Section 6.1 Areas Between Curves

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Section 6.1 Areas Between Curves

554 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION

8. The curves intesect when ²− 4 = 2 ⇒ ²− 6 = 0 ⇒ ( − 6) = 0 ⇒  = 0 or 6.

 =6

0[2 − (²− 4)] 

=6

0(6 − ²)  =

3²−¹3³6 0

=

3(6)²−¹3(6)³

− (0 − 0)

= 108 − 72 = 36

9.  =

 2 1

1

− 1

²



 =

 ln  +1



2

1

=

ln 2 +¹₂

− (ln 1 + 1)

= ln 2 −¹2 ≈ 019

10. By observation,  = sin  and  = 2 intersect at (0 0) and (2 1) for  ≥ 0.

 =

 2 0



sin  −2





 =



− cos  −1

²

2 0

= 0 −

4



− (−1) = 1 − 4

11. The curves intersect when 1 − ²= ²− 1 ⇔ 2 = 2² ⇔ ² = 1 ⇔  = ±1.

 =

 1

−1

(1 − ²) − (²− 1)



=

 1

−12(1 − ²) 

= 2 · 2

 1 0

(1 − ²) 

= 4

 −¹3³1 0= 4

1 −¹3

=⁸₃

12. 4 + ² = 12 ⇔ ( + 6)( − 2) = 0 ⇔

 = −6 or  = 2, so  = −6 or  = 2 and

 =

 2

−6

−¹4²+ 3

− 



=

−12¹³−¹2²+ 32

−6

=

−²3 − 2 + 6

− (18 − 18 − 18)

= 22 −²3 =⁶⁴₃

556 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 18. The curves intersect when√

 − 1 =  − 1 ⇒

 − 1 = ²− 2 + 1 ⇔ 0 = ²− 3 + 2 ⇔ 0 = ( − 1)( − 2) ⇔  = 1 or 2.

 =

 2 1

√ − 1 − ( − 1)



=

2

3( − 1)^32−¹2( − 1)²2 1=2

3−¹2

− (0 − 0) = ¹6

19. By inspection, the curves intersect at  = ±¹2.

 =

 12

−12[cos  − (4²− 1)] 

= 2

12 0

(cos  − 4²+ 1)  [by symmetry]

= 21

sin  −⁴3³+ 12 0 = 21

−¹6+¹₂

− 0

= 21

+¹₃

= ²_+²₃

20.  =√

2 −  ⇒ ²= 2 −  ⇔  = 2 − ², so the curves intersect when ⁴= 2 − ² ⇔ ⁴+ ²− 2 = 0 ⇔ (²+ 2)(²− 1) = 0 ⇔  = 1 [since  ≥ 0].

 =

 1

0 [(2 − ²) − ⁴)]  =

2 −¹3³−¹5⁵1 0

=

2 −¹3 −¹5

− 0 = ²²15

21. The curves intersect when tan  = 2 sin  (on [−3 3]) ⇔ sin  = 2 sin  cos  ⇔

2 sin  cos  − sin  = 0 ⇔ sin  (2 cos  − 1) = 0 ⇔ sin  = 0 or cos  = ¹2 ⇔  = 0 or  = ±^3.

 = 2

3

0 (2 sin  − tan )  [by symmetry]

= 2

−2 cos  − ln |sec |3 0

= 2 [(−1 − ln 2) − (−2 − 0)]

= 2(1 − ln 2) = 2 − 2 ln 2

22. The curves intersect when ³=  ⇔ ³−  = 0 ⇔

(²− 1) = 0 ⇔ ( + 1)( − 1) = 0 ⇔

 = 0or  = ±1.

