 2U1₀{ ₁₂5{2₁₂7{g{ = 2U1₀₁₂5{ ₁₂5{2g{ = 2U1₀ 5 12({  {2) g{ = 56 ₁ 2{213{3 1 0=56 ₁ 213  =5 6 ₁ 6  = 5 36

5.5

The Substitution Rule

1. Letx = {. Then gx =  g{> so g{ =  gx. Thus,Uh{g{ =Uhx(gx) = hx+ F = h{+ F. Don’t forget that it is often very easy to check an indenite integration by differentiating your answer. In this case,

g

g{(h{+ F) = [h{(1)] = h{, the desired result. 3. Letx = {3+ 1. Then gx = 3{2g{ and {2g{ =1₃gx, so

] {2s_{3_{+ 1 g{ =}] _x1 3gx  = 1₃x_3@23@2+ F = 1₃· 2₃x3@2_{+ F =} 2 9({3+ 1)3@2+ F.

5. Letx = cos . Then gx =  sin gand sin g = gx, so ]

cos3_{ sin g=}] _x3_{(gx) = x}4

4 + F = 14cos4 + F.

SECTION 5.5 THE SUBSTITUTION RULE ¤ 251 9.Letx = 3{  2. Then gx = 3 g{ and g{ = 1₃gx, soU(3{  2)20g{ =Ux201₃gx=1₃·₂₁1x21+ F = ₆₃1(3{  2)21+ F. 11.Letx = 2{ + {2. Thengx = (2 + 2{) g{ = 2(1 + {) g{ and ({ + 1) g{ = 1₂gx, so

] ({ + 1)s2{ + {2_{g{ =}] _x1 2gx  = 1₂ x_3@23@2+ F = 1 3  2{ + {23@2 + F.

Or: Letx =2{ + {2. Thenx2 = 2{ + {2  2x gx = (2 + 2{) g{  x gx = (1 + {) g{, so U

({ + 1)2{ + {2_{g{ =}U_{x · x gx =}U_x2_{gx =}1

3x3+ F = 13(2{ + {2)3@2+ F.

13.Letx = 5  3{. Then gx = 3 g{ and g{ = 1₃gx, so ] g{ 5  3{ = ] 1 x  1 3gx  = 1 3ln |x| + F = 13ln |5  3{| + F.

15.Letx = w. Then gx =  gw and gw =1gx, soUsin w gw =Usin x1gx= 1( cos x) + F = 1cos w + F. 17.Letx = 3d{ + e{3. Then gx = (3d + 3e{2) g{ = 3(d + e{2) g{, so

] _{d + e{}2  3d{ + e{3g{ = ] 1 3gx x1@2 = 1₃ ] x1@2_{gx =} 1 3· 2x2+ F = 23 s 3d{ + e{3_{+ F.} 19.Letx = ln {. Then gx = g{ {, so ] _{(ln {)}2 { g{ = U x2_{gx =}1 3x3+ F = 13(ln {)3+ F. 21.Letx =w. Then gx = gw 2wand 1  wgw = 2 gx, so ] cos_w w gw = U

cos x (2 gx) = 2 sin x + F = 2 sinw + F. 23.Letx = sin . Then gx = cos g, soUcos  sin6g=Ux6gx = 1₇x7+ F = 1₇sin7 + F.

25.Letx = 1 + h{. Thengx = h{g{, soUh{1 + h{g{ =U x gx =2₃x3@2+ F = 2₃(1 + h{)3@2+ F.

Or: Letx =1 + h{. Thenx2 = 1 + h{and2x gx = h{g{, so U

h{_{1 + h}{_{g{ =}U_{x · 2x gx =} 2

3x3+ F = 23(1 + h{)3@2+ F.

27.Letx = 1 + }3. Thengx = 3}2g} and }2g} = 1₃gx, so ] }2 3  1 + }3g} = ] x1@31 3gx  =1 3 ·32x2@3+ F = 12(1 + }3)2@3+ F.

29.Letx = tan {. Then gx = sec2{ g{, soUhtan {sec2{ g{ =Uhxgx = hx+ F = htan {+ F.

