Checkpoint
∑ i = 3 6 2 i = 2 3 + 2 4 + 2 5 + 2 6 = 120 ∑ i = 3 6 2 i = 2 3 + 2 4 + 2 5 + 2 6 = 120
15,550
440
The left-endpoint approximation is 0.7595. The right-endpoint approximation is 0.6345.
Upper sum=8.0313.Upper sum=8.0313.
A ≈ 1.125 A ≈ 1.125
6
18 square units
6
18
6 ∫ 1 3 x 3 d x − 4 ∫ 1 3 x 2 d x + 2 ∫ 1 3 x d x − ∫ 1 3 3 d x 6 ∫ 1 3 x 3 d x − 4 ∫ 1 3 x 2 d x + 2 ∫ 1 3 x d x − ∫ 1 3 3 d x
−7
3
Average value = 1.5 ; c = 3 Average value = 1.5 ; c = 3
c = 3 c = 3
g ′ ( r ) = r 2 + 4 g ′ ( r ) = r 2 + 4
F ′ ( x ) = 3 x 2 cos x 3 F ′ ( x ) = 3 x 2 cos x 3
F ′ ( x ) = 2 x cos x 2 − cos x F ′ ( x ) = 2 x cos x 2 − cos x
7 24 7 24
Kathy still wins, but by a much larger margin: James skates 24 ft in 3 sec, but Kathy skates 29.3634 ft in 3 sec.
− 10 3 − 10 3
Net displacement: e2−92≈−0.8055m;e2−92≈−0.8055m; total distance traveled: 4ln4−7.5+e22≈1.7404ln4−7.5+e22≈1.740 m
17.5 mi
64 5 64 5
∫ 3 x 2 ( x 3 − 3 ) 2 d x = 1 3 ( x 3 − 3 ) 3 + C ∫ 3 x 2 ( x 3 − 3 ) 2 d x = 1 3 ( x 3 − 3 ) 3 + C
( x 3 + 5 ) 10 30 + C ( x 3 + 5 ) 10 30 + C
− 1 sin t + C − 1 sin t + C
− cos 4 t 4 + C − cos 4 t 4 + C
91 3 91 3
2 3 π ≈ 0.2122 2 3 π ≈ 0.2122
∫ x 2 e −2 x 3 d x = − 1 6 e −2 x 3 + C ∫ x 2 e −2 x 3 d x = − 1 6 e −2 x 3 + C
∫ e x ( 3 e x − 2 ) 2 d x = 1 9 ( 3 e x − 2 ) 3 ∫ e x ( 3 e x − 2 ) 2 d x = 1 9 ( 3 e x − 2 ) 3
∫ 2 x 3 e x 4 d x = 1 2 e x 4 ∫ 2 x 3 e x 4 d x = 1 2 e x 4
1 2 ∫ 0 4 e u d u = 1 2 ( e 4 − 1 ) 1 2 ∫ 0 4 e u d u = 1 2 ( e 4 − 1 )
Q(t)=2tln2+8.557.Q(t)=2tln2+8.557. There are 20,099 bacteria in the dish after 3 hours.
There are 116 flies.
∫ 1 2 1 x 3 e 4 x −2 d x = 1 8 [ e 4 − e ] ∫ 1 2 1 x 3 e 4 x −2 d x = 1 8 [ e 4 − e ]
ln | x + 2 | + C ln | x + 2 | + C
x ln 3 ( ln x − 1 ) + C x ln 3 ( ln x − 1 ) + C
1 4 sin −1 ( 4 x ) + C 1 4 sin −1 ( 4 x ) + C
sin −1 ( x 3 ) + C sin −1 ( x 3 ) + C
1 10 tan −1 ( 2 x 5 ) + C 1 10 tan −1 ( 2 x 5 ) + C
1 4 tan −1 ( x 4 ) + C 1 4 tan −1 ( x 4 ) + C
π 8 π 8
Section 1.1 Exercises
a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting j=i−1.j=i−1. d. They are equal; the first sum factors the terms of the second.
