Now Work Problem PROBLEMS 14.9 In Problems 1–24, use a definite integral to find the area of the region bounded by the given curve, the x-axis, and the given lines. In each case, first sketch the region. Watch out for areas of regions that are below the x-axis. 2. y = x+ 5, x= 2, x= 4 3. y = 3x, x = 1, 4. y = x, x = 2, x = 3 5. y = x+x² +x, %3D %3D 6. y = x² – 2x, x = -3, x = –1 1. y = 5x + 2, x= 1, X = 4 7. y = 3x - 4x, x = -2, x = -1
Now Work Problem PROBLEMS 14.9 In Problems 1–24, use a definite integral to find the area of the region bounded by the given curve, the x-axis, and the given lines. In each case, first sketch the region. Watch out for areas of regions that are below the x-axis. 2. y = x+ 5, x= 2, x= 4 3. y = 3x, x = 1, 4. y = x, x = 2, x = 3 5. y = x+x² +x, %3D %3D 6. y = x² – 2x, x = -3, x = –1 1. y = 5x + 2, x= 1, X = 4 7. y = 3x - 4x, x = -2, x = -1
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
1 and 3
![(4, 2)
grap
X-y = 2
dy
27.
dx
func
(1, –1)
-x-y = 2
(а)
(b)
(c)
(1, –1)
(a)
(b)
FIGURE 14.36 Region of Example 7 with vertical and horizontal strips.
28.
Perhaps the use of horizontal strips can simplify our work. In Figure 14.36(b), the width
of the strip is Ay. The rightmost curve is always x – y = 2 (or x = y + 2), and the
leftmost curve is always y
is [(y + 2) – y²]Ay, so the total area is
func
(а)
= x (or x = y²). Therefore, the area of the horizontal strin
(b)
(c)
%3D
(d)
| v +2 - y) dy =
Clearly, the use of horizontal strips is the most desirable approach to solving the prob-
area =
9.
29.
%3D
fund
lem. Only a single integral is needed, and it is much simpler to compute.
(a)
(b)
(c)
(d)
PROBLEMS 14.9
Now Work Problem 57 4
30.
In Problems 1–24, use a definite integral to find the area of the
region bounded by the given curve, the x-axis, and the given lines.
In each case, first sketch the region. Watch out for areas of regions
2. y =x+5, x = 2, x = 4
%3D
3. y = 3x, x= 1, x= 3
that are below the x-axis.
4. y = x, x = 2, x = 3
%3D
1. y = 5x + 2, x= 1,
6. y = x² – 2x, x=-1
5. y =x+x² +x², x= 1
(b)
%3D
x = -3,
%3D
cert
%3D
7. y = 3x2 – 4x, x = -1
x = -2,
(c)
LEGO](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0e2d6f28-e3d7-414c-b1d4-a752131175c2%2F9794d706-5c16-4af5-b325-de07d231a3af%2Frjx4dku.jpeg&w=3840&q=75)
Transcribed Image Text:(4, 2)
grap
X-y = 2
dy
27.
dx
func
(1, –1)
-x-y = 2
(а)
(b)
(c)
(1, –1)
(a)
(b)
FIGURE 14.36 Region of Example 7 with vertical and horizontal strips.
28.
Perhaps the use of horizontal strips can simplify our work. In Figure 14.36(b), the width
of the strip is Ay. The rightmost curve is always x – y = 2 (or x = y + 2), and the
leftmost curve is always y
is [(y + 2) – y²]Ay, so the total area is
func
(а)
= x (or x = y²). Therefore, the area of the horizontal strin
(b)
(c)
%3D
(d)
| v +2 - y) dy =
Clearly, the use of horizontal strips is the most desirable approach to solving the prob-
area =
9.
29.
%3D
fund
lem. Only a single integral is needed, and it is much simpler to compute.
(a)
(b)
(c)
(d)
PROBLEMS 14.9
Now Work Problem 57 4
30.
In Problems 1–24, use a definite integral to find the area of the
region bounded by the given curve, the x-axis, and the given lines.
In each case, first sketch the region. Watch out for areas of regions
2. y =x+5, x = 2, x = 4
%3D
3. y = 3x, x= 1, x= 3
that are below the x-axis.
4. y = x, x = 2, x = 3
%3D
1. y = 5x + 2, x= 1,
6. y = x² – 2x, x=-1
5. y =x+x² +x², x= 1
(b)
%3D
x = -3,
%3D
cert
%3D
7. y = 3x2 – 4x, x = -1
x = -2,
(c)
LEGO
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