31.Letx = sin {. Then gx = cos { g{, so

] _{cos {} sin2_{g{ = ] ₁ x2gx = ] x2_{gx = x}1 1 + F =  1x+ F =  1sin {+ F [or csc { + F ].

33.Letx = cot {. Then gx =  csc2{ g{ and csc2{ g{ = gx, so ] 

cot { csc2_{{ g{ =}] _{x (gx) = x}3@2

3@2 + F = 23(cot {)3@2+ F.

35.

] _{sin 2{}

1 + cos2_{g{ = 2

] _{sin { cos {}

1 + cos2_{g{ = 2L. Let x = cos {. Then gx =  sin { g{, so

2L = 2] _{1 + x}x gx₂ = 2 ·1

2ln(1 + x2) + F =  ln(1 + x2) + F =  ln(1 + cos2{) + F. Or: Letx = 1 + cos2{.

37. ]

cot { g{ =] cos {_{sin {}g{. Let x = sin {. Then gx = cos { g{, so] cot { g{ =] _x1gx = ln |x| + F = ln |sin {| + F. 39. Letx = sec {. Then gx = sec { tan { g{, so

U

sec3_{{ tan { g{ =}U_sec2_{{ (sec { tan {) g{ =}U_x2_{gx =}1

3x3+ F = 13sec3{ + F.

41. Letx = sin1{. Then gx = 1

1  {2 g{, so ] _g{  1  {2_sin1_{= ] ₁ xgx = ln |x| + F = lnsin1{ +F. 43. Letx = 1 + {2. Thengx = 2{ g{, so ] 1 + { 1 + {2g{ = ] 1 1 + {2g{ + ] { 1 + {2 g{ = tan1{ + ] 1 2gx x = tan1{ +12ln|x| + F = tan1_{{ +}1 2ln1 +{2 +F = tan1{ +12ln  1 + {2_{+ F [since 1 + {}2_{A 0].} 45. Letx = { + 2. Then gx = g{, so ] { 4  { + 2g{ = ] x  2 4  x gx = ] (x3@4_{ 2x}1@4_{) gx =} 4 7x7@4 2 ·43x3@4+ F = 4 7({ + 2)7@483({ + 2)3@4+ F In Exercises 47–50, leti({)denote the integrand andI ({)its antiderivative (withF = 0).

47. i({) = {({2 1)3. x = {2 1  gx = 2{ g{> so U {({2_{ 1)}3_{g{ =}U_x31 2gx  =1 8x4+ F =18({2 1)4+ F

Wherei is positive (negative), I is increasing (decreasing). Where i changes from negative to positive (positive to negative),I has a local minimum (maximum).

49. i({) = sin3{ cos {. x = sin {  gx = cos { g{, so U

sin3{ cos { g{ =Ux3gx = 1

4x4+ F = 14sin4{ + F

Note that at{ = ₂,i changes from positive to negative and I has a local maximum. Also, bothi and I are periodic with period , so at { = 0 and at{ = , i changes from negative to positive and I has local minima.

51. Letx = {  1, so gx = g{. When { = 0, x = 1; when { = 2, x = 1. Thus,U₀2({  1)25g{ =U_11 x25gx = 0 by Theorem 7(b), sincei(x) = x25is an odd function.

SECTION 5.5 THE SUBSTITUTION RULE ¤ 253 53.Letx = 1 + 2{3, sogx = 6{2g{. When { = 0, x = 1; when { = 1, x = 3. Thus,

U1 0{2  1 + 2{35_{g{ =}U3 1x5 ₁ 6gx  =1 6 ₁ 6x6 3 1= 361(36 16) = 361(729  1) = 72836 =1829 .

55.Letx = w@4, so gx = 1₄gw. When w = 0, x = 0; when w = , x = @4. Thus, U 0 sec2(w@4) gw = U@4 0 sec2x (4 gx) = 4  tan x@4₀ = 4tan 4  tan 0  = 4(1  0) = 4. 57.U_@6@6 tan3g= 0 by Theorem 7(b), since i() = tan3 is an odd function.

