Higher Engineering Mathematics, Sixth Edition

Some applications of integration 377

2. Sketch the curvesy=x^2 +3andy= 7 − 3 x
   and determine the area enclosed by them.
   [20^56 square units]


3. Determine the area enclosed by the three
   straight linesy= 3 x,2y=xandy+ 2 x=5.
   [2^12 square units]

38.3 Mean and r.m.s. values

With reference to Fig. 38.5,

mean value,y=

1

b−a

∫b

a

ydx

and r.m.s. value=

√

√

√

√

{

1

b−a

∫b

a

y^2 dx

}

0 x 5 ax 5 b

y

x

y

y 5 f(x)

Figure 38.5

Problem 4. A sinusoidal voltagev=100sinωt

volts. Use integration to determine over half a cycle

(a) the mean value, and (b) the r.m.s. value.

(a) Half a cycle means the limits are 0 toπradians.

Mean value,y=

1

π− 0

∫π

0

vd(ωt)

=

1

π

∫π

0

100sinωtd(ωt)

=

100

π

[−cosωt]π 0

=

100

π

[(−cosπ)−(−cos 0)]

=

100

π

[(+ 1 )−(− 1 )]=

200

π

=63.66volts

[Note that for a sine wave,

mean value=

2

π

×maximum value

In this case, mean value=

2

π

× 100 = 63 .66V]

(b) r.m.s. value

=

√{

1

π− 0

∫π

0

v^2 d(ωt)

}

=

√{

1

π

∫π

0

(100sinωt)^2 d(ωt)

}

=

√{

10000

π

∫π

0

sin^2 ωtd(ωt)

}

,

which is not a ‘standard’ integral.

It is shown in Chapter 17 that

cos2A= 1 −2sin^2 A and this formula is used

whenever sin^2 Aneeds to be integrated.

Rearranging cos2A= 1 −2sin^2 Agives

sin^2 A=

1

2

( 1 −cos2A)

Hence

√{

10000

π

∫π

0

sin^2 ωtd(ωt)

}

=

√{

10000

π

∫π

0

1

2

( 1 −cos2ωt)d(ωt)

}

=

√{

10000

π

1

2

[

ωt−

sin2ωt

2

]π

0

}

=

√ √ √ √ √ √ √

⎧

⎪⎪

⎨

⎪⎪

⎩

10000

π

1

2

[(

π−

sin2π

2

)

−

(

0 −

sin0

2

)]

⎫

⎪⎪

⎬

⎪⎪

⎭

=

√{

10000

π

1

2

[π]

}

=

√{

10000

2

}

=

100

√

2

=70.71volts

[Note that for a sine wave,

r.m.s. value=

1

√

2

×maximum value.