Some applications of integration 377
- Sketch the curvesy=x^2 +3andy= 7 − 3 x
and determine the area enclosed by them.
[20^56 square units] - 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.