59.Letx = 1@{, so gx = 1@{2g{. When { = 1, x = 1; when { = 2, x = 1₂. Thus, ] 2 1 h1@{ {2 g{ = ] 1@2 1 h x_{(gx) = }_hx1@2 1 = (h1@2 h) = h   h. 61.Letx = 1 + 2{, so gx = 2 g{. When { = 0, x = 1; when { = 13, x = 27. Thus,

] 13 0 g{ 3 t (1 + 2{)2 = ] 27 1 x 2@31 2gx  =k1 2· 3x1@3 l27 1 = 3 2(3  1) = 3.

63.Letx = {2+ d2, sogx = 2{ g{ and { g{ =1₂gx. When { = 0, x = d2; when{ = d, x = 2d2. Thus, ] d 0 { s {2_{+ d}2_{g{ =}]2d 2 d2 x 1@21 2gx  = 1 2 k 2 3x3@2 l2d2 d2 = k 1 3x3@2 l2d2 d2 = 1 3 k (2d2₎3@2_{ (d}2₎3@2l₌ 1 3  22  1d3

65.Letx = {  1, so x + 1 = { and gx = g{. When { = 1, x = 0; when { = 2, x = 1. Thus, ] 2 1 {  {  1 g{ =] 1 0 (x + 1)  x gx =] 1 0 (x 3@2_{+ x}1@2_{) gx =}k2 5x5@2+23x3@2 l1 0= 2 5+23 =1615. 67.Letx = ln {, so gx = g{

{ . When{ = h, x = 1; when { = h4;x = 4. Thus, ] h4 h g{ {ln {= ] 4 1 x 1@2_{gx = 2}k_x1@2l4 1= 2(2  1) = 2.

69.Letx = h}+ }, so gx = (h}+ 1) g}. When } = 0, x = 1; when } = 1, x = h + 1. Thus, ] 1 0 h}_{+ 1} h}_{+ }}g} = ] h+1 1 1 xgx = k ln |x|lh+1 1 = ln |h + 1|  ln |1| = ln(h + 1).

71.From the graph, it appears that the area under the curve is about

1 +a little more than1₂· 1 · 0=7, or about1=4. The exact area is given by D =U1

0



2{ + 1 g{. Let x = 2{ + 1, so gx = 2 g{. The limits change to 2 · 0 + 1 = 1 and 2 · 1 + 1 = 3, and D =U₁3x1 2gx  = 1 2 k 2 3x3@2 l3 1= 1 3  33  1=3 1 3  1=399.

73.First write the integral as a sum of two integrals: L =U_22 ({ + 3)4  {2_{g{ = L1+ L2=}U2 2{  4  {2_{g{ +}U2 23  4  {2_{g{. L1= 0 by Theorem 7(b), since}

i({) = {4  {2_{is an odd function and we are integrating from{ = 2 to { = 2. We interpret L2as three times the area of}

a semicircle with radius2, so L = 0 + 3 ·1₂ · 22= 6.

TX.10

75. First Figure Letx ={, so { = x2andg{ = 2x gx. When { = 0, x = 0; when { = 1, x = 1. Thus, D1=U₀1h  {_{g{ =}U1 0hx(2x gx) = 2 U1 0 xhxgx. Second Figure D2=U₀12{h{g{ = 2U₀1xhxgx.

Third Figure Letx = sin {, so gx = cos { g{. When { = 0, x = 0; when { = ₂,x = 1. Thus,

D3=U₀@2hsin {sin 2{ g{ =U₀@2hsin {(2 sin { cos {) g{ =U₀1hx(2x gx) = 2U₀1xhxgx.

SinceD1= D2= D3, all three areas are equal.

77. The rate is measured in liters per minute. Integrating fromw = 0 minutes to w = 60 minutes will give us the total amount of oil that leaks out (in liters) during the rst hour.