 = 2

1 0

( − ³)  [by symmetry]

= 21

2²−¹4⁴1 0= 21

2−¹4

=¹₂

SECTION 6.1 AREAS BETWEEN CURVES ¤ 557 23. The curves intersect when√³

2 =¹₈² ⇔ 2 = 1

(2³)³⁶ ⇔ 2¹⁰ = ⁶ ⇔ ⁶− 2¹⁰ = 0 ⇔

(⁵− 2¹⁰) = 0 ⇔  = 0 or ⁵= 2¹⁰ ⇔  = 0 or  = 2²= 4, so for 0 ≤  ≤ 6,

 =

 4 0

√₃

2 −¹8²

 +

 6 4

1 8²−√³

2

 =

3 4

√3

2 ^43−24¹³4 0+

1 24³−³4

√3

2 ^436 4

=3 4

√3

2 · 4√³ 4 −⁶⁴24

− (0 − 0) +216 24 −³4

√3

2 · 6√³ 6

−64 24−³4

√3

2 · 4√³ 4

= 6 −⁸3 + 9 −⁹2

√3

12 −⁸3+ 6 = ⁴⁷₃ −⁹2

√3

12

24. Notice that cos  = sin 2 = 2 sin  cos  ⇔ 2 sin  cos  − cos  = 0 ⇔ cos  (2 sin  − 1) = 0 ⇔ 2 sin  = 1or cos  = 0 ⇔  =^6 or^₂.

 =

 6

0 (cos  − sin 2)  +

2

6 (sin 2 − cos ) 

=

sin  +¹₂cos 26

0 +

−¹2cos 2 − sin 2

6

= ¹₂+¹₂·¹2−

0 +¹₂· 1 +1

2− 1

−

−¹2·¹2−¹2

= ¹₂

25. By inspection, we see that the curves intersect at  = ±1 and that the area of the region enclosed by the curves is twice the area enclosed in the ﬁrst quadrant.

 = 2

 1

0 [(2 − ) − ⁴]  = 2

2 −¹2²−¹5⁵1 0

= 2

2 −¹2−¹5

− 0

= 2₁₃

10

= ¹³₅

560 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 36.  =

 1

−1|3^− 2^|  =

 0

−1

(2^− 3^)  +

 1 0

(3^− 2^) 

=

2^ ln² − 3^

ln³

0

−1

+

 3^ ln 3− 2^

ln 2

1 0

=

 1 ln 2− 1

ln 3



−

 1

2 ln 2− 1 3 ln 3

 +

 3 ln 3− 2

ln 2



−

 1 ln 3− 1

ln 2



=2 − 1 − 4 + 2

2 ln 2 + −3 + 1 + 9 − 3

3 ln 3 = 4

3 ln 3− 1 2 ln 2

37. From the graph, we see that the curves intersect at  = 0 and  =  ≈ 0896, with

 sin(²)  ⁴on (0 ). So the area  of the region bounded by the curves is

 =

  0

 sin(²) − ⁴

 =

−¹2cos(²) −¹5⁵^

0

= −¹2cos(²) −¹5⁵+¹₂ ≈ 0037

38. From the graph, we see that the curves intersect at  =  ≈ −132 and

 =  ≈ 054, with 2 − ² ^on ( ). So the area  of the region bounded by the curves is

 =





(2 − ²) − ^

 =

2 −¹3³− ^^

≈ 145

39. From the graph, we see that the curves intersect at

 =  ≈ −111  =  ≈ 125 and  =  ≈ 286 with

³− 3 + 4  3²− 2 on ( ) and 3²− 2  ³− 3 + 4 on ( ). So the area of the region bounded by the curves is

 =

 



(³− 3 + 4) − (3²− 2)

 +

 



(3²− 2) − (³− 3 + 4)



=

 



(³− 3²−  + 4)  +

 

 (−³+ 3²+  − 4) 

=₁

4⁴− ³−¹2²+ 4

+

−¹4⁴+ ³+¹₂²− 4

≈ 838

40. From the graph, we see that the curves intersect at  =  ≈ 029 and

 =  ≈ 608.  = 2√

is the upper curve, so the area of the region bounded by the curves is

 ≈

 



 2√

 − 13^

 =



4

3^32− 1 ln 1313^





≈ 511

55. To graph this function, we must ﬁrst express it as a combination of explicit functions of ; namely,  = ±√ + 3. We can see from the graph that the loop extends from  = −3 to  = 0, and that by symmetry, the area we seek is just twice the area under the top half of the curve on this interval, the equation of the top half being  = −√ + 3. So the area is  = 20

