﻿ Area under and between Curves Worksheet | Problems & Solutions

# Area under and between Curves Worksheet

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

#### 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.

2.
Find the area of the region bounded by $y$ = $x$2, $x$- axis and the lines $x$ = 0, $x$ = 2.
 a. $\frac{4}{3}$ b. 2 c. 4 d. 8 e. $\frac{8}{3}$

#### 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.

3.
Find the area of the region bounded by the curves $y$ = $x$3 and $y$ = $x$.
 a. $\frac{3}{2}$ sq. units b. 1 sq. unit c. $\frac{1}{2}$ 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.

4.
Find the area of the region bounded by the curves $y$ = $x$2 + 2 and $y$ = $x$ + 4.
 a. $\frac{9}{2}$ sq. units b. $\frac{7}{6}$ sq. units c. $\frac{10}{3}$ 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.

5.
Find the area of the region bounded by the curves $y$ = $x$2 - 5$x$ and $y$ = 4 - 2$x$.
 a. $\frac{16}{3}$sq. units b. $\frac{125}{6}$sq. units c. $\frac{56}{3}$sq. units d. $\frac{64}{3}$sq. units e. $\frac{13}{6}$sq. 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.

6.
Find the area of the region bounded by the curves $y$ = $x$2 and $y$ = $x$4.
 a. $\frac{1}{15}$sq. units b. $\frac{16}{15}$sq. units c. $\frac{2}{3}$sq. units d. $\frac{4}{15}$sq. units e. $\frac{2}{5}$sq. 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.

7.
Find the area of the region bounded by the curves $y$ = 1 + 4$x$ - $x$2 and $y$ = 1 + $x$2.
 a. $\frac{8}{3}$Sq.units b. 8 Sq.units c. $\frac{3}{8}$Sq.units d. 3 Sq.units e. $\frac{16}{3}$Sq.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.