• 沒有找到結果。

# 2 Fundamental Theorem of Calculus

N/A
N/A
Protected

Share "2 Fundamental Theorem of Calculus"

Copied!
23
0
0

(1)

### 1 Riemann Sum

1. Find∫12x−2dx. Hint: Choose xi to be the geometric mean of xi−1and xi(that is, xi = √xi−1xi) and use the identity

1

m(m+1)= m1m1+1 Solution:

504 ¤ CHAPTER 5 INTEGRALS (b)2

0  () sin 2  ≤2

0 |() sin 2|  [by part (a)] =2

0 |()| |sin 2|  ≤2

0 |()|  by Property 7

since |sin 2| ≤ 1 ⇒ |()| |sin 2| ≤ |()|.

71.Suppose that  is integrable on [0 1]  that is, lim

→∞

=1

 () ∆exists for any choice of  in [−1 ]. Let n denote a

positive integer and divide the interval [0 1] into n equal subintervals

 01

 ,

1

2

 ,  ,

 − 1

  1

. If we choose to be

a rational number in the ith subinterval, then we obtain the Riemann sum

=1

 () · 1

 = 0, so

lim→∞

=1

 () ·1

 = lim

→∞0 = 0. Now suppose we choose  to be an irrational number. Then we get

=1

 () · 1

 =

=11 ·1

 =  · 1

 = 1for each , so lim

→∞

=1

 () · 1

 = lim

→∞1 = 1. Since the value of

lim→∞

=1

 () ∆depends on the choice of the sample points , the limit does not exist, and  is not integrable on [0 1].

72.Partition the interval [0 1] into n equal subintervals and choose 1 = 1

2. Then with () = 1

,

=1

 () ∆ ≥ (1)∆ = 1 12·1

 =  Thus,

=1

 () ∆can be made arbitrarily large and hence,  is not integrable on [0 1].

73. lim

→∞

=1

4

5 = lim

→∞

=1

4

4 ·1

 = lim

→∞

=1



4

1

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

lim→∞

=1

 () ∆, where ∆ = (1 − 0) = 1, = 0 +  ∆ = , and () = 4. Thus, the deﬁnite integral is1

04.

74. lim

→∞

1

=1

1

1 + ()2 = lim

→∞

=1

1 1 + ()2 ·1

 = lim

→∞

=1

 () ∆, where ∆ = (1 − 0) = 1,

= 0 +  ∆ = , and () = 1

1 + 2. Thus, the deﬁnite integral is

1 0



1 + 2.

75.Choose = 1 + 

 and  = √−1=



1 + − 1



1 + 

 . Then

2

1−2 = lim

→∞

1

=1

 1

1 +− 1 

1 + = lim

→∞

=1

1 ( +  − 1)( + )

= lim

→∞

=1

 1

 +  − 1− 1

 + 

[by the hint] = lim

→∞

−1

=0

1

 + −

=1

1

 + 

= lim

→∞

1

+ 1

 + 1+ · · · + 1 2 − 1

 1

 + 1+ · · · + 1 2 − 1+ 1

2



= lim

→∞

1

 − 1 2

= lim

→∞

1 −12

= 12

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

2. Evaluate the limit

nlim→∞

n

i=1

i n√

2n2− i2 .

Solution:

Notice that the sum

n

i=1

i n√

2n2− i2 =∑n

i=1

i n2

2−(ni)2 =∑n

i=1 i

n

2−(ni)2⋅ 1 n

can be identified as the Riemann sum for the function f(x) ∶= 2x−x2 , 0≤ x ≤ 1 , with respect to the uniform partition of[0, 1] into n subintervals of length ∆x ∶= 1n with the right end-point ni chosen as the sample point on each subinterval[i−1n ,ni] for i = 1, 2, . . . , n. Therefore, the limit can be realised as the Riemann integral

nlim→∞

n

i=1 i

n

2−(ni)2⋅ 1 n = lim

n→∞

n

i=1

f(i

n) ∆x = ∫01f(x) dx

= ∫01 x

√2− x2 dx (u=2−x

2)

= −1 2 ∫

1 2

√du u

= −1 2[ 2√

u]12=√ 2− 1 .

### 2 Fundamental Theorem of Calculus

3. Differentiate

ex

ln x x sin(t2)dt Concepts: Fundamental theorem of calculus

Solution:

d dx ∫

ex ln x

x sin(t2)dt = d dx(x ∫ e

x

ln x

sin(t2)dt) = ∫ e

x

ln x

sin(t2)dt + x sin(e2x)ex− sin((ln x)2).

4. If f is a continuous function such that

x

1 f(t)dt = (x − 1)e2x+ ∫x1e−tf(t)dt (1)

(2)

for all x, find an explicit formula for f(x).

