The Integral - Penn Math

[Pages:10]CHAPTER

5

The Integral

5.1 Approximating and Computing Area

Preliminary Questions

1. The interval [2, 5] is divided into 6 subintervals in order to calculate R6 for some function. What are the right-endpoints of those subintervals? What are the left-endpoints?

2. If f (x) = x-2 on [3, 7], which is larger: RN or L N ?

3. Which of the following pairs of sums are not equal?

(a)

4 i =1

i,

4 =1

(b)

4 j =1

j2,

5 k=2

k2

(c)

4 j =1

j,

5 i =2

(i

-

1)

(d)

4 i =1

i (i

+

1),

5j=2( j - 1) j

4. The interval [1, 5] is divided into 16 subintervals.

(a) What are the left endpoints of the first and last subintervals? (b) What are the right endpoints of the first two subintervals?

5. True or False: (a) The right-endpoint rectangles lie below the graph of an increasing function. (b) If f is monotonic, then the area under the graph lies in between RN and L N . (c) If f is not monotonic, then L N and RN may converge to different limits as N . (d) If f (x) is constant, then the right-endpoint rectangles all have the same height.

Exercises

1. An athlete runs with velocity 4 mph for half an hour, 6 mph for the next hour, and 5 mph for another half-hour. Compute the total distance traveled and indicate on a graph how this quantity can be interpreted as an area.

The

total

distance

traveled

is

(4)

(

1 2

)

+

(6)

(1)

+

(5)

(

1 2

)

=

10

1 2

miles.

1

2 Chapter 5 The Integral

6 5 4 mph 3 2 1

0

0.5

1 1.5

2

hours

2.

Figure 1 shows the velocity of an object over a 3-minute interval. Determine the distance

3. tArasvsuelmede othvaetrtthheevienltoercvitaylso[f0a,n3]obanjedct[1is, 23.25t]f(tr/es.mUesmebEeqr.to(?c?o)ntvoedrtefterorminmepthhetodmistialensceper

mtraivneulted).by the object over the time intervals (in seconds) [0, 2] and [2, 5].

The total distance traveled is given by the area under the graph of v = 32t.

During the interval [0, 2], the object tFraivgeulsre12 (12)(64) = 64 ft.

During

the

interval

[2,

5],

the

object

travels

1 2

(3)(160

-

64)

+

(3)(64)

=

336

ft.

160

140

120

100

80

60

40

20

0

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

4.

Consider f (x) = 2x + 3 on [0, 3]. 5. Let (fa()x)D=etexr2m+inxe -the2l.eft- and right-endpoints if [0, 3] is divided into 6 subintervals.

(a) C(ba)lcCuloamtepRut3eaRnd6 aLn3dfoLr6.the interval [2, 5]. (b) S(ck)etFchintdhethgereaxpahcotfarfeaanudsinthgegreeocmtanetgrlyesanthdactommapkueteuptheeaecrhroorf itnhethaeptpwrooxciamlcautiloantiso.ns of (b).

Let f (x) = x2 + x - 2 and set a = 2, b = 5, n = 3, h = x = (b - a) /n = (5 - 2) /3 = 1. (a) Let xk = a + kh, k = 0, 1, 2, 3.

Selecting left endpoints of subintervals, xk, k = 0, 1, 2, or {2, 3, 4}, we have

2

2

L3 = f (xk) x = h f (xk) = (1) (4 + 10 + 18) = 32

k=0

k=0

Selecting right endpoints of subintervals, xk, k = 1, 2, 3, or {3, 4, 5}, we have

3

3

R3 = f (xk) x = h f (xk) = (1) (10 + 18 + 28) = 56

k=1

k=1

(b) Here are figures of the three sets of rectangles that approximate the area under the curve f (x) over the interval [2, 5].

5.1 Approximating and Computing Area

3

x2+x-2 30

25

x2+x-2 30

25

20

20

15

15

10

10

5

5

6.

2

2.5

3

3.5

4

4.5

5

x

2

2.5

3

3.5

4

4.5

5

x

Let f (x) = cos x.

7.

Cons(iad)erCfal(cxu)la=te5Rx4oannd[0L, 34 ]f.or the interval [0,

2

].