385 − 30 = 355 385 − 30 = 355
15 − ( −12 ) = 27
5 ( 15 ) + 4 ( −12 ) = 27
∑ j = 1 50 j 2 − 2 ∑ j = 1 50 j = ( 50 ) ( 51 ) ( 101 ) 6 − 2 ( 50 ) ( 51 ) 2 = 40 , 375
4 ∑ k = 1 25 k 2 − 100 ∑ k = 1 25 k = 4 ( 25 ) ( 26 ) ( 51 ) 6 − 50 ( 25 ) ( 26 ) = −10 , 400
R 4 = -0.25
R 6 = 0.372
L 4 = 2.20
L 8 = 0.6875
L6=9.000=R6. ਤ੍ਰਿਕੋਣ ਦਾ ਖੇਤਰਫਲ 9 ਹੈ।
L6=13.12899=R6. ਉਹ ਬਰਾਬਰ ਹਨ।
L 10 = 4 10 ∑ i = 1 10 4 − ( −2 + 4 ( i − 1 ) 10 ) 2
R 100 = e − 1 100 ∑ i = 1 100 ln ( 1 + ( e − 1 ) i 100 )
R100=0.33835,L100=0.32835. ਪਲਾਟ ਦਰਸਾਉਂਦਾ ਹੈ ਕਿ ਖੱਬਾ ਰੀਮੈਨ ਜੋੜ ਘੱਟ ਅੰਦਾਜ਼ਾ ਹੈ ਕਿਉਂਕਿ ਫੰਕਸ਼ਨ ਵੱਧ ਰਿਹਾ ਹੈ। ਇਸੇ ਤਰ੍ਹਾਂ, ਸੱਜਾ ਰੀਮੈਨ ਜੋੜ ਵੱਧ ਅੰਦਾਜ਼ਾ ਹੈ। ਖੇਤਰਫਲ ਖੱਬੇ ਅਤੇ ਸੱਜੇ ਰੀਮੈਨ ਜੋੜਾਂ ਦੇ ਵਿਚਕਾਰ ਪਿਆ ਹੈ। ਦ੍ਰਿਸ਼ਟੀਗਤ ਸਪੱਸ਼ਟਤਾ ਲਈ ਦਸ ਆਇਤਾਕਾਰ ਦਿਖਾਏ ਗਏ ਹਨ। ਇਹ ਵਿਵਹਾਰ ਹੋਰ ਆਇਤਾਕਾਰਾਂ ਲਈ ਵੀ ਜਾਰੀ ਰਹਿੰਦਾ ਹੈ।
L100=3.980,R100=4.020,R100>L100. ਖੱਬਾ ਅੰਤ ਬਿੰਦੂ ਜੋੜ ਘੱਟ ਅੰਦਾਜ਼ਾ ਹੈ ਕਿਉਂਕਿ ਫੰਕਸ਼ਨ ਵੱਧ ਰਿਹਾ ਹੈ। ਇਸੇ ਤਰ੍ਹਾਂ, ਸੱਜਾ ਅੰਤ ਬਿੰਦੂ ਅਨੁਮਾਨ ਵੱਧ ਅੰਦਾਜ਼ਾ ਹੈ। ਖੇਤਰਫਲ ਖੱਬੇ ਅਤੇ ਸੱਜੇ ਅੰਤ ਬਿੰਦੂ ਅਨੁਮਾਨਾਂ ਦੇ ਵਿਚਕਾਰ ਪਿਆ ਹੈ।
L100=3.555,R100=3.670 ਪਲਾਟ ਦਰਸਾਉਂਦਾ ਹੈ ਕਿ ਖੱਬਾ ਰੀਮੈਨ ਜੋੜ ਘੱਟ ਅੰਦਾਜ਼ਾ ਹੈ ਕਿਉਂਕਿ ਫੰਕਸ਼ਨ ਵੱਧ ਰਿਹਾ ਹੈ। ਦ੍ਰਿਸ਼ਟੀਗਤ ਸਪੱਸ਼ਟਤਾ ਲਈ ਦਸ ਆਇਤਾਕਾਰ ਦਿਖਾਏ ਗਏ ਹਨ। ਇਹ ਵਿਵਹਾਰ ਹੋਰ ਆਇਤਾਕਾਰਾਂ ਲਈ ਵੀ ਜਾਰੀ ਰਹਿੰਦਾ ਹੈ।
ਇਹ ਜੋੜ ਜਨਵਰੀ 2009 ਵਿੱਚ ਕੁੱਲ ਬਾਰਸ਼ ਨੂੰ ਦਰਸਾਉਂਦਾ ਹੈ।
ਕੁੱਲ ਮੀਲ 7×∑i=125(1+(i−1)10)=7×25+710×12×25=385mi ਹੈ।
8.1-ਇੰਚ ਦਾ ਸ਼ੁੱਧ ਵਾਧਾ ਪ੍ਰਾਪਤ ਕਰਨ ਲਈ ਸੰਖਿਆਵਾਂ ਨੂੰ ਜੋੜੋ।
309,389,957
L 8 = 3 + 2 + 1 + 2 + 3 + 4 + 5 + 4 = 24
L 8 = 3 + 5 + 7 + 6 + 8 + 6 + 5 + 4 = 44
L 10 ≈ 1.7604 , L 30 ≈ 1.7625 , L 50 ≈ 1.76265
R1=−1,L1=1,R10=−0.1,L10=0.1,L100=0.01, ਅਤੇ R100=−0.01। ਗ੍ਰਾਫ ਦੀ ਸਮਰੂਪਤਾ ਦੁਆਰਾ, ਸਹੀ ਖੇਤਰਫਲ ਸਿਫਰ ਹੈ।
R 1 = 0 , L 1 = 0 , R 10 = 2.4499 , L 10 = 2.4499 , R 100 = 2.1365 , L 100 = 2.1365
If [c,d][c,d] is a subinterval of [a,b][a,b] under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is f(c)(d−c).f(c)(d−c). But, f(c)≤f(x)f(c)≤f(x) for c≤x≤d,c≤x≤d, so the area under the graph of f between c and d is f(c)(d−c)f(c)(d−c) plus the area below the graph of f but above the horizontal line segment at height f(c),f(c), which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of f on [a,b].[a,b].
LN=b−aN∑i=1Nf(a+(b−a)i−1N)=b−aN∑i=0N−1f(a+(b−a)iN)LN=b−aN∑i=1Nf(a+(b−a)i−1N)=b−aN∑i=0N−1f(a+(b−a)iN) and RN=b−aN∑i=1Nf(a+(b−a)iN).RN=b−aN∑i=1Nf(a+(b−a)iN). The left sum has a term corresponding to i=0i=0 and the right sum has a term corresponding to i=N.i=N. In RN−LN,RN−LN, any term corresponding to i=1,2,…,N−1i=1,2,…,N−1 occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with RN−LN=b−aN(f(a+(b−a))NN)−(f(a)+(b−a)0N)=b−aN(f(b)−f(a)).RN−LN=b−aN(f(a+(b−a))NN)−(f(a)+(b−a)0N)=b−aN(f(b)−f(a)).