Concepts: Fundamental theorem of calculus

Solution:

Differentiate with respect to x on the both side.

f(x) = e2x+ 2(x − 1)e2x+ e−xf(x) = (2x − 1)e2x− e−xf(x). (2) Therefore,

(1 + e−x)f(x) = (2x − 1)e2x. (3)

We have

f(x) =(2x − 1)e2x

1+ e−x . (4)

5. Consider the function F ∶ [0, ∞] → R,

F(x) = ∫ x

2

x

1

t2+ sin2(t)dt.

(i) Find F(x).

(ii) Show limx→∞F(x) = 0.

(iii) Evaluate limx→∞xF(x).

Motivation. 本題為跨範圍的綜合題，重點是透過羅必達定理和微積分基本定理求極限。第一小題涉及微積分基本定理（課

Solution:

For (i), by the fundamental theorem of calculus and the chain rule, F(x) = 2x

x4+ sin2(x2)− 1 x2+ sin2(x). For (ii), since 0≤ t2+sin12(t)t12 for all t∈ R,

0≤ F (x) ≤ ∫ x

2

x

1 t2dt= 1

x− 1 x2. Since limx→∞1xx12 = 0, by the squeezing lemma limx→∞F(x) = 0.

For (iii), since limx→∞F(x) = 0, by l’Hospital’s rule,

xlim→∞xF(x) = lim

x→∞−x2F(x)

= lim

x→∞

−2x3

x4+ sin2(x2)+ x2 x2+ sin2(x)

= 1.

6. Find the minimum value of the area of the region under the curve y= x + 1/x from x = a to x = a + 1.5, for all a > 0.

Solution:

Page 2 of 23

(3)

### PROBLEMS PLUS

1. Differentiating both sides of the equation  sin  =2

0  () (using FTC1 and the Chain Rule for the right side) gives sin  +  cos  = 2 (2). Letting  = 2 so that (2) =  (4), we obtain sin 2 + 2 cos 2 = 4(4), so

 (4) =14(0 + 2 · 1) = 2.

2. The area  under the curve  =  + 1 from  =  to  =  + 15 is given by () =

+15

 +1



To find the minimum value of  we’ll differentiate  using FTC1 and set the derivative equal to 0.

0() = 



+15

 +1



= 



1

 +1

 + 



+15 1

 +1



= − 



1

 +1

 + 



+15 1

 +1



= −

 +1

 +

 + 15 + 1

 + 15

= 15 + 1

 + 15−1

0() = 0 ⇔ 15 + 1

 + 15−1

 = 0 ⇔ 15( + 15) +  − ( + 15) = 0 ⇔ 152+ 225 − 15 = 0 multiply by 43

⇔ 22+ 3 − 2 = 0 ⇔ (2 − 1)( + 2) = 0 ⇔  = 12 or

 = −2 Since   0,  = 12 00() = − 1

( + 15)2 + 1

2  0, so

1

2

=

2 12

 +1

 =

1

22+ ln ||2

12= (2 + ln 2) −1

8 − ln 2

= 158 + 2 ln 2is the minimum value of .

3. For  =4

0 (−2)4, let  =  − 2 so that  =  + 2 and  = . Then

 =2

−2( + 2)4 =2

−24 +2

−224 = 0 [by 5.5.7(b)] + 24

0(−2)4 = 2.

4. (a) From the graph of () =2 − 2

3 , it appears that the areas are equal; that is, the area enclosed is independent of .

(b) We first find the x-intercepts of the curve, to determine the limits of integration:  = 0 ⇔ 2 − 2= 0 ⇔  = 0 or  = 2. Now we integrate the function between these limits to find the enclosed area:

 =

2

0

2 − 2

3  = 1

3

21332

0 = 1

3

(2)213(2)3

= 1

3

43833

= 43, a constant.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c 545

7. The figure shows a semicircle with radius 1, horizontal diameter P Q, and tangent lines at P and Q. At what height above the diameter should the horizontal line be placed so as to minimize the shaded area?

583

## Problems Plus

1. Find the area of the region S − hsx, yd

### |

x > 0, y < 1, x21y2<4yj.

2. Find the centroid of the region enclosed by the loop of the curve y2− x32x4.

3. If a sphere of radius r is sliced by a plane whose distance from the center of the sphere is d, then the sphere is divided into two pieces called segments of one base (see the first figure).

The corresponding surfaces are called spherical zones of one base.

(a) Determine the surface areas of the two spherical zones indicated in the figure.

(b) Determine the approximate area of the Arctic Ocean by assuming that it is approxi- mately circular in shape, with center at the North Pole and “circumference” at 75°

north latitude. Use r − 3960 mi for the radius of the earth.