U60 0 u(w) gw = U60 0 100h0=01wgw [x = 0=01w, gx = 0=01gw] = 100U0=6₀ hx_{(100 gx) = 10,000}_hx0=6 0 = 10,000(h0=6 1)  4511=9  4512 liters

79. The volume of inhaled air in the lungs at timew is Y (w) =U₀wi(x) gx =U₀w1 2sin ₂ 5 x  gx =U₀2w@51 2sin y ₅ 2gy   substitutey =2₅ x, gy =2₅ gx = 5 4   cos y2w@5₀ = 5 4   cos2 5 w  + 1= 5 4  1  cos2 5 w  liters 81. Letx = 2{. Then gx = 2 g{, soU₀2i(2{) g{ =U₀4i(x)₂1gx= 1₂U₀4i(x) gx =1₂(10) = 5. 83. Letx = {. Then gx = g{, so Ue di({) g{ = Ue di(x)(gx) = Ud e i(x) gx = Ud e i({) g{

From the diagram, we see that the equality follows from the fact that we are reecting the graph ofi, and the limits of integration, about the |-axis. 85. Letx = 1  {. Then { = 1  x and g{ = gx, so

U₁ 0{d(1  {)eg{ = U₀ 1(1  x)dxe(gx) = U₁ 0xe(1  x)dgx = U₁ 0{e(1  {)dg{. 87. { sin {

1 + cos2_{ = { ·_{2  sin}sin {2_{ = { i(sin {), where i(w) = _{2  w}w 2. By Exercise 86,

] 0 { sin { 1 + cos2_{g{ = ]  0 { i(sin {) g{ = 2 ]  0 i(sin {) g{ = 2 ]  0 sin { 1 + cos2_{g{

Letx = cos {. Then gx =  sin { g{. When { = , x = 1 and when { = 0, x = 1. So  2 ]  0 sin { 1 + cos2_{g{ = ₂ ] 1 1 gx 1 + x2 = ₂ ] 1 1 gx 1 + x2 = ₂  tan1_x1 1 = ₂[tan1_{1  tan}1_{(1)] =} 2k4   ₄l= ₄2

CHAPTER 5 REVIEW ¤ 255

5

Review

1. (a)Sq_l=1i({_l) { is an expression for a Riemann sum of a function i. {

l is a point in thelth subinterval [{l1> {l] and { is the length of the subintervals.

(b) See Figure 1 in Section 5.2.

(c) In Section 5.2, see Figure 3 and the paragraph beside it. 2. (a) See Denition 5.2.2.

(b) See Figure 2 in Section 5.2.

(c) In Section 5.2, see Figure 4 and the paragraph by it (contains “net area”). 3.See the Fundamental Theorem of Calculus after Example 9 in Section 5.3. 4. (a) See the Net Change Theorem after Example 5 in Section 5.4.

(b)U_ww2

1 u(w) gw represents the change in the amount of water in the reservoir between time w1and timew2.

5. (a)U₆₀120y(w) gw represents the change in position of the particle from w = 60 to w = 120 seconds. (b)U₆₀120|y(w)| gw represents the total distance traveled by the particle from w = 60 to 120 seconds. (c)U₆₀120d(w) gw represents the change in the velocity of the particle from w = 60 to w = 120 seconds. 6. (a)Ui({) g{ is the family of functions {I | I0= i}. Any two such functions differ by a constant.

(b) The connection is given by the Net Change Theorem:U_dei({) g{ =U i({) g{e

difi is continuous.

7.The precise version of this statement is given by the Fundamental Theorem of Calculus. See the statement of this theorem and the paragraph that follows it at the end of Section 5.3.

8.See the Substitution Rule (5.5.4). This says that it is permissible to operate with theg{ after an integral sign as if it were a differential.

1.True by Property 2 of the Integral in Section 5.2. 3.True by Property 3 of the Integral in Section 5.2.

5.False. For example, leti({) = {2. ThenU₀1{2g{ =U₀1{ g{ = 1₂, buttU₀1{2g{ = t

1 3 =13.

7.True by Comparison Property 7 of the Integral in Section 5.2.

9.True. The integrand is an odd function that is continuous on[1> 1], so the result follows from Theorem 5.5.7(b).

TX.10

11. False. The functioni({) = 1@{4is not bounded on the interval[2> 1]. It has an innite discontinuity at { = 0, so it is not integrable on the interval. (If the integral were to exist, a positive value would be expected, by Comparison Property 6 of Integrals.)

