Section 6.1 Areas Between Curves
18. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.
4x + y2= 12, x = y Solution:
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) =
32−1336
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 ≈ 019
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 = 22 ⇔ 2= 1 ⇔ = ±1.
=
1
−1
(1 − 2) − (2− 1)
=
1
−12(1 − 2)
= 2 · 2
1
0 (1 − 2)
= 4
− 1331 0= 4
1 −13
= 83
12. 4 + 2= 12 ⇔ ( + 6)( − 2) = 0 ⇔
= −6 or = 2, so = −6 or = 2 and
=
2
−6
−142+ 3
−
=
−1213−122+ 32
−6
=
−23 − 2 + 6
− (18 − 18 − 18)
= 22 −23 = 643
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31. Sketch the region enclosed by the given curves and find its area.
y = x4, y = 2 − |x|
Solution:
SECTION 6.1 AREAS BETWEEN CURVES ¤ 557 23. The curves intersect when √3
2 = 182 ⇔ 2 = 1
(23)36 ⇔ 210 = 6 ⇔ 6− 210 = 0 ⇔
(5− 210) = 0 ⇔ = 0 or 5= 210 ⇔ = 0 or = 22= 4, so for 0 ≤ ≤ 6,
=
4 0
√3
2 −182
+
6 4
1
82−√3 2
=
3 4
√3
2 43− 24134 0+
1 243−34
√3
2 436 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 26
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 −122− 1551
0
= 2
2 − 12−15
− 0
= 213 10
= 135
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42. Use calculus to find the area of the triangle with the given vertices. (2, 0), (0, 2), (−1, 1).
Solution:
SECTION 6.1 AREAS BETWEEN CURVES ¤ 559
31.
1 + 2 = 2
1 + 3 ⇔ + 4 = 2+ 4 ⇔ = 2 ⇔ 0 = 2− ⇔ 0 = ( − 1) ⇔ = 0 or = 1.
=
1 0
1 + 2 − 2 1 + 3
=1
2ln(1 + 2) −13ln(1 + 3)1 0
=1
2ln 2 − 13ln 2
− (0 − 0) = 16ln 2
32. ln
= (ln )2
⇔ ln = (ln )2 ⇔ 0 = (ln )2− ln ⇔ 0 = ln (ln − 1) ⇔ ln = 0 or 1 ⇔ = 0or 1 [1 or ]
=
1
ln
− (ln )2
=
1
2(ln )2− 13(ln )3 1
=1 2−13
− (0 − 0) = 16
33. =
2 0
−45 + 5
−
−72 + 5
+
5 2
−45 + 5
− ( − 4)
=
2 0
27 10 +
5 2
−95 + 9
=27 2022
0+
−1092+ 95 2
=27 5 − 0
+
−452 + 45
−
−185 + 18
= 272
34. An equation of the line through (2 0) and (0 2) is = − + 2; through (2 0) and (−1 1) is = −13 +23; through (0 2) and (−1 1) is = + 2.
=
0
−1
( + 2) −
−13 + 23
+
2 0
(− + 2) −
−13 +23
=
0
−1
4 3 +43
+
2 0
−23 +43
=2
32+430
−1+
−132+432 0
= 0 −2
3−43
+
−43+83
− 0 = 2 35. The curves intersect when√
+ 2 = ⇒ + 2 = 2 ⇒ 2− − 2 = 0 ⇒ ( − 2)( + 1) = 0 ⇒
= −1 or 2. [−1 is extraneous]
=
4 0
√ + 2 − = 2 0
√ + 2 −
+
4 2
−√
+ 2
=
2
3( + 2)32−1222 0+
1
22− 23( + 2)324 2
=16
3 − 2
−2
3
2√ 2
− 0 +
8 −23
6√ 6
− 2 −163
= 4 +323 − 43
√2 − 4√
6 = 443 − 4√ 6 − 43
√2
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1
43. Evaluate the integral and interpret it as the area of a region. Sketch the region.
Z π2
0
| sin x − cos 2x|dx
Solution:
SECTION 6.1 AREAS BETWEEN CURVES ¤ 601 40. ln
= (ln )2
⇔ ln = (ln )2 ⇔ 0 = (ln )2− ln ⇔ 0 = ln (ln − 1) ⇔ ln = 0 or 1 ⇔ = 0or 1 [1 or ]
=
1
ln
−(ln )2
=1
2(ln )2−13(ln )3 1
=1 2 −13
− (0 − 0) = 16
41. An equation of the line through (0 0) and (3 1) is = 13; through (0 0) and (1 2) is = 2;
through (3 1) and (1 2) is = −12 +52.
=
1 0
2 −13
+
3 1
−12 +52
−13
=
1 0
5 3 +
3 1
−56 + 52
=5 621
0+
−1252+ 523
1
= 56 +
−154 +152
−
−125 +52
= 52
42. An equation of the line through (2 0) and (0 2) is = − + 2; through (2 0) and (−1 1) is = −13 +23; through (0 2) and (−1 1) is = + 2.