Graph 1: a. L(A)=0,B(A)=20;L(A)=0,B(A)=20; b. U(A)=20.U(A)=20. Graph 2: a. L(A)=9;L(A)=9; b. B(A)=11,U(A)=20.B(A)=11,U(A)=20. Graph 3: a. L(A)=11.0;L(A)=11.0; b. B(A)=4.5,U(A)=15.5.B(A)=4.5,U(A)=15.5.
Let A be the area of the unit circle. The circle encloses n congruent triangles each of area sin(2πn)2,sin(2πn)2, so n2sin(2πn)≤A.n2sin(2πn)≤A. Similarly, the circle is contained inside n congruent triangles each of area BH2=12cosπn+sinπntanπn2sin2πn,BH2=12cosπn+sinπntanπn2sin2πn, so A≤n2sin(2πn)(cos(πn)+sin(πn)tan(πn))2.A≤n2sin(2πn)(cos(πn)+sin(πn)tan(πn))2. As n→∞,n2sin(2πn)=sin(2πn)(2n)Α→π,n→∞,n2sin(2πn)=sin(2πn)(2n)Α→π, so we conclude π≤A.π≤A. Also, as n→∞,cos(πn)+sin(πn)tan(πn)→1,n→∞,cos(πn)+sin(πn)tan(πn)→1, so we also have A≤π.A≤π. By the squeeze theorem for limits, we conclude that A=π.A=π.
Section 1.2 Exercises
∫ 0 2 ( 5 x 2 − 3 x 3 ) d x ∫ 0 2 ( 5 x 2 − 3 x 3 ) d x
∫ 0 1 cos 2 ( 2 π x ) d x ∫ 0 1 cos 2 ( 2 π x ) d x
∫ 0 1 x d x ∫ 0 1 x d x
∫ 3 6 x d x ∫ 3 6 x d x
∫ 1 2 x log ( x 2 ) d x ∫ 1 2 x log ( x 2 ) d x
1 + 2 · 2 + 3 · 3 = 14 1 + 2 · 2 + 3 · 3 = 14
1 − 4 + 9 = 6 1 − 4 + 9 = 6
1 − 2 π + 9 = 10 − 2 π 1 − 2 π + 9 = 10 − 2 π
The integral is the area of the triangle, 1212
The integral is the area of the triangle, 9.
The integral is the area 12πr2=2π.12πr2=2π.
The integral is the area of the “big” triangle less the “missing” triangle, 9−12.9−12.
L = 2 + 0 + 10 + 5 + 4 = 21 , R = 0 + 10 + 10 + 2 + 0 = 22 , L + R 2 = 21.5 L = 2 + 0 + 10 + 5 + 4 = 21 , R = 0 + 10 + 10 + 2 + 0 = 22 , L + R 2 = 21.5
L = 0 + 4 + 0 + 4 + 2 = 10 , R = 4 + 0 + 2 + 4 + 0 = 10 , L + R 2 = 10 L = 0 + 4 + 0 + 4 + 2 = 10 , R = 4 + 0 + 2 + 4 + 0 = 10 , L + R 2 = 10
∫ 2 4 f ( x ) d x + ∫ 2 4 g ( x ) d x = 8 − 3 = 5 ∫ 2 4 f ( x ) d x + ∫ 2 4 g ( x ) d x = 8 − 3 = 5
∫ 2 4 f ( x ) d x − ∫ 2 4 g ( x ) d x = 8 + 3 = 11 ∫ 2 4 f ( x ) d x − ∫ 2 4 g ( x ) d x = 8 + 3 = 11
4 ∫ 2 4 f ( x ) d x − 3 ∫ 2 4 g ( x ) d x = 32 + 9 = 41 4 ∫ 2 4 f ( x ) d x − 3 ∫ 2 4 g ( x ) d x = 32 + 9 = 41
The integrand is odd; the integral is zero.
The integrand is antisymmetric with respect to x=3.x=3. The integral is zero.
1 − 1 2 + 1 3 − 1 4 = 7 12 1 − 1 2 + 1 3 − 1 4 = 7 12
∫ 0 1 ( 1 − 6 x + 12 x 2 − 8 x 3 ) d x = ( x - 3 x 2 + 4 x 3 - 2 x 4 ) = ( 1 - 3 + 4 - 2 ) ( 0 - 0 + 0 - 0 ) = 0 ∫ 0 1 ( 1 − 6 x + 12 x 2 − 8 x 3 ) d x = ( x - 3 x 2 + 4 x 3 - 2 x 4 ) = ( 1 - 3 + 4 - 2 ) ( 0 - 0 + 0 - 0 ) = 0
7 − 5 4 = 23 4 7 − 5 4 = 23 4
The integrand is negative over [−2,3].[−2,3].
x≤x2x≤x2 over [1,2],[1,2], so 1+x≤1+x21+x≤1+x2 over [1,2].[1,2].
cos(t)≥22.cos(t)≥22. Multiply by the length of the interval to get the inequality.
f ave = 0 ; c = 0 f ave = 0 ; c = 0
3232 when c=±32c=±32
f ave = 0 ; c = π 2 , 3 π 2 f ave = 0 ; c = π 2 , 3 π 2
L100=1.294,R100=1.301;L100=1.294,R100=1.301; the exact average is between these values.
L 100 × ( 1 2 ) = 0.5178 , R 100 × ( 1 2 ) = 0.5294 L 100 × ( 1 2 ) = 0.5178 , R 100 × ( 1 2 ) = 0.5294
L1=0,L10×(12)=8.743493,L100×(12)=12.861728.L1=0,L10×(12)=8.743493,L100×(12)=12.861728. The exact answer ≈26.799,≈26.799, so L100 is not accurate.