(c) A sphere of radius r is inscribed in a right circular cylinder of radius r. Two planes perpendicular to the central axis of the cylinder and a distance h apart cut off a spheri- cal zone of two bases on the sphere (see the second figure). Show that the surface area of the spherical zone equals the surface area of the region that the two planes cut off on the cylinder.

(d) The Torrid Zone is the region on the surface of the earth that is between the Tropic of Cancer (23.45° north latitude) and the Tropic of Capricorn (23.45° south latitude).

What is the area of the Torrid Zone?

r h

d

4. (a) Show that an observer at height H above the north pole of a sphere of radius r can see a part of the sphere that has area

2r2H r 1 H

(b) Two spheres with radii r and R are placed so that the distance between their centers is d, where d . r 1 R. Where should a light be placed on the line joining the centers of the spheres in order to illuminate the largest total surface?

5. Suppose that the density of seawater,  − szd, varies with the depth z below the surface.

(a) Show that the hydrostatic pressure is governed by the differential equation dP

dz − szdt

where t is the acceleration due to gravity. Let P0 and 0 be the pressure and density at z − 0. Express the pressure at depth z as an integral.

(b) Suppose the density of seawater at depth z is given by  − 0ezyH, where H is a posi- tive constant. Find the total force, expressed as an integral, exerted on a vertical circu- lar porthole of radius r whose center is located at a distance L . r below the surface.

6. The figure shows a semicircle with radius 1, horizontal diameter PQ, and tangent lines at P and Q. At what height above the diameter should the horizontal line be placed so as to minimize the shaded area?

7. Let P be a pyramid with a square base of side 2b and suppose that S is a sphere with its center on the base of P and S is tangent to all eight edges of P. Find the height of P. Then find the volume of the intersection of S and P.

P Q

FIGURE FoR pRoblEm 6

Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

796 ¤ CHAPTER 8 PROBLEMS PLUS

Thus, from (1) and (2), is not an allowable value of  if    + 

.

So  may have a maximum at  = , , or  − .

() = 22( −  − )

 −  and ( − ) = 22( −  − )

 − 

()  ( − ) ⇔ 2

 −   2

 −  ⇔ 2( − )  2( − ) ⇔ 2 − 3 2 − 3

2 − 2  3− 3 ⇔ ( − )( + )  ( − )(2+  + 2) ⇔   (2+  + 2)( + ) ⇔

  [( + )2− ]( + ) ⇔    +  − ( + ). Now  +  − ( + )   + , and we know that

   + , so we conclude that ()  ( − ).

In conclusion,  has an absolute maximum at  = provided  ≥  + 

; otherwise,  has its maximum at  = 

5. (a) Choose a vertical -axis pointing downward with its origin at the surface. In order to calculate the pressure at depth , consider  subintervals of the interval [0 ] by points and choose a point  ∈ [−1 ]for each . The thin layer of water lying between depth −1and depth has a density of approximately (), so the weight of a piece of that layer with unit cross-sectional area is () ∆. The total weight of a column of water extending from the surface to depth  (with unit cross-sectional area) would be approximately

=1

() ∆. The estimate becomes exact if we take the limit

as  → ∞; weight (or force) per unit area at depth  is  = lim

→∞

=1

() ∆. In other words,  () =

0 () .

More generally, if we make no assumptions about the location of the origin, then  () = 0+

0 () , where 0is the pressure at  = 0. Differentiating, we get  = ().

(b)  =

− ( + ) · 2√

2− 2

=

−

0++

00 

· 2√

2− 2

= 0

−2√

2− 2 + 0

−

(+)− 1

· 2√

2− 2

= (0− 0)

−2√

2− 2 + 0

−(+)· 2√

2− 2

= (0− 0)

2

+ 0

−· 2√

2− 2

6. The problem can be reduced to ﬁnding the line which minimizes the shaded area in the diagram. An equation of the circle in the ﬁrst quadrant is

 =

1 − 2. So the shaded area is

 () =

0

 1 −

1 − 2

 +

1

1 − 2

=

0

1 − 1 − 2

 −

1

1 − 2

0() = 1 −√

1 − 2−√

1 − 2 [by FTC] = 1 − 2√ 1 − 2

0= 0 ⇔ √

1 − 2=12 ⇒ 1 − 2= 14 ⇒ 2=34 ⇒  =23.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

CHAPTER 8 PROBLEMS PLUS ¤ 797

00() = −2 ·12(1 − 2)−12(−2) = 2

√1 − 2  0, so  =

√3

2 gives a minimum value of .