(a) F(bin)dSakfeotcrmh tuhlea gforar pRhNo(fusfeafnodrmthuelare(c?t?a)n)g. les that make up each of the approximations.

(b) I(sc)RNIsbthigegaerreoarusnmdaelrlethrethgarnapthhelaarrgeearuonrdsemr athlleergrthapanh?R4? Than L4? (c) Calculate limN RN .

(d) Calculate the area under the graph using geometry and verify your answer to (c).

Let f (x) = 5x on [0, 3]. Let n be a (symbolic) positive integer and set a = 0, b = 3, h = x = (b - a) /n = (3 - 0) /n = 3/n.

(a) Let xk = a + kh = 3k/n, k = 1, 2, . . . , n be the right endpoints of the n subintervals of [0, 3]. Then

n

Rn = h

k=1

f (xk) =

3 n

n

5

k=1

3k n

= 45

n2 n +

= 45 + 45

n2 2 2

2 2n

= 45 n k n2 k=1

(b) Since f (x) = 5x is increasing, the right endpoint approximation Rn is greater than the area under the graph.

45 45 45

(c)

The

area

under

the

graph

is

lim

n

Rn

= lim n

+ 2 2n

= = 22.5. 2

(d)

The area under the graph is the area of a triangle:

A

=

1 2

B

H

=

1 2

(3) (15)

=

45 2

,

which

agrees with the answer in (c).

8. Estimate R6 and L6 over [0, 1.5] for the function shown in Figure 2.

9. Estimate R6 and L6 for the graph in Figure 3.

Figure 2

1 .5

0 121452

33

33

Figure 3

Let

f (x) on [0, 2] be given by the figure in the text. For n

=

6, h

= (2 - 0)/6 =

1 3

,

{xk }6k=0 =

0,

1 3

,

2 3

,

1,

4 3

,

5 3

,

2

.

4 Chapter 5 The Integral

Therefore

L6

=

1 3

5 k=0

f (xk) =

1 3

R6

=

1 3

6 k=1

f (xk) =

1 3

1 + 5 + 7 + 9 + 1 + 3 = 1.625

2888

4

5 + 7 + 9 + 1 + 3 + 1 1.792

888

4

In Exercises 10?19, calculate the approximation for the given function and interval. 10.

R8, 7 - x, [3, 5] 11. R6, 2x2 - x + 2, [1, 4]

Let

f (x) =

2x2 - x

+ 2 on [1, 4]. For n

=

6, h

= (4 - 1)/6 =

1 2

,

{xk }6k=0 =

1,

1

1 2

,

2,

2

1 2

,

3,

3

1 2

,

4

. Therefore

R6

=

1 2

6

(2xk2

k=1

-

xk

+ 2)

=

47.5

12. L6, 2x2 - x + 2, [1, 4]

13. R4, x3 - 2x + 5, [0, 1]

Let

f (x) =

x3

- 2x

+ 5 on [0, 1]. For n

=

4, h

= (1 - 0)/4 =

1 4

,

{xk }4k=0 =

0,

1 4

,

1 2

,

3 4

,

1

. Therefore

R4

=

1 4

4

(xk3

k=1

- 2xk

+ 5)

4.140625

14.

R4, x3 - 2x + 5, [-2, 2]

15.

R5,

sin

x,

[

4

,

3 4

]

Let

f (x) = sin x

on

[

4

,

3 4

].

For

n

= 5, h

=

(3/4-/4) 5

=

10

,

{xk }5k=0 =

4

,

7 20

,

9 20

,

11 20

,

13 20

,

3 4

. Therefore

5 R5 = 10 k=1 sin xk 1.402563

16. 17.

L5, L4,

x -1, x -2,

[1, [1,

2] 3]

Let

f (x) = x-2 on [1, 3]. For n

= 4, h

= (3 - 1)/4 =

1 2

,

{xk }4k=0

=

1,

3 2

,

2,

5 2

,

3

.

Therefore

L4

=

1 2

3 k=0

xk-2

0.927222

18. 19.