L1×(1π)=1.352,L10×(1π)=−0.1837,L100×(1π)=−0.2956.L1×(1π)=1.352,L10×(1π)=−0.1837,L100×(1π)=−0.2956. The exact answer ≈−0.303,≈−0.303, so L100 is not accurate to first decimal.
Use tan2θ+1=sec2θ.tan2θ+1=sec2θ. Then, B−A=∫−π/4π/41dx=π2.B−A=∫−π/4π/41dx=π2.
∫02πcos2tdt=π,∫02πcos2tdt=π, so divide by the length 2π of the interval. cos2tcos2t has period π, so yes, it is true.
The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, A=−b−b2−4ac2aA=−b−b2−4ac2a and B=−b+b2−4ac2a.B=−b+b2−4ac2a.
If f(t0)>g(t0)f(t0)>g(t0) for some t0∈[a,b],t0∈[a,b], then since f−gf−g is continuous, there is an interval containing t0 such that f(t)>g(t)f(t)>g(t) over the interval [c,d],[c,d], and then ∫ddf(t)dt>∫cdg(t)dt∫ddf(t)dt>∫cdg(t)dt over this interval.
The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, ∫abf(t)dt=∫a0a1f(t)dt+∫a1a2f(t)dt+⋯+∫aN+1aNf(t)dt=∫a0a11dt+∫a1a21dt+⋯+∫aN+1aN1dt=(a1−a0)+(a2−a1)+⋯+(aN−aN−1)=aN−a0=b−a.∫abf(t)dt=∫a0a1f(t)dt+∫a1a2f(t)dt+⋯+∫aN+1aNf(t)dt=∫a0a11dt+∫a1a21dt+⋯+∫aN+1aN1dt=(a1−a0)+(a2−a1)+⋯+(aN−aN−1)=aN−a0=b−a. Dividing through by b−ab−a gives the desired identity.
∫ 0 N f ( t ) d t = ∑ i = 1 N ∫ i − 1 i f ( t ) d t = ∑ i = 1 N i 2 = N ( N + 1 ) ( 2 N + 1 ) 6 ∫ 0 N f ( t ) d t = ∑ i = 1 N ∫ i − 1 i f ( t ) d t = ∑ i = 1 N i 2 = N ( N + 1 ) ( 2 N + 1 ) 6
L10=1.815,R10=1.515,L10+R102=1.665,L10=1.815,R10=1.515,L10+R102=1.665, so the estimate is accurate to two decimal places.
The average is 1/2,1/2, which is equal to the integral in this case.
a. The graph is antisymmetric with respect to t=12t=12 over [0,1],[0,1], so the average value is zero. b. For any value of a, the graph between [a,a+1][a,a+1] is a shift of the graph over [0,1],[0,1], so the net areas above and below the axis do not change and the average remains zero.
Yes, the integral over any interval of length 1 is the same.
Section 1.3 Exercises
Yes. It is implied by the Mean Value Theorem for Integrals.
F′(2)=−1;F′(2)=−1; average value of F′F′ over [1,2][1,2] is −1/2.−1/2.
e cos x e cos x
1 16 − x 2 1 16 − x 2
x d d x x = 1 2 x d d x x = 1 2
− 1 − cos 2 x d d x cos x = | sin x | sin x − 1 − cos 2 x d d x cos x = | sin x | sin x
2 x | x | 1 + x 2 2 x | x | 1 + x 2
ln ( e 2 x ) d d x e x = 2 x e x ln ( e 2 x ) d d x e x = 2 x e x
a. f is positive over (1,2)(1,2) and (5,6),(5,6), negative over (0,1)(0,1) and (3,4),(3,4), and zero over (2,3)(2,3) and (4,5).(4,5). b. The maximum value is 2 and the minimum is −3. c. The average value is 0.
a. ℓ is positive over (0,1)(0,1) and (3,6),(3,6), and negative over (1,3).(1,3). b. It is increasing over (0,1)(0,1) and (3,5),(3,5), and it is constant over (1,3)(1,3) and (5,6).(5,6). c. Its average value is 13.13.