Note: Another strategy is to use the angle  as the variable (see the diagram above) and show that

 =  + cos  −412sin 2, which is minimized when  =6.

7. To ﬁnd the height of the pyramid, we use similar triangles. The ﬁrst ﬁgure shows a cross-section of the pyramid passing through the top and through two opposite corners of the square base. Now || = , since it is a radius of the sphere, which has diameter 2 since it is tangent to the opposite sides of the square base. Also, || =  since 4 is isosceles. So the height is || =√

2+ 2=√2 .

We ﬁrst observe that the shared volume is equal to half the volume of the sphere, minus the sum of the four equal volumes (caps of the sphere) cut off by the triangular faces of the pyramid. See Exercise 6.2.49 for a derivation of the formula for the volume of a cap of a sphere. To use the formula, we need to ﬁnd the perpendicular distance  of each triangular face from the surface of the sphere. We ﬁrst ﬁnd the distance  from the center of the sphere to one of the triangular faces. The third ﬁgure shows a cross-section of the pyramid through the top and through the midpoints of opposite sides of the square base. From similar triangles we ﬁnd that

 = ||

|| =

√2 

2+√

2 2 ⇒  =

√2 2

√32 =

√6 3 

So  =  −  =  −36 =336. So, using the formula  = 2( − 3) from Exercise 6.2.49 with  = , we ﬁnd that the volume of each of the caps is 

3 6 3 2

 −33· 36

=15− 696·6 +963=2

3277

√6

3. So, using our ﬁrst observation, the shared volume is  =124

33

− 42 3277

√6

3=28 27

√6 − 2

3. 8. Orient the positive -axis as in the ﬁgure.

Suppose that the plate has height  and is symmetric about the -axis. At depth  below the water (2 ≤  ≤ 2 + ), let the width of the plate be 2().

Now each of the  horizontal strips has height 

and the th strip (1 ≤  ≤ ) goes from

 = 2 +

 − 1

to  = 2 +



. The hydrostatic force on the th strip is  () =

2+()

2+[(−1)]

625[2 ()] .

Page 3 of 23

(4)

### 3 Techniques of Integration

8. Find min

x∈[0,2π]0π(cos t)(cos(x − t))dt

Solution:

cos α cos β= frac12 (cos(α − β) + cos(α + β)) ⇒ cos t ⋅ cos(x − t) =12[cos(2t − x) + cos x]

I(x) = ∫0π(cos t)(cos(x − t))dt =1 2[1

2sin(2t − x) + t cos x] ∣π

0 =1 2{[1

2sin(2π − x) + π cos x] − [1

2sin(−x)]}

= 1 2[1

2(sin(2π) cos x − cos(2π) sin x) + π cos x +1

2sin x] =π 2cos x critical point

I(x) = −π2sin x= 0 ⇒ x = 0, π, 2π, min

x∈[0,2π]I(x) = −π2, x= π

9. Find

∫ 1

sin(x) +√

3 cos(x)dx.

Motivation. 本題可變換為有理函數的積分（涉及課本7.1），採用高中學到的「正餘弦疊合」可更快得到答案。

Solution:

(Method 1) By the substitution t= tan(x2) we obtain

√−2

3∫ 1

(t −13)2+23dt. (5)

The trigonometric substitution t−13= 23tan(u) yields

−√

2∫ du.

Thus the antiderivative function is

−√

2 tan−1(

√3

√2tan(x 2) − 1

√2) + C.

(Method 2) Since sin(x) +√

3 cos(x) = 2 sin(x +π3),

∫ 1

sin(x) +√

3 cos(x)dx=1 2 ∫

1

sin(x +π3)dx

=1

2 ∫ csc(x +π 3) dx

= −1

2ln∣ cot (x +π

3) + csc (x +π 3)∣ + C.

10. (i) Let f∶ [−a, a] → R be a continuous function with a > 0 such that f(−x) = −f(x) for all x ∈ [−a, a]. Prove

a

−af(x) dx = 0.

(ii) Evaluate

1

−1

1+ x2018√ sin−1(x) x2+ 1 dx.

Motivation. 本題為跨範圍的綜合題，重點是利用函數的奇偶性質算出定積分。第一小題涉及變數變換（課本5.5），第二

Page 4 of 23

(5)

Solution:

For (i), since f(−x) = −f(x), the substitution u = −x yields

a

−af(x) dx = − ∫−aaf(u) du and so∫−aa f(x) dx = 0.