L L

4, 5,

cos x x2 +

,

[

4

3|x

, |,

2

]

[-3,

2]

Let f (x) = x2 + 3 |x| on [-3, 2]. For n = 5, h = (2 - (-3))/5 = 1, {xk}5k=0 = {-3, -2, -1, 0, 1, 2}. Therefore

4

L5 = 1 (xk2 + 3 |x |) = 36

k=0

5.1 Approximating and Computing Area

5

20. R & W Compute the average of R5 and L5 to approximate the area under the graph of

21. Cf a(xlc)ul=atext-h1e osvuemrs[:3.5, 5]. Explain why the average is more accurate than either endpoint a(ap)pro5xim3 ation.

i =1

5

(b) 3

i =0

4

(c) k3

k=2

4

(d) sin j

j =3

2

When the number of summands is small, we may compute the sum directly. Where

applicable (especially when the number of summands is large), we may use summation

laws. For illustration, both ways are shown in this exercise.

5

(a) Do this directly: 3 = 3 + 3 + 3 + 3 + 3 = 15. Or use the law for summing a

i =1 5

constant: 3 = (3)(5) = 15.

i =1

5

5

(b) Here 3 = 3 + 3 + 3 + 3 + 3 + 3 = 18 or 3 = (3)(6) = 18.

i =0

i =0

4

(c) Again, k3 = 23 + 33 + 43 = 99 or

k=2

4

k3 =

k=2

=

4

1

k3 -

k3

k=1

k=1

44 43 42 ++ -

424

14 13 12 ++

424

= 99.

4

j

(d) Finally, sin

= 1 + 0 + (-1) + 0 = 0.

j =1

2

In Exercises 22?34, use the Linearity Rules and formulas (??)?(??) to rewrite and evaluate the

sums.

22.

6

10

j3

23. j4=1j 3

j =1

We have 24.

10

10 j =1

4 j3

=

4

20 2 j 25. j=1 (2k + 1)

k=1

10 j =1

j3 = 4

104 4

+

103 2

+

102 4

= 12100.

We have

20 k=1

(2k

+

1)

=

2

26.

15

(6 - 3i)

i =1

20 k=1

k

+

20 k=1

1

=

2

202 + 20

2

2

+ 20 = 440.

6 Chapter 5 The Integral

10

27. ( 3 - 2 2)

=1

We have

10 =1

(

3-2

2) =

28.

30

18 (3s2 - 4s - 1)

29. s=1 (4i - 2i 2 + 9)

i =1

10 =1

3-2

10 =1

2=

104 + 103 + 102

4

2

4

We have

18 i =1

(4i

-

2i 2

+

9)

=

4

18 i =1

i

-

2

18 i =1

i

2

+

9

4

182 2

+

18 2

-2

183 3

+

182 2

+

18 6

+ (9)(18) = -3372.

30.

10

200 j 2 Hint: write as a difference of two sums.

31. j=5j

j =101

-2

103 + 102 + 10

3

2

6

18 i =1

1

=

= 2255.

We have 32.

50

200 j =101

j=

20 (7 j - 9) 33. j=4 6 j 2

j =10

200 j =1

j-

100 j =1

j=

+ 2002 200

2

2

-

+ 1002

100

2

2

= 15050.

We have

20 j =10

6 j2

=

6

20 j =1

j2 - 6

9 j =1

j2 = 6

203 3

+

202 2

+

20 6

-6

93 3

+

92 2

+

9 6

= 15510.

34.

10

35. Writ(e6tjh3e-sujm2)of the cubes of the whole numbers from 100 to 250 in summation notation.

j =3

250

The sum of the first 100 cubes may be written in summation notation as k3.

36.

k=100

Write the sum

In Exercises 37?41, calculate the1su+m1, 3a+ssum2in+g 2th3a+t a?1? =? +-1,n +i1=n013 ai = 10, and

in summation notation. 10

37. 2ai

i =1

10

10

We have 2ai = 2 ai = (2)(10) = 20.

38.

i =1

i =1

10

10 (ai - bi ) 39. i=1 (3a + 4b )

=1

We have

10 i =1

bi

=

7.

10

10

10

(3ai + 4bi ) = 3 ai + 4 bi

i =1

i =1

i =1

= (3)(10) + (4)(7) = 58.

