Basic Engineering Mathematics, Fifth Edition

Introduction to integration 333

x 0 1.0 1.5 2.0 2.5 3.0 3.5 4.0

y 4 7 10.75 16 22.75 31 40.75 52

01234 x

4

50

40

30

20

10

y

y 5 3 x^2 1 4

Figure 35.6

Selecting 6 intervals each of width 0.5 gives

area=( 0. 5 )

[

1

2

( 7 + 52 )+ 10. 75 + 16

+ 22. 75 + 31 + 40. 75

]

=75.375 square units

(b) By the mid-ordinate rule

area=(width of interval)(sum of mid-ordinates)

Selecting 6 intervals, each of width 0.5, gives the

mid-ordinates as shown by the broken lines in

Figure 35.6. Thus,

area=( 0. 5 )( 8. 7 + 13. 2 + 19. 2 + 26. 7

+ 35. 7 + 46. 2 )

=74.85 square units

(c) By Simpson’s rule

area=

1

3

(

width of

interval

)[(

first+last

ordinates

)

+ 4

(

sum of even

ordinates

)

+ 2

(

sum of remaining

odd ordinates

)]

Selecting 6 intervals, each of width 0.5, gives

area=

1

3

( 0. 5 )[( 7 + 52 )+ 4 ( 10. 75 + 22. 75

+ 40. 75 )+ 2 ( 16 + 31 )]

=75 square units

(d) By integration

shaded area=

∫ 4

1

ydx

=

∫ 4

1

( 3 x^2 + 4 )dx=

[

x^3 + 4 x

] 4

1

=( 64 + 16 )−( 1 + 4 )

=75 square units

Integration gives the precise value for the area under

a curve. In this case, Simpson’s rule is seen to be the

most accurate of the three approximate methods.

Problem 30. Find the area enclosed by the curve

y=sin2x,thex-axis and the ordinatesx=0and

x=

π

3

Asketchofy=sin2xis shown in Figure 35.7. (Note

thaty=sin2xhas a period of

2 π

2

i.e.,πradians.)

1

0 /3 /2 

y 5 sin 2x

x

y

Figure 35.7