Basic Engineering Mathematics, Fifth Edition

330 Basic Engineering Mathematics

=

{

−

3

2

(− 1 )

}

−

{

−

3

2

( 1 )

}

=

3

2

+

3

2

= 3

Problem 23. Evaluate

∫ 2

1

4cos3tdt

∫ 2

1

4cos3tdt=

[

( 4 )

(

1

3

)

sin3t

] 2

1

=

[

4

3

sin3t

] 2

1

=

{

4

3

sin6

}

−

{

4

3

sin3

}

Note that limits of trigonometric functions are always

expressed inradians– thus, for example, sin 6 means

the sine of 6 radians=− 0. 279415 ...Hence,

∫ 2

1

4cos3tdt=

{

4

3

(− 0. 279415 ...)

}

−

{

4

3

( 0. 141120 ...)

}

=(− 0. 37255 )−( 0. 18816 )

=−0.5607

Problem 24. Evaluate

∫ 2

1

4 e2xdxcorrect to

4 significant figures

∫ 2

1

4 e^2 xdx=

[

4

2

e^2 x

] 2

1

= 2

[

e^2 x

] 2

1

=2[e^4 −e^2 ]

=2[54. 5982 − 7 .3891]

=94.42

Problem 25. Evaluate

∫ 4

1

3

4 u

ducorrect to 4

significant figures

∫ 4

1

3

4 u

du=

[

3

4

lnu

] 4

1

=

3

4

[ln4−ln 1]

=

3

4

[1. 3863 −0]= 1. 040

Now try the following Practice Exercise

PracticeExercise 140 Definite integrals

(answers on page 355)

In problems 1 to 10, evaluate the definite integrals

(where necessary, correct to 4 significant figures).

1. (a)

∫ 2

1

xdx (b)

∫ 2

1

(x− 1 )dx

2. (a)

∫ 4

1

5 x^2 dx (b)

∫ 1

− 1

−

3

4

t^2 dt

3. (a)

∫ 2

− 1

( 3 −x^2 )dx (b)

∫ 3

1

(x^2 − 4 x+ 3 )dx

4. (a)

∫ 2

1

(x^3 − 3 x)dx (b)

∫ 2

1

(x^2 − 3 x+ 3 )dx

5. (a)

∫ 4

0

2

√

xdx (b)

∫ 3

2

1

x^2

dx

6. (a)

∫π

0

3

2

cosθdθ (b)

∫π/ 2

0

4cosθdθ

7. (a)

∫π/ 3

π/ 6

2sin2θdθ (b)

∫ 2

0

3sintdt

8. (a)

∫ 1

0

5cos3xdx

(b)

∫π/ 2

π/ 4

(3sin2x−2cos3x)dx

9. (a)

∫ 1

0

3 e^3 tdt (b)

∫ 2

− 1

2

3 e^2 x

dx

10. (a)

∫ 3

2

2

3 x

dx (b)

∫ 3

1

2 x^2 + 1

x

dx

35.5 The area under a curve

The area shown shaded in Figure 35.1 may be deter-

mined using approximate methods such as the trape-

zoidal rule, the mid-ordinate rule or Simpson’s rule (see

Chapter 28) or, more precisely, by using integration.

0 x 5 ax 5 b

y 5 f(x)

y

x

Figure 35.1