13. False. For example, the function| = |{| is continuous on R, but has no derivative at { = 0.

15. False. Ue_di({) g{ is a constant, so g g{

Ue di({) g{



= 0, not i({) [unless i({) = 0]. Compare the given statement carefully with FTC1, in which the upper limit in the integral is{.

1. (a) O₆= S6

l=1i({l1) {

[{ =6  0₆ = 1]

= i({0) · 1 + i({1) · 1 + i({2) · 1 + i({3) · 1 + i({4) · 1 + i({5) · 1

 2 + 3=5 + 4 + 2 + (1) + (2=5) = 8

The Riemann sum represents the sum of the areas of the four rectangles above the{-axis minus the sum of the areas of the two rectangles below the {-axis. (b) P6= 6 S l=1i({l) { [{ = 6  0 6 = 1]

= i({1) · 1 + i({2) · 1 + i({3) · 1 + i({4) · 1 + i({5) · 1 + i({6) · 1

= i(0=5) + i(1=5) + i(2=5) + i(3=5) + i(4=5) + i(5=5)  3 + 3=9 + 3=4 + 0=3 + (2) + (2=9) = 5=7

3. U₀1{ +1  {2g{ =U₀1{ g{ +U₀11  {2g{ = L1+ L2. L1can be interpreted as the area of the triangle shown in the gure andL2can be interpreted as the area of the quarter-circle. Area=1₂(1)(1) +1₄()(1)2 =1₂+₄.

5. U₀6i({) g{ =U₀4i({) g{ +U₄6i({) g{  10 = 7 +U₄6i({) g{  U₄6i({) g{ = 10  7 = 3

7. First note that eitherd or e must be the graph ofU₀{i(w) gw, sinceU₀0i(w) gw = 0, and f(0) 6= 0. Now notice that e A 0 when f is increasing, and thatf A 0 when d is increasing. It follows that f is the graph of i({), e is the graph of i0({), and d is the graph ofU₀{i(w) gw.

9. U₁28{3+ 3{2g{ =8 ·1₄{4+ 3 ·1₃{32₁=2{4+ {32₁=2 · 24+ 23 (2 + 1) = 40  3 = 37 11. U₀11  {9g{ ={ ₁₀1 {101₀=1 ₁₀1 0 =₁₀9

CHAPTER 5 REVIEW ¤ 257 13. ] 9 1 _{x  2x}2 x gx = ] 9 1 (x 1@2_{ 2x) gx =}k_2x1@2_{ x}2l9 1= (6  81)  (2  1) = 76

15.Letx = |2+ 1, so gx = 2| g| and | g| = 1₂gx. When | = 0, x = 1; when | = 1, x = 2. Thus, U1 0|(|2+ 1)5 g| = U2 1x5 ₁ 2gx  = 1 2 ₁ 6x6 2 1=121(64  1) =6312 =214. 17. ] 5 1 gw

(w  4)2 does not exist because the functioni(w) =_{(w  4)}1 2 has an innite discontinuity atw = 4;

that is,i is discontinuous on the interval [1> 5].

19.Letx = y3, sogx = 3y2gy. When y = 0, x = 0; when y = 1, x = 1. Thus, U1 0 y2cos(y3) gy = U1 0 cos x ₁ 3gx  = 1 3  sin x1₀=1 3(sin 1  0) = 13sin 1. 21. ] @4 @4 w4_{tan w}

2 + cos wgw = 0 by Theorem 5.5.7(b), since i(w) = w

4_{tan w}

2 + cos wis an odd function. 23. ]  1  { { 2 g{ =] 1_{ 1 2 g{ =] _{1₂  2_{+ 1  g{ =  1_{ 2 ln |{| + { + F 25.Letx = {2+ 4{. Then gx = (2{ + 4) g{ = 2({ + 2) g{, so ] { + 2  {2_{+ 4{}g{ = ] x1@21 2gx  = 1 2· 2x1@2+ F =  x + F =s{2_{+ 4{ + F.}

27.Letx = sin w. Then gx =  cos w gw, soUsin w cos w gw =Ux1gx= 1·1₂x2+ F =₂1 (sin w)2+ F.

