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

13.3 Definite and indefinite integrals 445
(b) Because interchanging thelimitsofintegration changes thesignoftheintegral,we
(c)
find
∫ 1
x 2 +1dx=−
∫ π
0
∫ 2
2
1
x 2 +1dx=− 10 3
sinx dx = [−cosx] π 0 =(−cosπ)−(−cos0)=1−(−1)=2
Figure 13.10 illustratesthisarea.
sin x
0 p
x
Figure13.10
The area isgiven byadefinite integral.
Note that:
(1) Theintegratedfunctionisevaluatedattheupperandlowerlimits,andthedifference
found.
(2) No constantofintegration is needed.
(3) Any anglesaremeasured inradians.
Example13.8 Findthe area underz(t) = e 2t fromt = 1 tot = 3.
∫ 3 ∫ 3
[ e
Solution Area = zdt= e 2t 2t
dt =
1 1 2
[ ] [ ] e
6 e
2
= − = 198
2 2
] 3
1
If the evaluation of an area by integration yields a negative quantity this means that
some or all of the corresponding area is below the horizontal axis. This is illustrated in
Example 13.9.
Example13.9 Findthe area bounded byy =x 3 and thexaxisfromx = −3tox = −2.
Solution Figure 13.11illustratesthe required area.
∫ −2
[ ] x
x 3 4 −2
dx=
−3 4
−3
{ } { }
(−2)
4 (−3)
4
= − = − 65 4 4 4
The area is65/4 square units; the negative sign indicates thatitisbelow thexaxis.