T 10 = 49.08 , ∫ −4 3 ( x 3 + 6 x 2 + x − 5 ) d x = 48 T 10 = 49.08 , ∫ −4 3 ( x 3 + 6 x 2 + x − 5 ) d x = 48
T 10 = 260.836 , ∫ 1 9 ( x + x 2 ) d x = 260 T 10 = 260.836 , ∫ 1 9 ( x + x 2 ) d x = 260
T 10 = 3.058 , ∫ 1 4 4 x 2 d x = 3 T 10 = 3.058 , ∫ 1 4 4 x 2 d x = 3
F ( x ) = x 3 3 + 3 x 2 2 − 5 x , F ( 3 ) − F ( −2 ) = − 35 6 F ( x ) = x 3 3 + 3 x 2 2 − 5 x , F ( 3 ) − F ( −2 ) = − 35 6
F ( x ) = − t 5 5 + 13 t 3 3 − 36 t , F ( 3 ) − F ( 2 ) = 62 15 F ( x ) = − t 5 5 + 13 t 3 3 − 36 t , F ( 3 ) − F ( 2 ) = 62 15
F ( x ) = x 100 100 , F ( 1 ) − F ( 0 ) = 1 100 F ( x ) = x 100 100 , F ( 1 ) − F ( 0 ) = 1 100
F ( x ) = x 3 3 + 1 x , F ( 4 ) − F ( 1 4 ) = 1125 64 F ( x ) = x 3 3 + 1 x , F ( 4 ) − F ( 1 4 ) = 1125 64
F ( x ) = x , F ( 4 ) − F ( 1 ) = 1 F ( x ) = x , F ( 4 ) − F ( 1 ) = 1
F ( x ) = 4 3 t 3 / 4 , F ( 16 ) − F ( 1 ) = 28 3 F ( x ) = 4 3 t 3 / 4 , F ( 16 ) − F ( 1 ) = 28 3
F ( x ) = − cos x , F ( π 2 ) − F ( 0 ) = 1 F ( x ) = − cos x , F ( π 2 ) − F ( 0 ) = 1
F ( x ) = sec x , F ( π 4 ) − F ( 0 ) = 2 − 1 F ( x ) = sec x , F ( π 4 ) − F ( 0 ) = 2 − 1
F ( x ) = − cot ( x ) , F ( π 2 ) − F ( π 4 ) = 1 F ( x ) = − cot ( x ) , F ( π 2 ) − F ( π 4 ) = 1
F ( t ) = − 1 t + 1 2 t 2 , F ( −1 ) − F ( −2 ) = 7 8 F ( t ) = − 1 t + 1 2 t 2 , F ( −1 ) − F ( −2 ) = 7 8
F ( x ) = e x − e F ( x ) = e x − e
F ( x ) = 0 F ( x ) = 0
∫ −2 −1 ( t 2 − 2 t − 3 ) d t − ∫ −1 3 ( t 2 − 2 t − 3 ) d t + ∫ 3 4 ( t 2 − 2 t − 3 ) d t = 46 3 ∫ −2 −1 ( t 2 − 2 t − 3 ) d t − ∫ −1 3 ( t 2 − 2 t − 3 ) d t + ∫ 3 4 ( t 2 − 2 t − 3 ) d t = 46 3
− ∫ − π / 2 0 sin t d t + ∫ 0 π / 2 sin t d t = 2 − ∫ − π / 2 0 sin t d t + ∫ 0 π / 2 sin t d t = 2
a. The average is 11.21×10911.21×109 since cos(πt6)cos(πt6) has period 12 and integral 0 over any period. Consumption is equal to the average when cos(πt6)=0,cos(πt6)=0, when t=3,t=3, and when t=9.t=9. b. Total consumption is the average rate times duration: 11.21×12×109=1.35×101111.21×12×109=1.35×1011 c. 109(11.21−16∫39cos(πt6)dt)=109(11.21+2π)=11.84x109109(11.21−16∫39cos(πt6)dt)=109(11.21+2π)=11.84x109
If f is not constant, then its average is strictly smaller than the maximum and larger than the minimum, which are attained over [a,b][a,b] by the extreme value theorem.
a. d2θ=(acosθ+c)2+b2sin2θ=a2+c2cos2θ+2accosθ=(a+ccosθ)2;d2θ=(acosθ+c)2+b2sin2θ=a2+c2cos2θ+2accosθ=(a+ccosθ)2; b. d–=12π∫02π(a+ccosθ)dθ=ad–=12π∫02π(a+ccosθ)dθ=a
Mean gravitational force = GmM2π ∫02π 1 (a+2a2−b2cosθ)2 dθ. GmM2π ∫02π 1 (a+2a2−b2cosθ)2 dθ.
Section 1.4 Exercises
∫ ( x − 1 x ) d x = ∫ x 1 / 2 d x − ∫ x −1 / 2 d x = 2 3 x 3 / 2 + C 1 − 2 x 1 / 2 + C 2 = 2 3 x 3 / 2 − 2 x 1 / 2 + C ∫ ( x − 1 x ) d x = ∫ x 1 / 2 d x − ∫ x −1 / 2 d x = 2 3 x 3 / 2 + C 1 − 2 x 1 / 2 + C 2 = 2 3 x 3 / 2 − 2 x 1 / 2 + C
∫ d x 2 x = 1 2 ln | x | + C ∫ d x 2 x = 1 2 ln | x | + C
∫ 0 π sin x d x − ∫ 0 π cos x d x = − cos x | 0 π − ( sin x ) | 0 π = ( − ( −1 ) + 1 ) − ( 0 − 0 ) = 2 ∫ 0 π sin x d x − ∫ 0 π cos x d x = − cos x | 0 π − ( sin x ) | 0 π = ( − ( −1 ) + 1 ) − ( 0 − 0 ) = 2
P(s)=4s,P(s)=4s, so dPds=4dPds=4 and ∫244ds=8.∫244ds=8.
∫ 1 2 N d s = N ∫ 1 2 N d s = N
With p as in the previous exercise, each of the 12 pentagons increases in area from 2p to 4p units so the net increase in the area of the dodecahedron is 36p units.
18 s 2 = 6 ∫ s 2 s 2 x d x 18 s 2 = 6 ∫ s 2 s 2 x d x
12 π R 2 = 8 π ∫ R 2 R r d r 12 π R 2 = 8 π ∫ R 2 R r d r
d(t)=∫0tv(s)ds=4t−t2.d(t)=∫0tv(s)ds=4t−t2. The total distance is d(2)=4m.d(2)=4m.
d(t)=∫0tv(s)ds.d(t)=∫0tv(s)ds. For t<3,d(t)=∫0t(6−2t)dt=6t−t2.t<3,d(t)=∫0t(6−2t)dt=6t−t2. For t>3,d(t)=d(3)+∫3t(2t−6)dt=9+(t2−6t)|36.t>3,d(t)=d(3)+∫3t(2t−6)dt=9+(t2−6t)|36. The total distance is d(6)=18m.d(6)=18m.
v(t)=40−9.8tm/sec;h(t)=1.5+40t−4.9t2v(t)=40−9.8tm/sec;h(t)=1.5+40t−4.9t2 m/s
The net increase is 1 unit.
