Section 5.1 The Area and Distance Problems

Section 5.1 The Area and Distance Problems

18. Use Definition 2 to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

f (x) = x + ln x, 3 ≤ x ≤ 8.

Solution:

SECTION 5.1 THE AREA AND DISTANCE PROBLEMS ¤ 521

13. For a decreasing function, using left endpoints gives us an overestimate and using right endpoints results in an underestimate.

We will use 6to get an estimate. ∆ = 1, so

6= 1[(05) + (15) + (25) + (35) + (45) + (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.

14. For an increasing function, using left endpoints gives us an underestimate and using right endpoints results in an overestimate.

We will use 6to get an estimate. ∆ =³⁰₆^{− 0} = 5s = ₃₆₀₀⁵ h = ₇₂₀¹ h.

6= ₇₂₀¹ [(25) + (75) + (125) + (175) + (225) + (275)]

= ₇₂₀¹ (3125 + 66 + 88 + 1035 + 11375 + 11925) = ₇₂₀¹ (52175) ≈ 0725 km

For a very rough check on the above calculation, we can draw a line from (0 0) to (30 120) and calculate the area of the triangle: ¹₂(30)(120) = 1800. Divide by 3600 to get 05, which is clearly an underestimate, making our midpoint estimate of 0725seem reasonable. Of course, answers will vary due to different readings of the graph.

15. () = −( − 21)( + 1) and ∆ = ¹²6^{− 0} = 2

6= 2 · (1) + 2 · (3) + 2 · (5) + 2 · (7) + 2 · (9) + 2 · (11)

= 2 · 40 + 2 · 216 + 2 · 480 + 2 · 784 + 2 · 1080 + 2 · 1320

= 7840(infected cellsmL) · days

Thus, the total amount of infection needed to develop symptoms of measles is about 7840 infected cells per mL of blood plasma.

16. () = ²^, 0 ≤  ≤ 4. ∆ = (4 − 0) = 4 and ^ = 0 +  ∆ = 4.

 = lim

→∞= lim

→∞



=1

 () ∆ = lim

→∞



=1

(4)²^4· 4

.

17. () = 2 + sin², 0 ≤  ≤ . ∆ = ( − 0) =  and ^ = 0 +  ∆ = .

 = lim

→∞= lim

→∞



=1

 () ∆ = lim

→∞



=1

2 + sin²()

· 

.

18. () =  + ln , 3 ≤  ≤ 8. ∆ = (8 − 3) = 5 and ^= 3 +  ∆ = 3 + 5.

 = lim

→∞= lim

→∞



=1

 () ∆ = lim

→∞



=1[(3 + 5) + ln(3 + 5)] · 5

. 19. () = √

³+ 8, 1 ≤  ≤ 5. ∆ = (5 − 1) = 4 and ^= 1 +  ∆ = 1 + 4.

 = lim

→∞= lim

→∞



=1

 () ∆ = lim

→∞



=1

(1 + 4)

(1 + 4)³+ 8

· 4

.

20. lim

→∞



=1

1







3

can be interpreted as the area of the region lying under the graph of  = ³on the interval [0 1],

since for  = ³on [0 1] with ∆ = 1 − 0

 = 1

, = 0 +  ∆ = 1

 = 

, and ^∗ = , the expression for area is

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22. Determine a region whose area is equal to the given limit. Do not evaluate the limit. lim

n→∞

n

P

i=1 3 n

q 1 + ³ⁱ_n. Solution:

SECTION 5.1 AREAS AND DISTANCES ¤ 491

19. () = −( − 21)( + 1) and ∆ = ¹²6^{− 0}= 2

6= 2 · (1) + 2 · (3) + 2 · (5) + 2 · (7) + 2 · (9) + 2 · (11)

= 2 · 40 + 2 · 216 + 2 · 480 + 2 · 784 + 2 · 1080 + 2 · 1320

= 7840(infected cellsmL) · days

Thus, the total amount of infection needed to develop symptoms of measles is about 7840 infected cells per mL of blood plasma.

20. (a) Use ∆ = 14 days. The number of people who died of SARS in Singapore between March 1 and May 24, 2003, using left endpoints is

6= 14(00079 + 00638 + 01944 + 04435 + 05620 + 04630) = 14(17346) = 242844 ≈ 24 people Using right endpoints,

6= 14(00638 + 01944 + 04435 + 05620 + 04630 + 02897) = 14(20164) = 282296 ≈ 28 people (b) Let  be the number of days since March 1, 2003, () be the number of deaths per day on day , and the graph of  = ()

be a reasonable continuous function on the interval [0 84]. Then the number of SARS deaths from  =  to  =  is approximately equal to the area under the curve  = () from  =  to  = .