29.Letx ={. Then gx = g{ 2{, so ] _h_{  {g{ = 2 ] hx_{gx = 2h}x_{+ F = 2h}{_{+ F.}

31.Letx = ln(cos {). Then gx =  sin {

cos { g{ =  tan { g{, so U tan { ln(cos {) g{ = Ux gx = 1 2x2+ F = 12[ln(cos {)]2+ F. 33.Letx = 1 + {4. Thengx = 4{3g{, so ] {3 1 + {4g{ = 1₄ ] 1 xgx = 14ln|x| + F =14ln  1 + {4_{+ F.}

35.Letx = 1 + sec . Then gx = sec  tan g, so ]

sec  tan  1 + sec  g =

] 1

1 + sec (sec  tan g) = ]

1

xgx = ln |x| + F = ln |1 + sec | + F.

37.Since{2 4 ? 0 for 0  { ? 2 and {2 4 A 0 for 2 ? {  3, we have{2 4 = ({2 4) = 4  {2for0  { ? 2 and {2 4 = {2_{ 4 for 2 ? {  3. Thus,} ]3 0 { 2 4g{ =] 2 0 (4  { 2_{) g{ +}] 3 2 ({ 2_{ 4) g{ =}_{4{  {}3 3 2 0+  {3 3  4{ 3 2 =8 8 3   0 + (9  12) 8 3 8  = 16 3  3 +163 =323 93 =233

TX.10

In Exercises 39 and 40, leti({)denote the integrand andI ({)its antiderivative (withF = 0). 39. Letx = 1 + sin {. Then gx = cos { g{, so

] _{cos { g{} 

1 + sin { = U

x1@2_{gx = 2x}1@2_{+ F = 2}_{1 + sin { + F.}

41. From the graph, it appears that the area under the curve| = {{ between { = 0 and{ = 4 is somewhat less than half the area of an 8 × 4 rectangle, so perhaps about13 or 14. To nd the exact value, we evaluate

U4 0 {  { g{ =U₀4{3@2_{g{ =}k2 5{5@2 l4 0= 2 5(4)5@2=645 = 12=8. 43. I ({) = ] { 0 w2 1 + w3gw  I0({) = g_g{ ] { 0 w2 1 + w3gw = { 2 1 + {3 45. Letx = {4. Then gx g{ = 4{3. Also, gjg{= gjgxgxg{, so j0_{({) = g} g{ ] {4 0 cos(w 2_{) gw = g} gx ] x 0 cos(w 2_{) gw · gx} g{ = cos(x2) gxg{ = 4{3cos({8). 47. | = ] {  { hw w gw = ] 1  { hw w gw + ] { 1 hw w gw =  ] { 1 hw w gw + ]{ 1 hw w gw  g| g{=  gg{ #] _{ 1 hw w gw $ + g g{ ] { 1 hw w gw  . Letx ={. Then g g{ ]_{ 1 hw w gw = gg{ ]x 1 hw w gw = ggx ] _x 1 hw w gw  gx g{= h x x · 1₂_{= h _{  { · 12{= h _{ 2{ , sog| g{= h  { 2{ + h { {. 49. If1  {  3, then12+ 3 {2+ 3 32+ 3  2 {2+ 3  23, so 2(3  1) U3 1  {2_{+ 3 g{  2}_{3(3  1); that is, 4 }U3 1  {2_{+ 3 g{  4}_3.

51. 0  {  1  0  cos {  1  {2cos {  {2  U₀1{2cos { g{ U₀1{2g{ =1₃{31₀=1₃ [Property 7]. 53. cos {  1  h{cos {  h{  U₀1h{cos { g{ U₀1h{g{ = [h{]1₀= h  1

55. { = (3  0)@6 =1₂, so the endpoints are0,1₂,1,3₂,2, 5₂, and3, and the midpoints are1₄,3₄,5₄, 7₄, 9₄, and11₄. The Midpoint Rule gives

U3 0sin({3) g{  6 S l=1i({l) { = 1 2 k sin1 4 3 + sin3 4 3 + sin5 4 3 + sin7 4 3 + sin9 4 3 + sin11 4 3l  0=280981.