At t=5,t=5, the height of water is x=(15π)1/3m..x=(15π)1/3m.. The net change in height from t=5t=5 to t=10t=10 is (30π)1/3−(15π)1/3(30π)1/3−(15π)1/3 m.
The total daily power consumption is estimated as the sum of the hourly power rates, or 911 gW-h.
17 kJ
a. 54.3%; b. 27.00%; c. The curve in the following plot is 2.35(t+3)e−0.15(t+3).2.35(t+3)e−0.15(t+3).
In dry conditions, with initial velocity v0=30v0=30 m/s, D=64.3D=64.3 and, if v0=25,D=44.64.v0=25,D=44.64. In wet conditions, if v0=30,v0=30, and D=180D=180 and if v0=25,D=125.v0=25,D=125.
225 cal
E ( 150 ) = 28 , E ( 300 ) = 22 , E ( 450 ) = 16 E ( 150 ) = 28 , E ( 300 ) = 22 , E ( 450 ) = 16
a.
b. Between 600 and 1000 the average decrease in vehicles per hour per lane is −0.0075. Between 1000 and 1500 it is −0.006 per vehicles per hour per lane, and between 1500 and 2100 it is −0.04 vehicles per hour per lane. c.
The graph is nonlinear, with minutes per mile increasing dramatically as vehicles per hour per lane reach 2000.
1 37 ∫ 0 37 p ( t ) d t = 0.07 ( 37 ) 3 4 + 2.42 ( 37 ) 2 3 − 25.63 ( 37 ) 2 + 521.23 ≈ 2037 1 37 ∫ 0 37 p ( t ) d t = 0.07 ( 37 ) 3 4 + 2.42 ( 37 ) 2 3 − 25.63 ( 37 ) 2 + 521.23 ≈ 2037
Average acceleration is − 0.7 3 t3 + 1.44 2 t2 + 10.44 t |05 = − 0.7 3 (53) + 10.44 (5)≈ 41 − 0.7 3 t3 + 1.44 2 t2 + 10.44 t |05 = − 0.7 3 (53) + 10.44 (5)≈ 41 mph/s
s(t) = v(t)dt = ∫ ( − 7 30 t 3 + 0.72 t 2 + 10.44t + 23.967 ) d t = − 7 120 t 4 + 0.72 3 t 3 + 10.44 2 t 2 + 23.967t + C = 174.24 feet s(t) = v(t)dt = ∫ ( − 7 30 t 3 + 0.72 t 2 + 10.44t + 23.967 ) d t = − 7 120 t 4 + 0.72 3 t 3 + 10.44 2 t 2 + 23.967t + C = 174.24 feet
1 40 ∫ 0 40 ( −0.068 t + 5.14 ) d t = − 0.068 ( 40 ) 2 + 5.14 = 3.78 m/sec 1 40 ∫ 0 40 ( −0.068 t + 5.14 ) d t = − 0.068 ( 40 ) 2 + 5.14 = 3.78 m/sec
Section 1.5 Exercises
u = h ( x ) u = h ( x )
f ( u ) = ( u + 1 ) 2 u f ( u ) = ( u + 1 ) 2 u
d u = 8 x d x ; f ( u ) = 1 8 u d u = 8 x d x ; f ( u ) = 1 8 u
1 5 ( x + 1 ) 5 + C 1 5 ( x + 1 ) 5 + C
− 1 12 ( 2 x − 3 ) 6 + C − 1 12 ( 2 x − 3 ) 6 + C
x 2 + 1 + C x 2 + 1 + C
1 8 ( x 2 − 2 x ) 4 + C 1 8 ( x 2 − 2 x ) 4 + C
sin θ − sin 3 θ 3 + C sin θ − sin 3 θ 3 + C
( 1 − x ) 101 101 − ( 1 − x ) 100 100 + C ( 1 − x ) 101 101 − ( 1 − x ) 100 100 + C
∫(11x-7)−3dx= −122(11x−7)2+C ∫ ( 11 x - 7 ) −3 dx = − 1 22 ( 11 x − 7 ) 2 + C
− cos 4 θ 4 + C − cos 4 θ 4 + C
− cos 3 ( π t ) 3 π + C − cos 3 ( π t ) 3 π + C
− 1 4 cos 2 ( t 2 ) + C − 1 4 cos 2 ( t 2 ) + C
− 1 3 ( x 3 − 3 ) + C − 1 3 ( x 3 − 3 ) + C
− 2 ( y 3 − 2 ) 3 1 − y 3 + C − 2 ( y 3 − 2 ) 3 1 − y 3 + C
1 33 ( 1 − cos 3 θ ) 11 + C 1 33 ( 1 − cos 3 θ ) 11 + C
1 12 ( sin 3 θ − 3 sin 2 θ ) 4 + C 1 12 ( sin 3 θ − 3 sin 2 θ ) 4 + C
L50=−8.5779.L50=−8.5779. The exact area is −818−818
L50=−0.006399L50=−0.006399 … The exact area is 0.
u = 1 + x 2 , d u = 2 x d x , 1 2 ∫ 1 2 u −1 / 2 d u = 2 − 1 u = 1 + x 2 , d u = 2 x d x , 1 2 ∫ 1 2 u −1 / 2 d u = 2 − 1
u = 1 + t 3 , d u = 3 t 2 d t , 1 3 ∫ 1 2 u −1 / 2 d u = 2 3 ( 2 − 1 ) u = 1 + t 3 , d u = 3 t 2 d t , 1 3 ∫ 1 2 u −1 / 2 d u = 2 3 ( 2 − 1 )
u = cos θ , d u = − sin θ d θ , ∫ 1 / 2 1 u −4 d u = 1 3 ( 2 2 − 1 ) u = cos θ , d u = − sin θ d θ , ∫ 1 / 2 1 u −4 d u = 1 3 ( 2 2 − 1 )
The antiderivative is y=sin(ln(2x)).y=sin(ln(2x)). Since the antiderivative is not continuous at x=0,x=0, one cannot find a value of C that would make y=sin(ln(2x))−Cy=sin(ln(2x))−C work as a definite integral.