For (ii), since g(x) = x2018xsin2+1−1(x) is continuous and satisfies g(−x) = −g(x) for all x ∈ [−1, 1], from the item (i) we have∫−11 g(x) dx = 0. Hence it suffices to evaluate

1

−1

√ 1

x2+ 1dx. (6)

By the trigonometric substitution x= tan(u) we obtain

π 4

−π 4

sec(u) du = ln ∣ tan(u) + sec(u)∣∣

π 4

−π 4

= ln ∣

√2+ 1

√2− 1∣.

11. Use differentiation rules to prove that a) ∫ √

1− x2dx= x1−x22−cos−1x+ C b) ∫ √

y2− 1 dy = yy2−1−ln ∣y+2 y2−1∣ + D

Solution:

12. Find the integral of∫ x√

4x2+ 4xdx.

Solution:

∫ x√

4x2+ 4xdx = ∫ (x +1212)√

4x2+ 4xdx = 23(x2+ x)3212∫ √

(2x + 1)2− 1dx To integrate I= ∫ √

(2x + 1)2− 1dx

denote sec θ= 2x + 1, 0 ≤ θ <π2 or π≤ θ <32π⇒ sec θdθ = 2dx and tan θ ≥ 0 in the region

∴ I = ∫ tanθ ⋅sec θ tan θ 2 dθ= 1

2[tan θ sec θ − ∫ sec3θdθ] =1

2[tan θ sec θ − ∫ secθ(tan2θ+ 1)dθ]

= −I +1

2(tan θ sec θ − ln ∣ sec θ + tan θ∣) + c

⇒ I =1

4(tan θ sec θ − ln ∣ sec θ + tan θ∣) + C = 1 4)(√

4x2+ 4x(2x + 1) − ln ∣2x + 1 +√

4x2+ 4x∣) + C

∴ ∫ x√

4x2+ 4xdx =23(x2+ x)3218(√

4x2+ 4x(2x + 1) − ln ∣2x + 1 +√

4x2+ 4x∣) + C

13. ∫0(x+1)(xdx2+1)

Solution:

A

x+1+Bxx2+C+1 = (x+1)(x1 2+1) ⇒ A(x2+ 1) + (Bx + C)(x + 1) = 1 ⇒ (A + B)x2+ (B + C)x + A + C = 1

⇒⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

A+ B = 0 B+ C = 0 A+ C = 1

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎨⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎩ A= 1

2 C=1 2 B= −1

2

(6)

⇒ ∫ (x+1)(xdx2+1) = ∫ [x12+1+ 12(−x+1)x2+1 ] dx =12ln∣x + 1∣ −14ln∣x2+ 1∣ +12tan−1x+ c

0 dx

(x + 1)(x2+ 1) = lim

b→∞

b 0

dx

(x + 1)(x2+ 1)= lim

b→∞[1

2ln∣x + 1∣ −1

4ln∣x2+ 1∣ +1

2tan−1x+ c] ∣b

0

= lim

b→∞[ln ∣(b + 1)12 (b2+ 1)14∣ +1

2tan−1b− 0]

= 0 +1 2⋅π

2 = π 4

14. Compute∫01 x 1+x3dx

Solution:

1+ x3= (x + 1)(x2− x + 1) assume 1+xx3 = xA+1+xBx2−x+1+C

x= A(x2− x + 1) + (Bx + C)(x + 1) holds for all x by continuity.

plugging x= −1 ⇒ −1 = 3A ⇒ A = −13 x= 0 ⇒ 0 = A + C ⇒ C =13

compare the coefficient of x2 ⇒ 0 = A + B ⇒ B = 13

1+xx3 = 13(x−1+1+x2x−x+1+1 )

∴ ∫011+xx3dx= 1301(−x1+1+xx2−x+112+32) dx = −13ln∣x + 1∣ + 16ln∣x2− x + 1∣∣1

0+1201 dx (x−12)2+34

01 dx (x−12)2+34

denote u=x−12

ÔÔÔÔÔÔÔ ∫1212 udu2+34 =23tan−1 u3

2

1 2

12

=√

3 tan−1 1 3 =√

3⋅π6

∴ the integral = −13ln 2+63π

### 4 Improper Integrals

15. Determine whether the integral∫1 tan−1x

x2 dx converges and compute the value of it converges.

Solution:

1tanx−12 xdx= lim

b→∞1btanx−12 xdx

denote u= tan−1x tan u= x ⇒ dx = sec2udu

∴ ∫1btan−1x

x2 dx= ∫ tan

−1b

π 4

u sec2u

tan2u du= ∫ tan

−1b

π 4

u csc2udu

= u cot u∣tan

−1b

π

4 + ∫ tan

−1b

π 4

cot udu= −u cot u + ln ∣ sin u∣∣tan

−1b

π 4

= −tan−1b

b + ln b

√1+ b2+π 4 − ln 1

√2

lim

b→∞

tan−1b

b = 0 because lim

b→∞tan−1b=π2 lim

b→∞lnb

1+b2 = ln 1 = 0

∴ ∫1tanx−12 xdx= lim

b→∞1b tan−1x

x2 dx= π4+ln 22

16. Let the curve be the graph of y= x23. −1 < x < 8.

(a) Write down the equation which describe the length of this curve.