40. 10

41. Canayi ou calculate

10 i =1

ai bi

from

the

information

given?

i =2

10

From the information given, ai bi cannot be computed.

i =1

5.1 Approximating and Computing Area

7

42.

ni

43.

Evaluate lim Evaluate nlim

n i =1

in22 -. i

+1 .

n i =1

n3

Now

sn

=

n i =1

i2 - i + 1 n3

=

1 n3

=1 n3

n3 + n2 + n 3 26

n

i2 -

i =1

- n2 + n 22

n

n

i+ 1

i =1

i =1

+ (n) = 1 + 2 . 3 3n2

Therefore,

lim

n

sn

=

1 .

3

In Exercises 44?59, use formulas (??)?(??) to find a formula for RN for the given function and interval. Then compute the area under the graph as a limit. 44. x; [0, 3] 45. x; [2, 7]

Let f (x) = x on the interval [2, 7]. Then x = 7 - 2 = 5 and a = 2. Hence, NN

RN =

N

x

f (2 + j

x) = 5

N

2+ j 5

j =1

N j=1

N

= 10 N 1 + 25 N j

N j=1

N 2 j=1

=

10 N

+

25

N2 + N

= 10 + 25 + 25

N

N2 2 2

2 2N

and

25 25 45

lim

N

RN

=

lim

N

10 + + 2 2N

= = 22.5. 2

46. 3 - x; [1, 2] 47. 2x + 7; [3, 6]

Let f (x) = 2x + 7 on the interval [3, 6]. Then

x = 6 - 3 = 3 and a = 3. Hence, NN

RN =

N

x

f (3 + j

x) = 3

N

2 3+ j 3

j =1

N j=1

N

+7

= 39

N 1 + 18

N

j

=

39 N

+

18

N2 + N

= 39 + 9 + 9

N j=1

N 2 j=1

N

N2 2 2

N

and

lim

N

RN

=

lim

N

48 + 9 N

= 48.

48. x2; [0, 1]

49. x2; [2, 4]

8 Chapter 5 The Integral

Let f (x) = x2 on the interval [2, 4]. Then

x

=

4-2 N

=

2 N

and a

= 2. Hence,

RN =

N

x f (2 + j

j =1

2N x) =

N j=1

4 + j 8 + j2 4

N

N2

= 8 N 1 + 16 N j + 8 N j 2 = 8 N + 16

N j=1

N 2 j=1

N 3 j=1

N

N2

+ 8 N3 + N2 + N N3 3 2 6

=8+8+ 8 + 8 + 4 + 4 N 3 N 3N2

N2 + N 22

and

lim RN = lim

N

N

56 + 12 + 4 3 N 3N2

18.6667.

50. 4 - x2; [0, 2]

51. 3x2 - x + 4; [0, 1] Let f (x) = 3x2 - x + 4 on the interval [0, 1]. Then

x = 1 - 0 = 1 and a = 0. Hence, NN

RN =

N

x

f (0 + j

x) = 1

N

j =1

N j=1

3j2 1 - j 1 + 4 N2 N

=3

N j2 - 1

N

4N

j+

1

N 3 j=1

N 2 j=1

N j=1

=3

N3 N2 N ++

-1

N2 N +

+ 4N

N3 3 2 6

N2 2 2

N

=1+ 3 + 1 - 1 - 1 +4 2N 2N2 2 2N

and

11

lim

N

RN

=

lim

N

4.5 + + N 2N2

= 4.5.

52. 3x2 - x + 4; [1, 5]

53. 4x3 - 3x; [0, 2] Let f (x) = 4x3 - 3x on the interval [0, 2]. Then

x = 2 - 0 = 2 and a = 0. Hence, NN

N

2N

RN =

x f (0 + j x) =

j =1

N j=1

= 64 N j 3 - 12 N j = 64

N 4 j=1

N 2 j=1

N4

32 16

6

= 16 + + - 6 -

N N2

N

4j3 8 - 3j 2

N3

N

N4 + N3 + N2 424

- 12 N2

N2 + N 22

and

lim

N

R

N

=

lim

N

10 + 26 + 16 N N2

= 10.

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