134 Chapter 5Integration
(iv)
(v)
(vi)
(vii)
0 Exercises 16–25
Average value of a function
Because the definite integral is identified as the area under the curve it follows from
equation (5.10) that the average value of the functiony 1 = 1 f(x)in the intervala 1 ≤ 1 x 1 ≤ 1 bis
(5.13)
EXAMPLE 5.5Find the average value ofsin 1 θin the interval 01 ≤ 1 θ 1 ≤ 1 π 22.
By Example 5.4(v),
0 Exercises 26–28
Three properties of the definite integral
Letf(x) 1 = 1 F′(x)so that, by equation (5.11),
1.The value of the integral does not depend on the symbol used for the variable of
integration:
(5.14)
In each case, the value of the integral isF(b) 1 − 1 F(a).
ZZZ
abababf x dx() == =f t dt() f u du()
Z
abfxdx Fb Fa() =−() ()
sinθθ==sin θ
12
02ππ
π2
Z d
y
A
ba
fxdx
dx
abab=
−
=
Z
Z
()
Z
232332
3
2
dx
x
= x
ln =− =ln ln ln
Z
−+−−+−−=− =−
11211221
2
1
2
edt e e
tt−−− = −
1
2
1
2
222eee
Z
02022
πππ
sinθθd =−cosθ cos cos
=−
−− 0 00 11
()=−−=() ( )
ZZ
abababdx==dx x b a
1 =−