(b) This integration is an improper integration. Express it as limit of integration over a proper domain. Determine whether the integration converge.

(c) If the integration converges, evaluate the length.

Concepts: Length of a graph of a function, improper integral, substitution

Page 6 of 23

(7)

Solution:

(a) Let L be the length of this curve. We have

L= ∫−18

√ 1+ (dy

dx)2dx= ∫−18

√ 1+4

9x−23dx. (7)

(b) As x goes to 0, the integrand goes to infinity. We should express this equation as

8

−1

√ 1+4

9x−23 dx= lim

→0



−1

√ 1+4

9x−23 dx+ lim

→0+

8



√ 1+4

9x−23dx. (8)

Compute the indefinite integral, let t= x13, 3t2dt= dx

√ 1+4

9x−23 dx= ∫

√ 1+4

9t−2⋅ 3t2dt= ∫ sgn(t)

√ t2+4

9 ⋅ 3tdt

=3

2 ∫ sgn(t)

√ t2+4

9dt2= sgn(x)

√ x23+4

9

3

.

(9)

Therefore,

lim→0



−1

√ 1+4

9x−23dx= lim

→0

√ x23 +4

9

3

−1=

√13 9

3

− 8

27 (10)

and

lim→0+

8



√ 1+4

9x−23dx= lim

→0+

√ x23 +4

9

3

8=

√40 9

3

− 8

27. (11)

The integration converges.

(c) The length is given by

L=

√13 9

3

− 8 27+

√40 9

3

− 8 27=

√13 9

3

+

√40 9

3

−16

27. (12)

Remark: It will be easier if we only use the positive side or negative side.

Remark: It is possible to calculate the area of surface of revolution.

17. Consider the function f∶ (0, 1] → R,

f(x) = sin (1 x) (i) Show that the area between the curves y= f(x) and y = 0 is finite.

(ii) Show that the arclength of the curve y= f(x) is infinite.

Motivation. 本題為跨範圍的綜合題，給出面積有限但弧長無限的例子。重點是面積跟弧長公式（課本6.1和8.1），以及估

Solution:

For (i), the area is given by

1 0 ∣ sin (1

x)∣ dx. (13)

Since∣ sin(1x)∣ ≤ 1 and ∫01dx is finite, by the comparison test the area is finite.

For (ii), the arclength is given by

1 0

1+cos2(1x)

x4 dx. (14)

Since for all x∈ [0, 1] √

1+cos2(1x)

x4 ≥ cos2(1) x2

and∫01cosx22(1)dx is infinite, by the comparison test the arclenth is infinite.

(8)

### 5 Applications

18. Find the average value of y= 2e−tcos t over the interval 0≤ t ≤ 2π.

Solution:

Average value ¯y=02e−tcos tdt

consider I= ∫ e−tcos tdt, let u= cos t, du = − sin tdt, dv = e−t, v= −e−t

I= (−e−t) cos t − ∫ (−e−t)(− sin t)dt = (−e−t) cos t − ∫ e−tsin tdt Let u= sin t, du = cos tdt, dv = e−tdt, v= −e−t

I= (−e−t) cos t − [(−e−t) sin t − ∫ (−e−t) cos tdt] ⇒ 2I = (−e−t)(cos t − sin t) Thus∫0e−tcos tdt= 12[−e−t(cos t − sin t)] ∣

0 = 12(−e−2π+ 1) Hence ¯y=212(−e−2π+1)= 1(1 − e−2π)

19. A torus is generated by rotating the circle(x − 2)2+ y2= 12about the y-axis. Find the volume enclosed by the torus.

Solution:

Shell method: V = ∫12(2πx ⋅ 2y)dx = ∫122πx⋅ 2√

1− (x − 2)2dx= 4π ∫12x√

1− (x − 2)2dx Let u= x − 2, V = 4π ∫−11(u + 2)√

1− u2du= 4π2

20. The solid lies between planes perpendicular to the x-axis at x= −1 and x = 1. The cross-sections perpendicular to the x-axis are circular disks whose diameters run from the parabola y= x2 to the parabola y= 2 − x2.

Solution:

V = ∫−11π(1 − x2)2dx= 16π/15.

21. A cylindrical glass of radius r and height L is filled with and then tiled until the water exactly cover half the base.

Find the volume of water in the glass.

Solution:

A(x) = L

2r(r2− x2).

