WEEK 7 Area Under the Curve

APPLICATIONS OF INTEGRATION

UNIT 4 : APPLICATIONS OF INTEGRATION

INTRODUCTION

Integration plays an important role in the study of physical situations. It can be used
to find areas under the curves which lead to various interpretations. For example,
the graph of power against time represents the total energy used in a particular time
period. Integration can also be used to find volumes, surface areas and arc length
which has various interpretation.
In this unit, we will use our knowledge of integration to determine areas under the
curves and volumes of revolution of solids. This will follow with solving applied
problems on motion of a point.

LEARNING OUTCOMES

After completing this unit, students should be able to:
1. determine area bounded by a curve by using definite integral.
2. determine volume of revolution of solid by using definite integral.
3. determine displacement, velocity and acceleration by using differentiation and

integration.

1 DUM 20032

APPLICATIONS OF INTEGRATION

4.1 AREA BOUNDED BY A CURVE

4.1.1 AREA UNDER THE CURVE AND THE X-AXIS

Figure 4.1 shows AB which is part the curve y = f(x) between x = a
and x = b.
Take a narrow stripe in the shaded region in the figure on the right
which has a width of . The shape of the stripe becomes a rectangle
if is small.

Area of the strip = Area of rectangle = length x width

y
y = f(x)

y x
a
Figure 4.1 b

When we add up the areas of all the strips from x = a to x = b , we get
Area =

y

y = f(x) Area bounded by the curve y = f(x), x-axis,

x = a and x = b is given by

A x
a
b

2 DUM 20032

APPLICATIONS OF INTEGRATION

Example 4.1 :
Find the area of the region bounded by the curve :
a) y = x2 , x–axis, x = 2 and x = 5.
y y = x2

A

0 25 x

3 DUM 20032

APPLICATIONS OF INTEGRATION

b) y = x2 – 4x + 3 and the x-axis
y

y = x2 – 4x + 3

01 A3

4 DUM 20032

APPLICATIONS OF INTEGRATION

c) y = x(x - 1)(x + 2) and the x-axis.

5 DUM 20032

APPLICATIONS OF INTEGRATION Area is measured in square unit.

Solution :
a) Area =
=
=
=
= 39 unit2

b) Area = Area cannot be negative. The
= negative value implies area is below
= x-axis. So, modulus the answer to
ignore negative sign.
=
=
=
= unit2

c) Area,A = P + Q +
=
=
=
=

=

=

=+ Ignore negative sign as it
= only implies that area is
= unit2 below x-axis

6 DUM 20032

APPLICATIONS OF INTEGRATION

UNIT EXERCISE 4.1.1

1. yy == 44xx −− xx 2
y 2

x
03

Find the area of shaded region bounded by the curve, y, x-axis, x = 0
and x = 3 .

2.
y

y = 5x − x2

x
03

Find the area of shaded region bounded by the curve, y, x-axis, x = 0
and x = 3 .

7 DUM 20032

APPLICATIONS OF INTEGRATION

3. y

y = x2 - 2x - 3

-1 3 x

Find the area of shaded region bounded by the curve, y, x-axis, x = -1
and x = 3 .

4.

Find the area of shaded region bounded by the curve, y, x-axis, x = 2
and x = 4 .

5.
y

y = x2 – 5x + 4

1 4x

8 DUM 20032

APPLICATIONS OF INTEGRATION

Find the area of shaded region bounded by the curve, y, x-axis, x = 1
and x = 4.

y y = x2 – 3x – 10

6. x
6
-4

Find the area of shaded region bounded by the curve, y, x-axis, x = -4
and x = 6 .

7. Find the area of the region bounded by the curve,
(a) y = 2x, the x-axis and the lines x = 4 and x = 8.

(b) y = x3 , the x-axis and the line x = 2.

9 DUM 20032

APPLICATIONS OF INTEGRATION

(c) y = x2, x – axis , x = –3 and x = 3.

4.1.2 AREA UNDER THE CURVE AND THE Y-AXIS

To find the area bounded by the curve x =f(y), the y-axis and the

lines y = c and y = d divide the area into small stripes parallel to the

axis each of width . Take a typical stripe distance y from the axis.

Area of stripe (shaded) .