The antiderivative is y=12sec2x.y=12sec2x. You should take C=−2C=−2 so that F(−π3)=0.F(−π3)=0.
The antiderivative is y=13(2x3+1)3/2.y=13(2x3+1)3/2. One should take C=−13.C=−13.
No, because the integrand is discontinuous at x=1.x=1.
u=sin(t2);u=sin(t2); the integral becomes 12∫00udu.12∫00udu.
u=(1+(t−12)2);u=(1+(t−12)2); the integral becomes −∫5/45/41udu.−∫5/45/41udu.
u=1−t;u=1−t; the integral becomes∫1−1ucos(π(1−u))du=∫1−1u[cosπcosπu−sinπsinπu]du=−∫1−1ucosπudu=∫−11ucosπudu=0∫1−1ucos(π(1−u))du=∫1−1u[cosπcosπu−sinπsinπu]du=−∫1−1ucosπudu=∫−11ucosπudu=0since the integrand is odd.
Setting u=cxu=cx and du=cdxdu=cdx gets you 1bc−ac∫a/cb/cf(cx)dx=cb−a∫u=au=bf(u)duc=1b−a∫abf(u)du.1bc−ac∫a/cb/cf(cx)dx=cb−a∫u=au=bf(u)duc=1b−a∫abf(u)du.
∫0xg(t)dt=12∫u=1−x21duua=12(1−a)u1−a|u=1−x21=12(1−a)(1−(1−x2)1−a).∫0xg(t)dt=12∫u=1−x21duua=12(1−a)u1−a|u=1−x21=12(1−a)(1−(1−x2)1−a). As x→1x→1 the limit is 12(1−a)12(1−a) if a<1,a<1, and the limit diverges to +∞ if a≥1.a≥1.
∫ t = π 0 b 1 − cos 2 t × ( − a sin t ) d t = ∫ t = 0 π a b sin 2 t d t ∫ t = π 0 b 1 − cos 2 t × ( − a sin t ) d t = ∫ t = 0 π a b sin 2 t d t
f ( t ) = 2 cos ( 3 t ) − cos ( 2 t ) ; ∫ 0 π ( 2 cos ( 3 t ) − cos ( 2 t ) ) d t = 0 f ( t ) = 2 cos ( 3 t ) − cos ( 2 t ) ; ∫ 0 π ( 2 cos ( 3 t ) − cos ( 2 t ) ) d t = 0
Section 1.6 Exercises
−1 3 e −3 x + C −1 3 e −3 x + C
− 3 − x ln 3 + C − 3 − x ln 3 + C
2 ln x + C or ln x 2 +C 2 ln x + C or ln x 2 +C
2 x + C 2 x + C
− 1 ln x + C − 1 ln x + C
ln ( ln ( ln x ) ) + C ln ( ln ( ln x ) ) + C
ln ( x cos x ) + C ln ( x cos x ) + C
− 1 2 ( ln ( cos ( x ) ) ) 2 + C − 1 2 ( ln ( cos ( x ) ) ) 2 + C
− e − x 3 3 + C − e − x 3 3 + C
e tan x + C e tan x + C
t + C t + C
2 9 x 3 ( ln ( x 3 ) − 1 ) + C 2 9 x 3 ( ln ( x 3 ) − 1 ) + C
2 x ( ln x − 2 ) + C 2 x ( ln x − 2 ) + C
∫ 0 ln x e t d t = e t | 0 ln x = e ln x − e 0 = x − 1 ∫ 0 ln x e t d t = e t | 0 ln x = e ln x − e 0 = x − 1
- 1 3 ln sin ( 3 x ) + cos ( 3 x ) +C - 1 3 ln sin ( 3 x ) + cos ( 3 x ) +C
− 1 2 ln | csc ( x 2 ) + cot ( x 2 ) | + C − 1 2 ln | csc ( x 2 ) + cot ( x 2 ) | + C
− 1 2 ( ln ( csc x ) ) 2 + C − 1 2 ( ln ( csc x ) ) 2 + C
1 3 ln ( 26 7 ) 1 3 ln ( 26 7 )
ln ( 3 − 1 ) ln ( 3 − 1 )
1 2 ln 3 2 1 2 ln 3 2
y − 2 ln | y + 1 | + C y − 2 ln | y + 1 | + C
ln | sin x − cos x | + C ln | sin x − cos x | + C
− 1 3 ( 1 − ( ln x ) 2 ) 3 / 2 + C − 1 3 ( 1 − ( ln x ) 2 ) 3 / 2 + C
Exact solution: e−1e,R50=0.6258.e−1e,R50=0.6258. Since f is decreasing, the right endpoint estimate underestimates the area.
Exact solution: 2ln(3)−ln(6)2,R50=0.2033.2ln(3)−ln(6)2,R50=0.2033. Since f is increasing, the right endpoint estimate overestimates the area.
Exact solution: −1ln(4),R50=−0.7164.−1ln(4),R50=−0.7164. Since f is negative and increasing on the interval, the right-endpoint estimate overestimates the value of the integral, but it underestimates the area of the region (absolute value of the integral).
