082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

486 Chapter 16 Further topics in integration
Example16.8 Evaluate
∫ ∞
1
sintdt.
Solution
∫ ∞
1
sintdt =[−cost] ∞ 1
Nowlim t→∞
(−cost)doesnotexist,thatisthefunctioncost doesnotapproachalimit
ast → ∞, and so the integral cannot be evaluated. We say the integraldiverges.
∫ 1
1
Example16.9 Evaluate √ dx.
0 x
1
Solution Theintegrand, √ ,becomesinfinitewhenx = 0,whichisintheintervalofintegration.
x
The point x = 0 is ‘removed’ from the interval. We consider
slightly greater than 0,and then letb→0 + . Now,
∫ 1
b
Then,
∫ 1
0
1
√ x
dx = [2 √ x] 1 b =2−2√ b
∫
1 1
√ dx = lim x b→0 + b
1
√ x
dx = lim
b→0 +(2−2√ b)=2
The improper integral exists and has value 2.
∫ 1
b
1
√ x
dx where b is
∫ 2
1
Example16.10 Determinewhether the integral dx exists ornot.
0 x
Solution As in Example 16.9 the integrand is not defined atx = 0, so we consider
b >0andthenletb →0 + .
So,
∫ 2
b
1
x dx = [ln|x|]2 b =ln2−lnb
(∫ 2
)
1
lim
b→0 b x dx = lim(ln2−lnb)
b→0
Since lim b→0
lnbdoes notexistthe integral diverges.
∫ 2
b
1
dx for
x
∫ 2
1
Example16.11 Evaluate dx ifpossible.
−1 x
Solution We ‘remove’
∫
the point x = 0 where the integrand becomes infinite and consider two
b
∫
1
2
integrals:
−1 x dx wherebis slightly smaller than 0, and 1
dx wherecis slightly
c x ∫ 2
largerthan0.Iftheseintegralsexistasb→0 − andc → 0 + 1
then
x dxconverges.If
−1