Thus

V = ∫−rrA(x)dx = L 2r ∫

r

−r(r2− x2)dx =L

r(r2x−x3

3 )∣r−r= 2 3r2L.

468

1. (a) Find a positive continuous function f such that the area under the graph of f from 0 to t is Astd − t3 for all t . 0.

(b) A solid is generated by rotating about the x-axis the region under the curve y − fsxd, where f is a positive function and x > 0. The volume generated by the part of the curve from x − 0 to x − b is b2 for all b . 0. Find the function f.

2. There is a line through the origin that divides the region bounded by the parabola y − x 2 x2 and the x-axis into two regions with equal area. What is the slope of that line?

3. The figure shows a horizontal line y − c intersecting the curve y − 8x 2 27x3. Find the number c such that the areas of the shaded regions are equal.

4. A cylindrical glass of radius r and height L is filled with water and then tilted until the water remaining in the glass exactly covers its base.

(a) Determine a way to “slice” the water into parallel rectangular cross-sections and then set up a definite integral for the volume of the water in the glass.

(b) Determine a way to “slice” the water into parallel cross-sections that are trapezoids and then set up a definite integral for the volume of the water.

(c) Find the volume of water in the glass by evaluating one of the integrals in part (a) or part (b).

(d) Find the volume of the water in the glass from purely geometric considerations.

(e) Suppose the glass is tilted until the water exactly covers half the base. In what direction can you “slice” the water into triangular cross-sections? Rectangular cross-sections?

Cross-sections that are segments of circles? Find the volume of water in the glass.

### MasterID: 00769

r

L L

r

5. (a) Show that the volume of a segment of height h of a sphere of radius r is V −13h2s3r 2 hd

(See the figure.)

(b) Show that if a sphere of radius 1 is sliced by a plane at a distance x from the center in such a way that the volume of one segment is twice the volume of the other, then x is a solution of the equation

3x329x 1 2 − 0

where 0 , x , 1. Use Newton’s method to find x accurate to four decimal places.

(c) Using the formula for the volume of a segment of a sphere, it can be shown that the depth x to which a floating sphere of radius r sinks in water is a root of the equation

x323rx214r3s − 0

where s is the specific gravity of the sphere. Suppose a wooden sphere of radius 0.5 m has specific gravity 0.75. Calculate, to four-decimal-place accuracy, the depth to which the sphere will sink.

(d) A hemispherical bowl has radius 5 inches and water is running into the bowl at the rate of 0.2 in3ys.

(i) How fast is the water level in the bowl rising at the instant the water is 3 inches deep?

(ii) At a certain instant, the water is 4 inches deep. How long will it take to fill the bowl?

6. Archimedes’ Principle states that the buoyant force on an object partially or fully sub- merged in a fluid is equal to the weight of the fluid that the object displaces. Thus, for an FiGURE FOR PROblEm 3

0 x

y y=8x-27˛

y=c

r

h FiGURE FOR PROblEm 5

## Problems Plus

Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

22. Let R be the region bounded by y= 2x ln x, y = x ln x, x = 1 and x = 2.

(a) Find the area of R.

(b) Find the volume of the solid obtained from rotating R around y-axis.

Page 8 of 23

(9)

(c) Find the volume of the solid obtained from rotating R around x-axis.

Solution:

(a)

2

1 (2x ln x − x ln x)dx = ∫12x ln xdx= 1

2x2ln x∣2

1− ∫121

2xdx= 2 ln 2 −1 4x22

1= 2 ln 2 − 1 +1

4 = 2 ln 2 −3 4 (b)

2 1

2πx⋅ x ln xdx = 2π [x3 3 ln x∣2

1− ∫12x2

3 dx] = 2π (8

3ln 2−8 9+1

9) = 2π (8 3ln 2−7

9) (c)

2 1

π[4x2(ln x)2− x2(ln x)2]dx = ∫12π⋅ 3x2(ln x)2dx= π[x3(ln x)22

1− ∫12x3⋅ 2 ln x ⋅ 1 xdx]

= π(x3(ln x)2−2

3x3ln x+2 9x3)∣2

1= π(8(ln 2)2−16

3 ln 2+16 9 −2

9)

= π(8(ln 2)2−16

3 ln 2+14 9 )

23. (1). Let S be the solid obtained by rotating the region bounded by the parabola y= x−x2and y= 0 about the y-axis.

Find the volume of S.

(2). Denote Cm be the cone by rotating the line y= mx about the y-axis. Find m such that Cm divides S into two solids of equal volume.

Solution:

1.) The solid obtained by rotating the region about the y-axis has the volume:

2π∫

1 0

x(x − x2)dx = 2π(1 3x3−1

4x410) = π 6.

