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

446 Chapter 13 Integration
y
y = x 3
y
y = sin x
–3
–2
0 x
–p
p
x
Figure13.11
Areas below thexaxisare classed asnegative.
Figure13.12
Thepositive and negative areas cancel
each other out.
Example13.10 (a) Sketchy = sinx forx = −πtox = π.
(b) Calculate
∫ π
−π
sinxdx and comment onyourfindings.
(c) Calculatetheareaenclosedbyy = sinxandthexaxisbetweenx = −πandx = π.
Solution (a) Agraph ofy = sinx betweenx = −πandx = π isshown inFigure 13.12.
(b)
∫ π
−π
sinx dx = [−cosx] π −π = −cos(π) +cos(−π) = 0
Examining Figure 13.12 we see that the positive and negative contributions have
cancelledeachotherout;thatis,theareaabovethexaxisisequalinsizetothearea
below thexaxis.
(c) From(b)theareaabovethexaxisisequalinsizetotheareabelowthexaxis.From
Example 13.7(c) the area above the x axis is 2. Hence the total area enclosed by
y=sinxandthexaxisfromx=−πtox=πis4.
Ifanareacontainspartsbothaboveandbelowthehorizontalaxisthencalculatingan
integral will give the net area. If the total area is required, then the relevant limits must
first befound.Asketchofthe function oftenclarifiesthe situation.
Example13.11 Findthe area containedbyy = sinx fromx = 0 tox = 3π/2.
Solution Figure13.13illustratestherequiredarea.Fromthisweseethattherearepartsbothabove
and below thexaxisand the crossover pointoccurswhenx = π.
∫ π
0
∫ 3π/2
π
sinx dx = [−cosx] π 0
=−cosπ+cos0=2
sinx dx = [−cosx] 3π/2
π
( ) 3π
=−cos +cosπ=−1
2