−3

−√

 + 3. We substitute  =  + 3, so  =  and the limits change to 0 and 3, and we get

 = −23

0 [( − 3)√

 ]  = −23

0(^32− 3^12) 

= −2

2

5^52− 2^323 0= −2₂

5

3²√ 3

− 2 3√

3

= ²⁴₅ √ 3

56. We start by ﬁnding the equation of the tangent line to  = ²at the point (1 1):

⁰= 2, so the slope of the tangent is 2(1) = 2, and its equation is

 − 1 = 2( − 1), or  = 2 − 1. We would need two integrals to integrate with respect to , but only one to integrate with respect to .

 =1 0

1

2( + 1) −√

 =

1

4²+¹₂ −²3^321 0

=¹₄ +¹₂ −²3 = ₁₂¹

57. By the symmetry of the problem, we consider only the ﬁrst quadrant, where

 = ² ⇒  =

. We are looking for a number  such that

  0

  =

 4



  ⇒ ²3

^32 0=²₃

^324

 ⇒

^32= 4^32− ^32 ⇒ 2^32= 8 ⇒ ^32= 4 ⇒  = 4^23≈ 252.

58. (a) We want to choose  so that

  1

1

² =

 4



1

²  ⇒

−1



 1

=

−1



4



⇒ −1

 + 1 = −1 4+1

 ⇒ 5

4 =2

 ⇒  = 8

5. (b) The area under the curve  = 1²from  = 1 to  = 4 is³₄ [take  = 4 in the ﬁrst integral in part (a)]. Now the line

 = must intersect the curve  = 1√ and not the line  = 4, since the area under the line  = 14²from  = 1 to

 = 4is only₁₆³, which is less than half of³₄. We want to choose  so that the upper area in the diagram is half of the total area under the curve  = 1²from  = 1 to  = 4. This implies that

1



1√ − 1

 =¹₂ ·³4 ⇒ 

2 √ − 1

= ³₈ ⇒ 1 − 2√

 +  = ³₈ ⇒

 − 2√

 +⁵₈ = 0. Letting  =√

, we get ²− 2 +⁵8 = 0 ⇒ 8²− 16 + 5 = 0. Thus,  =¹⁶^±^√²⁵⁶16 ^{− 160} = 1 ±^√4⁶. But  =√

  1 ⇒

 = 1 −^√4⁶ ⇒  = ² = 1 +³₈−^√2⁶ =¹₈

11 − 4√ 6

≈ 01503.

1

(2)

55. To graph this function, we must ﬁrst express it as a combination of explicit functions of ; namely,  = ±√ + 3. We can see from the graph that the loop extends from  = −3 to  = 0, and that by symmetry, the area we seek is just twice the area under the top half of the curve on this interval, the equation of the top half being  = −√ + 3. So the area is  = 20

−3

−√

 + 3. We substitute  =  + 3, so  =  and the limits change to 0 and 3, and we get

 = −23

0 [( − 3)√

 ]  = −23

0(^32− 3^12) 

= −2

2

5^52− 2^323 0= −22

5

3²√ 3

− 2 3√

3

= ²⁴₅ √ 3

56. We start by ﬁnding the equation of the tangent line to  = ²at the point (1 1):

⁰= 2, so the slope of the tangent is 2(1) = 2, and its equation is

 − 1 = 2( − 1), or  = 2 − 1. We would need two integrals to integrate with respect to , but only one to integrate with respect to .

 =1 0

₁

2( + 1) −√

 =

1

4²+¹₂ −²3^321 0

=¹₄ +¹₂ −²3 = ₁₂¹

57. By the symmetry of the problem, we consider only the ﬁrst quadrant, where

 = ² ⇒  =

. We are looking for a number  such that

  0

  =

 4



  ⇒ ²3

^32 0=²₃

^324

 ⇒

^32= 4^32− ^32 ⇒ 2^32= 8 ⇒ ^32= 4 ⇒  = 4^23≈ 252.