=
0
−1
( + 2) −
−13 +23
+
2 0
(− + 2) −
−13 +23
=
0
−1
4
3 + 43
+
2 0
−23 +43
=2
32+ 430
−1+
−132+ 432 0
= 0 −2 3 −43
+
−43 +83
− 0 = 2
43. The curves intersect when sin = cos 2 (on [0 2]) ⇔ sin = 1 − 2 sin2 ⇔ 2 sin2 + sin − 1 = 0 ⇔ (2 sin − 1)(sin + 1) = 0 ⇒ sin = 12 ⇒ = 6.
=
2 0
|sin − cos 2|
=
6
0 (cos 2 − sin ) +
2
6 (sin − cos 2)
=1
2sin 2 + cos 6
0 +
− cos −12sin 22
6
=1 4
√3 + 12√ 3
− (0 + 1) + (0 − 0) −
−12
√3 −14
√3
= 32√ 3 − 1
44. =
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
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69. The figure shows a horizontal line y = c intersecting the curve y = 8x − 27x3. Find the number c such that the areas of the shaded regions are equal.
64. Find the area of the region bounded by the parabola y = x2,
the tangent line to this parabola at (1,1), and the x-axis.
65. Find the number b such that the line y = b divides the region bounded by the curves y = x2 and y = 4 into two regions with equal area.
66. (a) Find the number a such that the line x = a bisects the area under the curve y = 1/x2, 1 三 x ~ 4.
(b) Find the number b such that the line y = b bisects the area in part (a).
67. Find the values of c such that the area of the region bounded by the parabolas y = x2 - c2 and y = c2 - x2 is 576.
68. Suppose that 0 < c < 7r /2. For what value of c is the area of the region enclosed by the curves y = cos x, y = cos(x - c), and x = 0 equal to the area of the region enclosed by the curves y = cos(x - c)，χ= 宵，andy = O?
A叫 ED PROJECT I THE GINIINDEX
y
(0.8,0.485)
01 0.2 0.4 0.6 0.8 1 x
FIGURE 1
Lorenz curve for the United States in 2016
APPLlED PROJE仁 T The Gini Index 445
69. The figure shows a horizontalline y = c intersecting the curve y = 8x - 27x3. Find the number c such that the areas of the shaded regions are equal.
y y=8x- 27x3
x
70. For what values of m do the line y = mx and the curve y = x/(x2 + 1) enclose a region? Find the area of the reglOn.
How is it possible to measure the distribution of income among the inhabitants of a given country? One such measure is the Gini index, named after the 1talian economist Corrado Gini, who first devised it in 1912.
We first rank al1 households in a country by income and then we compute the percentage of households whose total income is a given percentage of the country's total income. We de自 ne
a Lorenz curve y = L(x) on the interval [0, 1] by plotting the point (α/100， b/IOO) on the curve if the bottom a% of households receive b% of the total income. For instance, in Figure 1 the point (0.4, 0.114) is on the Lorenz curve for the United States in 2016 because the poorest 40% of the population received just 11.4% of the total income. Likewise, the bottom 80% of the population received 48.5% of the total income, so the point (0.8, 0.485) lies on the Lorenz curve. (The Lorenz curve is named after the American economist Max Lorenz.)
Figure 2 shows some typical Lorenz curves. They all pass through the points (0, 0) and (L, 1) and are concave upward. 1n the extreme case L(x) = 几 societyis perfectly egalitarian:
the poorest a% of the population receives α% of the total income and so everybody receives the same income. The area between a Lorenz curve y = L(x) and the line y = x measures how much the income distribution di旺ersfrom absolute equality. The Gini index (sometimes called the Gini coefficient or the coefficient of inequality) is the area between the Lorenz curve and the line y = x (shaded in Figure 3) divided by the area under y = x.
y
lncome fraction
) ' ! ,i
( y
。
Population fraction FIGURE 2
1 x
FIGURE 3
(continued) Solution:
SECTION 6.1 AREAS BETWEEN CURVES ¤ 607
67. 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 −133 0= 4
3−133
= 42 33
= 833 So = 576 ⇔ 833= 576 ⇔ 3= 216 ⇔ =√3216 = 6.
Note that = −6 is another solution, since the graphs are the same.
68. 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.
69. Let and be the coordinates of the points where the line intersects the curve. From the figure, 1= 2 ⇒
0
−
8 − 273
=
8 − 273
−
− 42+ 2744
0=
42−2744−
− 42+2744=
42−2744−
−
42−2744− 0 = 42−2744− = 42−2744−
8 − 273
= 42−2744− 82+ 274= 8144− 42
= 281 42− 4
So for 0, 2= 1681 ⇒ = 49. Thus, = 8 − 273= 84 9
− 2764 729
= 329 −6427 = 3227.
70. 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 ⇒
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