CHAPTER 5 REVIEW ¤ 259 57.Note thatu(w) = e0(w), where e(w) = the number of barrels of oil consumed up to time w. So, by the Net Change Theorem,

U8

0 u(w) gw = e(8)  e(0) represents the number of barrels of oil consumed from Jan. 1, 2000, through Jan. 1, 2008.

59.We use the Midpoint Rule withq = 6 and w = 24  0₆ = 4. The increase in the bee population was U24

0 u(w) gw  P6= 4[u(2) + u(6) + u(10) + u(14) + u(18) + u(22)]

 4[50 + 1000 + 7000 + 8550 + 1350 + 150] = 4(18,100) = 72,400 61.Letx = 2 sin . Then gx = 2 cos gand when  = 0, x = 0; when  = ₂,x = 2. Thus,

U@2

0 i(2 sin ) cos g=

U2 0i(x) ₁ 2gx  = 1 2 U2 0i(x) gx = 12 U2 0i({) g{ =12(6) = 3.

63.Area under the curve| = sinh f{ between { = 0 and { = 1 is equal to 1  U1

0 sinh f{ g{ = 1  1f



cosh f{1₀= 1  1

f(cosh f  1) = 1 

cosh f  1 = f  cosh f = f + 1. From the graph, we get f = 0 and f  1=6161, but f = 0 isn’t a solution for this problem since the curve | = sinh f{ becomes | = 0 and the area under it is 0. Thus, f  1=6161.

65.Using FTC1, we differentiate both sides of the given equation,U₀{i(w) gw = {h2{+U₀{hwi(w) gw, and get i({) = h2{_{+ 2{h}2{_{+ h}{_{i({)  i({)}_{1  h}{_{= h}2{_{+ 2{h}2{ _{ i({) = h}2{(1 + 2{)

1  h{ .

67.Letx = i({) and gx = i0({) g{. So 2U_dei({)i0({) g{ = 2U_i(d)i(e)x gx =x2i(e)_i(d)= [i(e)]2 [i(d)]2.

69.Letx = 1  {. Then gx = g{, soU₀1i(1  {) g{ =U₁0i(x)(gx) =U₀1i(x) gx =U₀1i({) g{.

71.The shaded region has areaU₀1i({) g{ =1₃. The integralU₀1i1(|) g| gives the area of the unshaded region, which we know to be1 1₃ = 2₃. SoU₀1i1(|) g| =2₃.

PROBLEMS PLUS

1. Differentiating both sides of the equation{ sin { =U₀{2i(w) gw (using FTC1 and the Chain Rule for the right side) gives sin { + { cos { = 2{i({2_{). Letting { = 2 so that i({}2_{) = i(4), we obtain sin 2 + 2 cos 2 = 4i(4), so}

i(4) =1

4(0 + 2 · 1) = 2.

3. Differentiating the given equation,U₀{i(w) gw = [i({)]2, using FTC1 givesi({) = 2i({) i0({)  i({)[2i0_{({)  1] = 0, so i({) = 0 or i}0_{({) =} 1

2. Sincei({) is never 0, we must have i0({) = 12andi0({) = 12 

i({) = 1

2{ + F. To nd F, we substitute into the given equation to get

U{ 0 ₁ 2w + F  gw =1 2{ + F 2  1

4{2+ F{ = 14{2+ F{ + F2. It follows thatF2 = 0, so F = 0, and i({) =12{.

5. i({) = ] j({) 0 1  1 + w3gw, where j({) = ] cos { 0 [1 + sin(w

2_{)] gw. Using FTC1 and the Chain Rule (twice) we have}

i0_{({) = s} 1 1 + [j({)]3j 0_{({) = s} 1 1 + [j({)]3[1 + sin(cos 2_{{)]( sin {). Now j} 2  = ] ₀ 0 [1 + sin(w 2_{)] gw = 0, so} i0 2  = 1 1 + 0(1 + sin 0)(1) = 1 · 1 · (1) = 1.