11 2 ln 2 11 2 ln 2
1 ln ( 65 , 536 ) 1 ln ( 65 , 536 )
∫NN+1xe−x2dx=12(e−N2−e−(N+1)2).∫NN+1xe−x2dx=12(e−N2−e−(N+1)2). The quantity is less than 0.01 when N=2.N=2.
∫ a b d x x = ln ( b ) − ln ( a ) = ln ( 1 a ) − ln ( 1 b ) = ∫ 1 / b 1 / a d x x ∫ a b d x x = ln ( b ) − ln ( a ) = ln ( 1 a ) − ln ( 1 b ) = ∫ 1 / b 1 / a d x x
23
We may assume that x>1,so1x<1.x>1,so1x<1. Then, ∫11/xdtt.∫11/xdtt. Now make the substitution u=1t,u=1t, so du=−dtt2du=−dtt2 and duu=−dtt,duu=−dtt, and change endpoints: ∫11/xdtt=−∫1xduu=−lnx.∫11/xdtt=−∫1xduu=−lnx.
Answers will vary.
x=E(ln(x)).x=E(ln(x)). Then, 1=E'(lnx)xorx=E'(lnx).1=E'(lnx)xorx=E'(lnx). Since any number t can be written t=lnxt=lnx for some x, and for such t we have x=E(t),x=E(t), it follows that for any t,E'(t)=E(t).t,E'(t)=E(t).
R10=0.6811,R100=0.6827R10=0.6811,R100=0.6827
Section 1.7 Exercises
sin −1 x | 0 3 / 2 = π 3 sin −1 x | 0 3 / 2 = π 3
tan −1 x | 3 1 = − π 12 tan −1 x | 3 1 = − π 12
sec −1 | 2 3 2 = π 4 - π 6 = π 12 sec −1 | 2 3 2 = π 4 - π 6 = π 12
sin −1 ( x 3 ) + C sin −1 ( x 3 ) + C
1 3 tan −1 ( x 3 ) + C 1 3 tan −1 ( x 3 ) + C
1 3 sec −1 ( x 3 ) + C 1 3 sec −1 ( x 3 ) + C
cos(π2−θ)=sinθ.cos(π2−θ)=sinθ. So, sin−1t=π2−cos−1t.sin−1t=π2−cos−1t. They differ by a constant.
1−t21−t2 is not defined as a real number when t>1.t>1.
The antiderivative is sin−1(x3)+C.sin−1(x3)+C. Taking C=π2C=π2 recovers the definite integral.
The antiderivative is 12tan−1(sinx2)+C.12tan−1(sinx2)+C. Taking C=12tan−1(sin(6)2)C=12tan−1(sin(6)2) recovers the definite integral.
1 2 ( sin −1 t ) 2 + C 1 2 ( sin −1 t ) 2 + C
1 4 ( tan −1 ( 2 t ) ) 2 + C 1 4 ( tan −1 ( 2 t ) ) 2 + C
1 4 ( sec −1 ( t 2 ) 2 ) + C 1 4 ( sec −1 ( t 2 ) 2 ) + C
The antiderivative is 12sec−1(x2)+C.12sec−1(x2)+C. Taking C=0C=0 recovers the definite integral over [2,6].[2,6].
The general antiderivative is tan−1(xsinx)+C.tan−1(xsinx)+C. Taking C=−tan−1(6sin(6))C=−tan−1(6sin(6)) recovers the definite integral.
The general antiderivative is tan−1(lnx)+C.tan−1(lnx)+C. Taking C=π2=tan−1∞C=π2=tan−1∞ recovers the definite integral.
sin −1 ( e t ) + C sin −1 ( e t ) + C
sin −1 ( ln t ) + C sin −1 ( ln t ) + C
− 1 4 ( cos −1 ( 2 t ) ) 2 + C − 1 4 ( cos −1 ( 2 t ) ) 2 + C
1 2 ln ( 4 3 ) 1 2 ln ( 4 3 )
1 − 2 5 1 − 2 5
2tan−1(A)→π2tan−1(A)→π as A→∞A→∞
Using the hint, one has ∫csc2xcsc2x+cot2xdx=∫csc2x1+2cot2xdx.∫csc2xcsc2x+cot2xdx=∫csc2x1+2cot2xdx. Set u=2cotx.u=2cotx. Then, du=−2csc2xdu=−2csc2x and the integral is −12∫du1+u2=−12tan−1u+C=12tan−1(2cotx)+C.−12∫du1+u2=−12tan−1u+C=12tan−1(2cotx)+C. If one uses the identity tan−1s+tan−1(1s)=π2,tan−1s+tan−1(1s)=π2, then this can also be written 12tan−1(tanx2)+C.12tan−1(tanx2)+C.
x≈±1.7321.x≈±1.7321. The left endpoint estimate with N=100N=100 is 4.781 and these decimals persist for N=500.N=500.
Review Exercises
False
True
L4=5.25,R4=3.25,L4=5.25,R4=3.25, exact answer: 4
L4=5.364,R4=5.364,L4=5.364,R4=5.364, exact answer: 5.870
− 4 3 − 4 3
1
− 1 2 ( x + 4 ) 2 + C − 1 2 ( x + 4 ) 2 + C
4 3 sin −1 ( x 3 ) + C 4 3 sin −1 ( x 3 ) + C
sin t 1 + t 2 sin t 1 + t 2
4 ln x x + 1 4 ln x x + 1
$6,328,113
$73.36
19117 12 ft/sec , or 1593 ft/sec 19117 12 ft/sec , or 1593 ft/sec