21. () = 2

²+ 1, 1 ≤  ≤ 3. ∆ = (3 − 1) = 2 and ^= 1 + ∆ = 1 + 2.

 = lim

→∞= lim

→∞



=1

 ()∆ = lim

→∞



=1

2(1 + 2) (1 + 2)²+ 1· 2

. 22. () = ²+√

1 + 2, 4 ≤  ≤ 7. ∆ = (7 − 4) = 3 and ^= 4 +  ∆ = 4 + 3.

 = lim

→∞= lim

→∞



=1

 () ∆ = lim

→∞



=1

(4 + 3)²+

1 + 2(4 + 3)

· 3

. 23. () =√

sin , 0 ≤  ≤ . ∆ = ( − 0) =  and ^= 0 +  ∆ = .

 = lim

→∞= lim

→∞



=1

 () ∆ = lim

→∞



=1

sin() ·

. 24. lim

→∞



=1

3



 1 +3

 can be interpreted as the area of the region lying under the graph of  =√

1 + on the interval [0 3],

since for  =√

1 + on [0 3] with ∆ = 3 − 0

 = 3

, = 0 +  ∆ = 3

, and ^∗ = , the expression for the area is

 = lim

→∞



=1

 (^∗) ∆ = lim

→∞



=1

 1 + 3

 3

. Note that this answer is not unique. We could use  = √ on [1 4] or, in general,  =√

 −  on [ + 1  + 4], where  is any real number.

25. lim

→∞



=1

 4tan

4can be interpreted as the area of the region lying under the graph of  = tan  on the interval 0^₄,

since for  = tan  on 0^₄

with ∆ = 4 − 0

 = 

4, = 0 +  ∆ = 

4, and ^∗ = , the expression for the area is

 = lim

→∞



=1

 (^∗) ∆ = lim

→∞



=1

tan



4

 

4. Note that this answer is not unique, since the expression for the area is the same for the function  = tan( − ) on the interval

  + ^₄

, where  is any integer.

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34. (a) Let An be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2π/n, show that

An =1

2nr²sin2π n (b) Show that lim

n→∞An = πr². [Hint: Use Equation 3.3.5]

Solution:

SECTION 5.2 THE DEFINITE INTEGRAL ¤ 493 31.  = () = cos . ∆ =  − 0

 = 

 and = 0 +  ∆ = 

.

 = lim

→∞= lim

→∞



=1

 () ∆ = lim

→∞



=1

cos







· 



CAS= lim

→∞







 sin





 1 2+ 1



2 sin

  2

 − 

2





^CAS= sin 

If  = ^₂, then  = sin^₂ = 1.

32. (a) The diagram shows one of the  congruent triangles, 4, with central angle 2.  is the center of the circle and  is one of the sides of the polygon.

Radius  is drawn so as to bisect∠. It follows that  intersects  at right angles and bisects . Thus, 4 is divided into two right triangles with legs of length ¹₂() =  sin()and  cos(). 4 has area

2 ·¹2[ sin()][ cos()] = ²sin() cos() = ¹₂²sin(2), so =  · area(4) = ¹2²sin(2).

(b) To use Equation 3.3.2, lim

→0

sin 

 = 1, we need to have the same expression in the denominator as we have in the argument of the sine function—in this case, 2.

lim→∞= lim

→∞

1

2²sin(2) = lim

→∞

1

2²sin(2) 2 ·2

 = lim

→∞

sin(2)

2 ². Let  = 2

 . Then as  → ∞,  → 0, so lim_

→∞

sin(2)

2 ²= lim

→0

sin 

 ² = (1) ²= ².

5.2 The Definite Integral

1. () =  − 1, −6 ≤  ≤ 4. ∆ =  − 

 = 4 − (−6)

5 = 2.

Since we are using right endpoints, ^∗ = .

5=

5

=1

 () ∆

= (∆)[ (1) +  (2) +  (3) +  (4) +  (5) +  (6)]

= 2[ (−4) + (−2) + (0) + (2) + (4)]

= 2[−5 + (−3) + (−1) + 1 + 3]

= 2(−5) = −10

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.

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