2.) Let S1be the solid above the cone Cm. Then its volume satisfies π

12 = 2π ∫01−mx(x − x2− mx)dx = 2π(1− m

3 (1 − m)3−(1 − m)4 4 ) = π

6(1 − m)4. Hence, m= 1 − 41

2.

24. a) Find the area of the region 0≤ y ≤√

x− x2, 0≤ x ≤ 1 b) Find the x-centroid of the above region

c) Find the y-centroid of the above region

d) Find the volume of revolving the above region about the x-axis e) Find the volume of revolving the above region about the y-axis

Solution:

a) The area A = ∫01

x(1 − x) dx = 14−11

√1− y2dy= π8

b) ¯x= A101x√

x(1 − x) dx = 8A1−11(1 − y)√

1− y2dy = 12 c) ¯y = 2A101x(1 − x) dx = 2

d) The volume about x-axis is V = π ∫01x(1 − x) dx = π6 e) The volume about y-axis is V = 2π ∫01x√

x(1 − x) dx = π82

(10)

25. a) Find the area of the region 0≤ y ≤√

x2− x, 1 ≤ x ≤ 2 b) Find the x-centroid of the above region

c) Find the y-centroid of the above region

d) Find the volume of revolving the above region about the x-axis e) Find the volume of revolving the above region about the y-axis

Solution:

a) The area A = ∫12

x(x − 1) dx = 1413

y2− 1 dy = 38−ln(3+8 8) b) ¯x= A112x√

x(x − 1) dx = 8A113(y + 1)√

y2− 1 dy

= 12+9 88 8−3 ln(3+

8)

c) ¯y = 2A112x(x − 1) dx = 98−3 ln(3+10 8)

d) The volume about x-axis is V = π ∫12x(x − 1) dx = 6 e) The volume about y-axis is V = 2π ∫12x√

x(x − 1) dx

= (258−3 ln(3+48 8))

26. Let R be the region tan2(x) ≤ y ≤ tan(x) sec(x), 0 ≤ x < π/2.

a) Find the area A of this regin R b) Find the centroid(¯x, ¯y) of R

c) Find the volume generated by revolving R about the x-axis d) Find the volume generated by revolving R about the y-axis

Solution:

a) A= ∫0π/2(tan(x) sec(x) − tan2(x)) dx = [sec x − tan x + x]π0/2

= π2− 1

b) ¯x= A10π/2x(tan(x) sec(x) − tan2(x)) dx

= A1[[x(sec x − tan x + x)]π0/2 − ∫0π/2(sec x − tan x + x) dx]

= A1[π42 − [ln ∣ cos(x)(tan(x) + sec(x))∣ +x22]π0/2]

= A1[π42 − (ln 2 + π82)] =A1[π82 − ln 2]

¯

y=2A10π/2tan2(x)(sec2x− tan2x) dx = 2A10π/2tan2x dx

= 2A1 [tan x − x]π0/2 = ∞

c) Volume about x-axis= π ∫0π/2tan2(x)(sec2x− tan2x) dx

= ∞

d) Volume about y-axis= 2π ∫0π/2x(tan(x) sec(x) − tan2(x)) dx

= 2π[π82 − ln 2]

27. Find the length of the curve

y= ∫

x 1

√t2− 1 dt for 1 ≤ x ≤ 2 .

Solution:

By the Fundamental Theorem of Calculus, we have dy

dx = 1 2√

x

√ (√

x)2− 1 = 1 2

√ 1−1

x.

Page 10 of 23

A subgroup N which is open in the norm topology by Theorem 3.1.3 is a group of norms N L/K L ∗ of a finite abelian extension L/K.. Then N is open in the norm topology if and only if

We explicitly saw the dimensional reason for the occurrence of the magnetic catalysis on the basis of the scaling argument. However, the precise form of gap depends

In an Ising spin glass with a large number of spins the number of lowest-energy configurations (ground states) grows exponentially with increasing number of spins.. It is in

Miroslav Fiedler, Praha, Algebraic connectivity of graphs, Czechoslovak Mathematical Journal 23 (98) 1973,

A convenient way to implement a Boolean function with NAND gates is to obtain the simplified Boolean function in terms of Boolean operators and then convert the function to

Each unit in hidden layer receives only a portion of total errors and these errors then feedback to the input layer.. Go to step 4 until the error is

Microphone and 600 ohm line conduits shall be mechanically and electrically connected to receptacle boxes and electrically grounded to the audio system ground point.. Lines in

The min-max and the max-min k-split problem are defined similarly except that the objectives are to minimize the maximum subgraph, and to maximize the minimum subgraph respectively..