Hence, the required area is .

y
x = f(y)

d

c
x

Area bounded by the curve x = f(y), x-axis,
y = c and y = d is given by

10 DUM 20032

APPLICATIONS OF INTEGRATION

Example 4.2 :

a) Find the area of the region bounded by the curve x=y2, y-axis ,
y = 0 and y = 5.

y x = y2
x
5

A
5
0

b) Find the area bounded by the curve y2 = 4x,y-axis and y=4.
y

4 y2 = 4x
2
0x

11 DUM 20032

APPLICATIONS OF INTEGRATION

UNIT EXERCISE 4.1.2

1. y
9 y = x2
x

Find the area of the region bounded by the curve y = x2, y-axis , y = 0
and y = 9.

2.

Find the area of the region bounded by the curve y2 = 3x, y-axis ,
y = 2 and y = 5.

12 DUM 20032

APPLICATIONS OF INTEGRATION

3. y
x = y2 - 4
x

Find the area of the region bounded by the curve x = y2 – 4 and
y-axis.

4. y
8 y = x3
x

Find the area of the region bounded by the curve y = x3, y-axis and
y = 8.

4.1.3 AREA UNDER THE CURVE AND THE STRAIGHT LINE

y y2 = g(x) Area ,
A =

y1 = f(x)

a bx

13 DUM 20032

APPLICATIONS OF INTEGRATION

Example 4.3 :
Find the area enclosed by the line y = x + 1 and the curve
y = x2 – 2x + 1.
y
y=x+1
y = x2 – 2x + 1
x
From the graph intersection point :
x2 – 2x + 1 = x + 1
x2 – 3x =0
x = 0 or x = 3
A=
=
=
=
=
= unit2

14 DUM 20032

APPLICATIONS OF INTEGRATION

4.1.4 AREA UNDER THE CURVE, STRAIGHT LINE AND THE AXIS

Example 4.4 : y
a)

0 y = 2x

y=

x
t5

The diagram shows part of the curve and the straight line

y = 2x which intersect at point P.
i. find the value of t.
ii. the area of the shaded region.

Solution : y = 2x ……………..
a) i. …………. 

Substitute  into 

ii. Area =
=+
=
=
==

15 DUM 20032

APPLICATIONS OF INTEGRATION

b) y = x2 − 2 and y = 2

Intersection points :

y = x2 − 2 (i)

y=2 (ii)

(i) − (ii) : x2 − 2 = 2
x2 − 4 = 0

x = 2

x = 2 ; x = −2

2

Area, A =  4 − x2 dx
−2

= 4x − x3 2
3

−2

=  − 23  − 4(−2) − (−2)3 
4(2) 3   
  3 

= 32 unit 2
3

16 DUM 20032

APPLICATIONS OF INTEGRATION

UNIT EXERCISE 4.1

1. Find the area of the region bounded by the curve y = x2, x–axis , x = -3
and x = 3 in diagram below.
y

-3 3 x

2. Find the areas of the regions bounded by the curves,
(a) x = y2, the y-axis and the lines y = 2 and y = 5.
(b) , the y-axis , the lines y = 0 and y = 1.
(c) , the y-axis , y = 0 and y = 8.
(d) , the y-axis , the lines y = 2 and y = 4.

3. The diagram below shows the curve y = -x2 + 5x and the straight line
y = 6. Find the area of the curve y = -x2 + 5x and the straight line y = 6.
y
y=6
y = -x2 + 5x
x
23

17 DUM 20032

APPLICATIONS OF INTEGRATION

5.

The graph above shows the curve y = 16 – x2 intersecting the x-axis
and y-axis at points Q and P respectively .
(a) Find the coordinates of points P and Q.
(b) Find the area of the shaded region.

6. y
y = x2 + 2

y = -x + 2

2x

Find the area of the region bounded by the curve
y = x2 + 2, y = − x + 2 and x = 2 .

7. y

y = 4x – x2
y = 2x

x DUM 20032

18

APPLICATIONS OF INTEGRATION

Find the area of the region bounded by the curve y = 4x – x2 and y = 2x.

8. y y = 4
x2

y=x-1

(2,1)

x

Find the area of the region bounded by the curve y= 4 and y = x -1.
x2

9.

Find the area of the region bounded by the curve y = 4 ,y = 2x, x = 3 and
x2

x-axis.

19 DUM 20032