Introduction to integration 331
The shaded area in Figure 35.1 is given by
shaded area =
b
a
y dx =
b
a
f (x) dx
Thus, determining the area under a curve by integration
merely involves evaluating a definite integral, as shown
in Section 35.4.
There are several instances in engineering and science
where the area beneath a curve needs to be accurately
determined. For example, the areas between the limits
of a
velocity/time graph gives distance travelled,
force/distance graph gives work done,
voltage/current graph gives power, and so on.
Should a curve drop below the x-axis then y(= f (x))
becomes negative and
f (x) dx is negative. When
determining such areas by integration, a negative sign
is placed before the integral. For the curve shown in
Figure 35.2, the total shaded area is given by (area E+
area F + area G).
E
0
F
G
y
a
b
c
d
x
y ϭ f (x)
Figure 35.2
By integration,
total shaded area =
b
a
f (x) dx −
c
b
f (x) dx
+
d
c
f (x) dx
(Note that this is not the same as
d
a
f (x) dx)
It is usually necessary to sketch a curve in order to check
whether it crosses the x-axis.
Problem 26. Determine the area enclosed by
y = 2x + 3, the x-axis and ordinates x = 1 and
x = 4
y = 2x + 3 is a straight line graph as shown in
Figure 35.3, in which the required area is shown shaded.
12
10
8
6
4
2
1
0
2
3
4
5
x
y 5 2x 1 3
y
Figure 35.3
By integration,
shaded area =
4
1
y dx =
4
1
(2x + 3)dx =
2x 2
2
+ 3x
4
1
= [(16 + 12) − (1 + 3)] = 24 square units
(This answer may be checked since the shaded area
is a trapezium: area of trapezium =
1
2
(sum of parallel sides)(perpendicular distance between parallel sides)
=
1
2
(5 + 11)(3) = 24 square units.)
Problem 27. The velocity v of a body t seconds
after a certain instant is given by v =
2t 2 + 5
m/s.
Find by integration how far it moves in the interval
from t = 0 to t = 4 s
Since 2t 2 + 5 is a quadratic expression, the curve
v = 2t 2 + 5 is a parabola cutting the v-axis at v = 5,
as shown in Figure 35.4.
The distance travelled is given by the area under the v/t
curve (shown shaded in Figure 35.4). By integration,
shaded area =
4
0
v dt =
4
0
(2t
2
+ 5) dt =
2t 3
3
+ 5t
4
0
=
2(4 3 )
3
+ 5(4)
− (0)
i.e. distance travelled = 62.67 m
The shaded area in Figure 35.1 is given by
shaded area =
b
a
y dx =
b
a
f (x) dx
Thus, determining the area under a curve by integration
merely involves evaluating a definite integral, as shown
in Section 35.4.
There are several instances in engineering and science
where the area beneath a curve needs to be accurately
determined. For example, the areas between the limits
of a
velocity/time graph gives distance travelled,
force/distance graph gives work done,
voltage/current graph gives power, and so on.
Should a curve drop below the x-axis then y(= f (x))
becomes negative and
f (x) dx is negative. When
determining such areas by integration, a negative sign
is placed before the integral. For the curve shown in
Figure 35.2, the total shaded area is given by (area E+
area F + area G).
E
0
F
G
y
a
b
c
d
x
y ϭ f (x)
Figure 35.2
By integration,
total shaded area =
b
a
f (x) dx −
c
b
f (x) dx
+
d
c
f (x) dx
(Note that this is not the same as
d
a
f (x) dx)
It is usually necessary to sketch a curve in order to check
whether it crosses the x-axis.
Problem 26. Determine the area enclosed by
y = 2x + 3, the x-axis and ordinates x = 1 and
x = 4
y = 2x + 3 is a straight line graph as shown in
Figure 35.3, in which the required area is shown shaded.
12
10
8
6
4
2
1
0
2
3
4
5
x
y 5 2x 1 3
y
Figure 35.3
By integration,
shaded area =
4
1
y dx =
4
1
(2x + 3)dx =
2x 2
2
+ 3x
4
1
= [(16 + 12) − (1 + 3)] = 24 square units
(This answer may be checked since the shaded area
is a trapezium: area of trapezium =
1
2
(sum of parallel sides)(perpendicular distance between parallel sides)
=
1
2
(5 + 11)(3) = 24 square units.)
Problem 27. The velocity v of a body t seconds
after a certain instant is given by v =
2t 2 + 5
m/s.
Find by integration how far it moves in the interval
from t = 0 to t = 4 s
Since 2t 2 + 5 is a quadratic expression, the curve
v = 2t 2 + 5 is a parabola cutting the v-axis at v = 5,
as shown in Figure 35.4.
The distance travelled is given by the area under the v/t
curve (shown shaded in Figure 35.4). By integration,
shaded area =
4
0
v dt =
4
0
(2t
2
+ 5) dt =
2t 3
3
+ 5t
4
0
=
2(4 3 )
3
+ 5(4)
− (0)
i.e. distance travelled = 62.67 m