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Area under and between Curves Worksheet

Area under and between Curves Worksheet
  • Page 1
 1.  
Find the area of the region bounded by y = x2 + 2x + 3, x-axis and the lines x = - 1, x = 1.
a.
7 3
b.
4
c.
2
d.
20 3
e.
13 3


Solution:


The region R bounded by y = f(x) = x2 + 2x + 3, x - axis and the lines x = - 1, x = 1 is the shaded region in the figure.

The area of the region R = -11 f(x) dx
[Definition.]

= -11 (x2 + 2x + 3) dx
[Substitute f(x) = x2 + 2x + 3.]

= [x33+x2+3x]-11

= [1 / 3 + 1 + 3] - [- 1 / 3 + 1 - 3]

= 13 / 3 + 7 / 3 = 20 / 3

So, the area of the region = 20 / 3 sq. units.


Correct answer : (4)
 2.  
Find the area of the region bounded by y = x2, x- axis and the lines x = 0, x = 2.
a.
4 3
b.
2
c.
4
d.
8
e.
8 3


Solution:


The region R bounded by f(x) = x2, x - axis and the lines x = 0, x = 2, is the shaded region in the figure.

The area of the region R = 02 f(x) dx

= 02x2 dx
[Substitute f(x) = x2.]

= [x33]02

= 8 / 3 - 0 = 8 / 3

So, the area of the region = 8 / 3 sq units.


Correct answer : (5)
 3.  
Find the area of the region bounded by the curves y = x3 and y = x.
a.
3 2 sq. units
b.
1 sq. unit
c.
1 2 sq. units
d.
4 sq. units
e.
2 sq. units


Solution:

On solving the two curves for points of intersection, we have x3 = x

x (x2 - 1) = 0

x (x + 1) (x - 1) = 0

x = - 1, 0, 1
[Solve for x.]

At x = - 1, y = - 1, at x = 0, y = 0 and at x = 1, y = 1

The points of intersection are (0, 0), (- 1, - 1) and (1, 1).


The region R bounded by the curves is the shaded region in the figure.

Area of the region = - -10 (x - x3) dx + 01x - x3 dx

= [x44-x22 ]-10 + [x22-x44 ]01

= - 1 / 4 + 1 / 2 + 1 / 2 - 1 / 4

= 1 - 1 / 2 = 1 / 2

So, the area of the region is 1 / 2 sq. units.


Correct answer : (3)
 4.  
Find the area of the region bounded by the curves y = x2 + 2 and y = x + 4.
a.
9 2 sq. units
b.
7 6 sq. units
c.
10 3 sq. units
d.
6 sq. units
e.
27 sq. units


Solution:

On solving the two curves for points of intersection, we have x2 + 2 = x + 4

x2 - x - 2 = 0

(x - 2) (x + 1) = 0
[Factor.]

x = 2, - 1
[Solve for x.]

At x = 2, y = 6 and at x = - 1, y = 3.

The points of intersection are (2, 6) and (- 1, 3).


The region R bounded by the curves is the shaded region in the figure.

Area of the region = -12 [(x + 4) - (x2 + 2)] dx
[x + 4 > x2 + 2 on (-1, 2).]

= -12 (- x2 + x + 2) dx

= [- x33+x22 + 2x ]-12

= [- 8 / 3 + 4 / 2 + 4] - [1 / 3 + 1 / 2 - 2]

= [10 / 3] - [- 7 / 6]

= 20+76 = 27 / 6 = 9 / 2

So, the area of the region is 9 / 2 sq. units.


Correct answer : (1)
 5.  
Find the area of the region bounded by the curves y = x2 - 5x and y = 4 - 2x.
a.
16 3sq. units
b.
125 6sq. units
c.
56 3sq. units
d.
64 3sq. units
e.
13 6sq. units


Solution:

On solving the two curves for the points of intersection we have,

x2 - 5x = 4 - 2x
[Equate y.]

x2 - 3x - 4 = 0

(x - 4) (x + 1) = 0

x = - 1, 4
[Solve for x.]

At x = - 1, y = 6 and at x = 4, y = - 4.

The points of intersection are (- 1, 6) and (4, - 4).


The region R bounded by the curves is the shaded region in the figure.

The area of the region = -14 [(4 - 2x) - (x2 - 5x)] dx
[Since (4 - 2x) ≥ (x2 - 5x) in [- 1, 4].]

= -14 (4 + 3x - x2) dx

= [4x + 3x22-x33 ]-14

= [ 16 + 24 - 64 / 3] - [- 4 + 3 / 2 + 1 / 3]

= [40 - 64 / 3] - [- 13 / 6]

= 56 / 3 + 13 / 6 = 125 / 6

So, the area of the region = 125 / 6 sq. units.


Correct answer : (2)
 6.  
Find the area of the region bounded by the curves y = x2 and y = x4.
a.
1 15sq. units
b.
16 15sq. units
c.
2 3sq. units
d.
4 15sq. units
e.
2 5sq. units


Solution:

On solving the two curves for the point of intersection we have,

x2 = x4
[Equate y.]

x2 (x2 - 1) = 0

x2 (x + 1) (x - 1) = 0

x = 0, 1, -1

At x = 0, y = 0, at x = 1, y = 1 and at x = - 1, y = 1.

The points of intersection are (0, 0), (1, 1) and (-1, 1).


The region R bounded by the curves is the shaded region in the figure.

The area of the region = -11 (x2 - x4) dx
[Since x2 > x4 in [-1, 1].]

= [x33-x55 ]-11

= [1 / 3 - 1 / 5] - [- 1 / 3 + 1 / 5]

= 2 / 3 - 2 / 5 = 4 / 15

So, the area of the given region = 4 / 15 sq units.


Correct answer : (4)
 7.  
Find the area of the region bounded by the curves y = 1 + 4x - x2 and y = 1 + x2.
a.
8 3Sq.units
b.
8 Sq.units
c.
3 8Sq.units
d.
3 Sq.units
e.
16 3Sq.units


Solution:

On solving the two curves for the points of intersection we have,

1 + 4x - x2 = 1 + x2
[Equate y.]

4x - 2x2 = 0

2x (2 - x) = 0

x = 0, 2

At x = 0, y = 1 and at x = 2, y = 5.

The points of intersection are (0, 1) and (2, 5).


The region R bounded by the curves is the shaded region in the figure.

The area of the region = 02(1 + 4x - x2) - (1 + x2) dx
[Since (1 + 4x - x2) ≥ (1 + x2) in [0, 2].]

= 02(4x - 2x2) dx

= [4x22-2x33 ]02

= 8 - 16 / 3 = 8 / 3

So, the area of the region = 8 / 3 sq units.


Correct answer : (1)
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