58. (a) We want to choose  so that

  1

1

² =

 4



1

²  ⇒

−1



 1

=

−1



4



⇒ −1

 + 1 = −1 4+1

 ⇒ 5

4 =2

 ⇒  = 8

5. (b) The area under the curve  = 1²from  = 1 to  = 4 is³₄ [take  = 4 in the ﬁrst integral in part (a)]. Now the line

 = must intersect the curve  = 1√ and not the line  = 4, since the area under the line  = 14²from  = 1 to

 = 4is only₁₆³, which is less than half of³₄. We want to choose  so that the upper area in the diagram is half of the total area under the curve  = 1²from  = 1 to  = 4. This implies that

1



1√ − 1

 =¹₂ ·³4 ⇒ 

2 √ − 1

= ³₈ ⇒ 1 − 2√

 +  = ³₈ ⇒

 − 2√

 +⁵₈ = 0. Letting  =√

, we get ²− 2 +⁵8 = 0 ⇒ 8²− 16 + 5 = 0. Thus,  =¹⁶^±^√²⁵⁶16 ^{− 160} = 1 ±^√4⁶. But  =√

  1 ⇒

 = 1 −^√4⁶ ⇒  = ² = 1 +³₈−^√2⁶ =¹₈

11 − 4√ 6

≈ 01503.

SECTION 6.1 AREAS BETWEEN CURVES ¤ 565

59. We ﬁrst assume that   0, since  can be replaced by − in both equations without changing the graphs, and if  = 0 the curves do not enclose a region. We see from the graph that the enclosed area  lies between  = − and  = , and by symmetry, it is equal to four times the area in the ﬁrst quadrant. The enclosed area is

 = 4

0 (²− ²)  = 4

² −¹3³ 0= 4

³−¹3³

= 42 3³

=⁸₃³ So  = 576 ⇔ ⁸3³ = 576 ⇔ ³= 216 ⇔  =√³

216 = 6.

Note that  = −6 is another solution, since the graphs are the same.

60. It appears from the diagram that the curves  = cos  and  = cos( − ) intersect halfway between 0 and , namely, when  = 2. We can verify that this is indeed true by noting that cos(2 − ) = cos(−2) = cos(2). The point where cos( − ) crosses the -axis is  =^2 + . So we require that

2

0 [cos  − cos( − )]  = −

2+cos( − )  [the negative sign on the RHS is needed since the second area is beneath the -axis] ⇔ [sin  − sin ( − )]^20 = − [sin ( − )]^2+ ⇒ [sin(2) − sin(−2)] − [− sin(−)] = − sin( − ) + sin

2 + 

− 

⇔ 2 sin(2) − sin  = − sin  + 1.

[Here we have used the oddness of the sine function, and the fact that sin( − ) = sin ]. So 2 sin(2) = 1 ⇔ sin(2) =¹₂ ⇔ 2 =^₆ ⇔  = ^3.

61. The curve and the line will determine a region when they intersect at two or more points. So we solve the equation (²+ 1) =  ⇒

 = (²+ ) ⇒ (²+ ) −  = 0 ⇒

(²+  − 1) = 0 ⇒  = 0 or ²+  − 1 = 0 ⇒

 = 0or ²=1 − 

 ⇒  = 0 or  = ±

1

− 1. Note that if  = 1, this has only the solution  = 0, and no region is determined. But if 1 − 1  0 ⇔ 1  1 ⇔ 0    1, then there are two solutions. [Another way of seeing this is to observe that the slope of the tangent to  = (²+ 1)at the origin is ⁰(0) = 1and therefore we must have 0    1.] Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin. Since  and (²+ 1)are both odd functions, the total area is twice the area between the curves on the interval

 0

1 − 1

. So the total area enclosed is

2

 √1−1 0

 

²+ 1− 



 = 2₁

2ln(²+ 1) −¹2²√1−1

0 = [ln(1 − 1 + 1) − (1 − 1)] − (ln 1 − 0)

= ln(1) − 1 +  =  − ln  − 1