1000
Chapter 34 ◆ Methods of Integration
34–3
We are now able to do two applications that usually require integration of a trigonometric function.
Average Value of a Function
The area A under the curve y f ( x ) (Fig. 34–1) between x a and b is, by Eq. 388, b
A
3
f ( x ) dx
Within the same interval, the average ordinate of that function, y avg
, is that value of y that will cause the rectangle abcd to have the same area as that under the curve, or b
( b a ) y avg
A
3
f ( x ) dx y y = f
( x
) y avg d
0 y
1 y
2 y
3 y
4 a
Δ x
FIGURE 34–1 Average ordinate.
Thus:
y n c b x
1 y avg b a
3 b
f ( x ) dx 417
◆◆◆ Example 9: Find the average ordinate of a half-cycle of the sinusoidal voltage
V sin (volts)
Solution: By Eq. 417, with a 0 and b ,
V avg
V sin d
0
3
V cos
0
V
( cos cos 0)
2
V 0.637
V (volts) ◆◆◆
Root Mean Square Value of a Function
The root mean square (rms) value of a function is the square root of the average of the squares of the ordinates. In Fig. 34–1, if we take n values of y spaced apart by a distance x , the rms value is approximately rms y 2
1 y 2
2 y 2
3 n
• • • y 2 n

Section 34–3 ◆ Average and Root Mean Square Values or, using summation notation, rms n a i 1
y i
2 n
Multiplying numerator and denominator of the fraction under the radical by x , we obtain rms n a y i
2 i 1 n x x
But n x is simply the width ( b a ) of the interval. If we now let n approach infi nity, we get n
lim n
→
y i 1 i
2 x
3 b
[ f ( x )] 2 dx
Therefore:
rms
1 b a 3 b
[ f ( x )]
2 dx 418
1001
◆◆◆ Example 10: Find the rms value for the sinusoidal voltage of Example 9.
Solution: We substitute into Eq. 417 with a 0 and b .
rms
1
0
3
V 2 sin 2 d
V 2
3 sin 2 d
But, by Rule 16,
3 sin 2 d
2 sin 2
4 0
2 sin 2
4 2
So rms
V 2
2
V
2
0.707V (volts)
◆
Find the average ordinate for each function in the given interval.
1. y x 2
3.
5. y y sin
from 0 to 6
1
2 x
2 x from 4 to 12
from 0 to 2
2. y x 3
from 5 to 5
4. y x
from 0 to 4
9 x 2
6. 2 y cos 2 x 1 from 0 to
Find the rms value for each function in the given interval.
7. y 2 x 1 from 0 to 6
9. y x 2 x 2 from 1 to 4
11. y 2 cos x from 6 to 2
8. y sin 2 x from 0 to 2
10. y 3 tan x from 0 to 4
12. y 5 sin 2 x from 0 to 6
◆◆◆
In electrical work, the rms value of an alternating current or voltage is also called the effective value. We see then that the effective value is 0.707 of the peak value.

1002
Chapter 34 ◆ Methods of Integration
34–4
Derivation of the Rule
Integration by parts is a useful method that enables us to split products into two parts for easier integration. We start with the rule for the derivative of a product (Eq. 346) in differential form.
d ( u ) u dv du
Rearranging, we get u d d ( u ) du
Integrating gives us
3 u d
3 d ( u )
3
du or the following:
3 u d u
3
du
This rule is listed as number 6 in our table of integrals.
◆◆◆ Example 11: Integrate 1 x cos 3 x dx .
Solution: The integrand is a product to which none of our previous rules apply. We use integration by parts. We separate x cos 3 x dx into two parts, one of which we call u and the other d .
But which part shall we call u and which d ?
A good rule of thumb is to take d as the more complicated part, but one that you can still easily integrate. So we let u x and d cos 3 x dx
Now our Rule 6 requires that we know du and in addition to u and d . We obtain du by differentiating u , and we fi nd by integrating d .
u x d cos 3 x dx du dx and
3 cos 3 x dx
1
3 3 cos 3 x (3 dx )
1
3
sin 3 x C
1
Applying Rule 6, we have
3 x cos 3 x dx x p
1
3
sin 3 x C
1 q
3 p
1
3
sin 3 x C
1 q dx
3 u u
3
du x
3
sin 3 x C
1 x
1
9
3 sin 3 x (3 dx )
3
C
1 dx x
3
sin 3 x C
1 x
1
9
cos 3 x C
1 x C x
3
sin 3 x
1
9
cos 3 x C
◆◆◆