integration

376 INTEGRAL CALCULUS
38.3 Mean and r.m.s. values
With reference to Fig. 38.5,
∫ b
mean value, y = 1 y dx
b − a a
{
∫ }
and r.m.s. value = √ 1 b
y 2 dx
b − a
a
r.m.s. value
√ { 1
=
π − 0
√ { ∫ 1 π
=
π
=
0
√ {10000
π
∫ π
0
}
v 2 d(ωt)
}
(100 sin ωt) 2 d(ωt)
∫ π
0
}
sin 2 ωt d(ωt) ,
which is not a ‘standard’ integral.
It is shown in Chapter 18 that cos 2A = 1−2 sin 2 A
and this formula is used whenever sin 2 A needs to
be integrated.
Rearranging cos 2A = 1 − 2 sin 2 A gives
sin 2 A = 1 (1 − cos 2A)
2
Figure 38.5
Problem 4. A sinusoidal voltage v = 100 sin ωt
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 to π radians.
Mean value, y = 1 ∫ π
v d(ωt)
π − 0
= 1 π
∫ π
0
0
100 sin ωt d(ωt)
= 100
π [−cos ωt]π 0
[Note that for a sine wave,
= 100 [(−cos π) − (−cos 0)]
π
= 100
200
[(+1) − (−1)] =
π π
= 63.66 volts
mean value = 2 × maximum value
π
(b)
In this case, mean value = 2 × 100 = 63.66V]
π
√ {10000
Hence
π
√ {10000
=
π 0
√ {10000
1
=
π 2
⎧
⎪⎨
=
√
⎪⎩
=
∫ π
∫ π
0
}
sin 2 ωt d(ωt)
}
1
2 (1 − cos 2ωt)d(ωt)
[
ωt −
[(
10000 1
π
2
π −
√ {10000
π
√ {10000 }
=
2
}
1
2 [π]
]
sin 2ωt π }
2 0
)
sin 2π
2
−
(
0 − sin 0
2
= 100 √
2
= 70.71 volts
⎫
⎪⎬
)]
⎪⎭
[Note that for a sine wave,
r.m.s. value = 1 √
2
× maximum value.
In this case,
r.m.s. value = 1 √
2
× 100 = 70.71V]