Higher Engineering Mathematics, Sixth Edition

376 Higher Engineering Mathematics

Problem 2. Determine the area enclosed between

the curvesy=x^2 +1andy= 7 −x.

At the points of intersection the curves are equal. Thus,

equating theyvalues of each curve gives:

x^2 + 1 = 7 −x

from which, x^2 +x− 6 = 0

Factorizing gives (x− 2 )(x+ 3 )= 0

from whichx=2andx=− 3

By firstly determining the points of intersection the

range ofx-values has been found. Tables of values are

produced as shown below.

x − 3 − 2 − 1 0 1 2

y=x^2 + 1 10 5 2 1 2 5

x − 3 0 2

y= 7 −x 10 7 5

A sketch of the two curves is shown in Fig. 38.3.

y

10

23 22 2 10 1 2 x

5

y 572 x

y 5 x^211

Figure 38.3

Shaded area=

∫ 2

− 3

( 7 −x)dx−

∫ 2

− 3

(x^2 + 1 )dx

=

∫ 2

− 3

[( 7 −x)−(x^2 + 1 )]dx

=

∫ 2

− 3

( 6 −x−x^2 )dx

=

[

6 x−

x^2

2

−

x^3

3

] 2

− 3

=

(

12 − 2 −

8

3

)

−

(

− 18 −

9

2

+ 9

)

=

(

7

1

3

)

−

(

− 13

1

2

)

= 20

5

6

square units

Problem 3. Determine by integration the area

bounded by the three straight linesy= 4 −x,

y= 3 xand 3y=x.

Each of the straight lines are shown sketched in

Fig. 38.4.

1234 x

y 53 x

y 542 x

y

4

2

0

3 y 5 x(or y 5 x 3 )

Figure 38.4

Shaded area

=

∫ 1

0

(

3 x−

x

3

)

dx+

∫ 3

1

[

( 4 −x)−

x

3

]

dx

=

[

3 x^2

2

−

x^2

6

] 1

0

+

[

4 x−

x^2

2

−

x^2

6

] 3

1

=

[(

3

2

−

1

6

)

−( 0 )

]

+

[(

12 −

9

2

−

9

6

)

−

(

4 −

1

2

−

1

6

)]

=

(

1

1

3

)

+

(

6 − 3

1

3

)

=4 square units

Now try the following exercise

Exercise 147 Further problems on areas

under and between curves

1. Find the area enclosed by the curve
   y=4cos3x,thex-axis and ordinatesx= 0
   andx=

π

6

[1^13 square units]