7. By l’Hospital’s Rule and the Fundamental Theorem, using the notationexp(|) = h|, lim {0 U{ 0(1  tan 2w)1@wgw { H

= lim_{0(1  tan 2{)₁ 1@{ = exp  lim {0 ln(1  tan 2{) {  H = exp  lim {0 2 sec2 _2{ 1  tan 2{  = exp  2 · 12 1  0  = h2

9. i({) = 2 + {  {2= ({ + 2)({ + 1) = 0  { = 2 or { = 1. i({)  0 for { [1> 2] and i({) ? 0 everywhere else. The integralU_de(2 + {  {2) g{ has a maximum on the interval where the integrand is positive, which is [1> 2]. So d = 1, e = 2. (Any larger interval gives a smaller integral since i({) ? 0 outside [1> 2]. Any smaller interval also gives a smaller integral sincei({)  0 in [1> 2].)

11. (a) We can split the integralU₀q[[{]] g{ into the sumSq

l=1

kUl

l1[[{]] g{

l

. But on each of the intervals[l  1> l) of integration, [[{]] is a constant function, namely l  1. So the ith integral in the sum is equal to (l  1)[l  (l  1)] = (l  1). So the original integral is equal toSq

l=1(l  1) = q1_S

l=1l = (q  1)q2 .

261

TX.10

(b) We can writeU_de[[{]] g{ =U₀e[[{]] g{ U₀d[[{]] g{.

NowU₀e[[{]] g{ =U₀[[e]][[{]] g{ +U_[[e]]e [[{]] g{. The rst of these integrals is equal to1₂([[e]]  1) [[e]], by part (a), and since[[{]] = [[e]] on [[[e]] > e], the second integral is just [[e]] (e  [[e]]). So

Ue

0 [[{]] g{ =12([[e]]  1) [[e]] + [[e]] (e  [[e]]) = 12[[e]] (2e  [[e]]  1) and similarly

Ud

0 [[{]] g{ =12[[d]] (2d  [[d]]  1).

Therefore,U_de[[{]] g{ =1₂[[e]] (2e  [[e]]  1) 1₂[[d]] (2d  [[d]]  1).

13. LetT({) = ] _{ 0 S (w) gw =  dw + e₂w2_{+ f} 3w3+ g4w4 { 0= d{ + e2{ 2_{+ f}

3{3+ g4{4. ThenT(0) = 0, and T(1) = 0 by the given condition,d + e

2+ f3+ g4 = 0. Also, T0({) = S ({) = d + e{ + f{2+ g{3by FTC1. By Rolle’s Theorem, applied to T on [0> 1], there is a number u in (0> 1) such that T0_{(u) = 0, that is, such that S (u) = 0. Thus, the equation S ({) = 0 has a}

root between0 and 1.

More generally, ifS ({) = d₀+ d₁{ + d₂{2+ · · · + d_q{qand ifd₀+ d1

2 + d32 + · · · + dq + 1q = 0> then the equation S({) = 0 has a root between 0 and 1= The proof is the same as before:

LetT({) = ] {

0 S (w) gw = d0{ + d 1

2 {2+ d32{3+ · · · + dq + 1q {q. ThenT(0) = T(1) = 0 and T0({) = S ({). By Rolle’s Theorem applied toT on [0> 1], there is a number u in (0> 1) such that T0(u) = 0, that is, such that S (u) = 0.

15. Note that g g{ ] _{ 0 ] _x 0 i(w) gw  gx  =] { 0 i(w) gw by FTC1, while g g{ ] _{ 0 i(x)({  x) gx  = g_g{  {]{ 0 i(x) gx   g_g{ ] _{ 0 i(x)x gx  =U{₀i(x) gx + {i({)  i({){ =U{₀i(x) gx Hence,U{₀i(x)({  x) gx =U{₀Ux₀i(w) gwgx + F. Setting { = 0 gives F = 0.

17. lim q  1  qq + 1+ 1  qq + 2 + · · · + 1 _q q + q  = lim q 1 q u q q + 1+ u _q q + 2+ · · · + u _q q + q  = lim q 1 q # 1 s 1 + 1@q+ 1 s 1 + 2@q+ · · · + 1  1 + 1 $ = lim q 1 q q [ l=1 i  l q   wherei({) =  1 1 + {  = ] 1 0 1  1 + {g{ =  21 + {1₀= 22  1