october11 - Area
between
2
curves.
...

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Unformatted text preview: Area
between
2
curves.
 Sec$on
6.1
(conclude)
 VERTICAL
 INTEGRATION
 Given
2
func2ons
of
x:
y=f(x)
and
y=g(x),
defined
over
[a,b],
 find
the
area
of
the
region
enclosed
between
the
2
graphs.
 upper
func)on
 A=?
 le<‐most
 value
of
x
 lower
func)on

 right‐most
 value
of
x
 HORIZONTAL
 INTEGRATION
 Given
2
func2ons
of
y:
x=f(y)
and
x=g(y),
defined
over
[c,d],
 find
the
area
of
the
region
enclosed
between
the
2
graphs.
 biggest
 value
of
y
 A=?
 LeD

func)on

 Right
func)on
 smallest
 value
of
y
 Find
the
area
bounded
by

y=2x2+10

and

y=4x+16

 It
is
faster
to
compute
the
area
by
means
of

 A. Horizontal Integration (in dy) B. Vertical Integration (in dx) Find
the
area
bounded
by

y=2x2+10

and

y=4x+16

 1)
graph…
to
see
which
kind
of
integra)on
is
more
convenient
 Upper
func)on
 Easy, by Vertical Integration Lower
func)on
 a‐=‐1
 2)
find
the
extremes
of
integra2on

 b=3
 a=‐1
b=3
 Find
the
area
bounded
by

y=2x2+10

and

y=4x+16

 Upper
func)on
 Lower
func)on
 a‐=‐1
 b=3
 )
 )
 Find
the
area
bounded
by

y=2x2+10,
y=4x+16,
x=‐2
and
x=5

 The
extremes
of

 integra8on
are
 given
to
us!
 It
is
faster
to
compute
the
area
by
means
of

 A. Horizontal Integration (in dy) B. Vertical Integration (in dx) Find
the
area
bounded
by

y=2x2+10,
y=4x+16,
x=‐2
and
x=5

 Same
curves
as
before
 






again,
ver$cal
integra$on
 The
upper
and
 
lower
curve
change

 3
8mes
in
our
 




interval
 Ver$cal
 lines
 The
extremes
of

 integra8on
are
 given
to
us!
 Find
the
area
bounded
by

y=2x2+10,
y=4x+16,
x=‐2
and
x=5

 3
≤
x
≤
5
 Upper:
parabola
 Lower:
line
 ‐2
≤
x
≤
‐1

 Upper:
parabola
 Lower:
line
 ‐1
≤
x
≤
3
 Upper:
line
 Lower:
parabola
 We’ll
need
to
 split
the
integral
 into
3
parts
 ‐2
≤
x
≤
‐1

 Upper:
parabola
 Lower:
line
 ‐1
≤
x
≤
3
 Upper:
line
 Lower:
parabola
 3
≤
x
≤
5
 Upper:
parabola
 Lower:
line
 3 1 2 Find
the
area
bounded
by
y=cos(x),
y=sin(x),
x=
π/2
and
x=0
 It
is
faster
to
compute
the
area
by
means
of

 A. Horizontal Integration (in dy) B. Vertical Integration (in dx) Find
the
area
bounded
by
y=cos(x),
y=sin(x),
x=
π/2
and
x=0
 1)
Graph
to
see
which
is
the
upper
func)on
and
which
is
the
lower
one
 2)
Find
the
intersec)on
of
the
two
curves:

sin(x)=cos(x)

 X
=
π/4
 0
≤
x
≤
π/4

 Upper:
cos(x)
 Lower:
sin(x)
 π/4
≤
x
≤
π/2

 Upper:
sin(x)
 Lower:
cos(x)
 Upper
func)on
 We’ll
need
to
 split
the
integral
 into
2
parts
 0
≤
x
≤
π/4

 Upper:
cos(x)
 Lower:
sin(x)
 π/4
≤
x
≤
π/2

 Upper:
sin(x)
 Lower:
cos(x)
 Find the area of the region enclosed by the curves x=(1/2)y2-3 and y=x-1. horizontal parabola with vertex at (-3,0) line Finding the intersection points y+1=(1/2)y2-3 2y+2=y2-6 y2--2y-8=0 (y-4)(y+2)=0 y=4 or y=-2. Plugging y=2 or 4 into the equation of the line, we see that x=5 if y=4, and x=-1 if y=-2. So the points of intersection are (-1,-2) and (5,4). Ques$on
 Can
the
area

 be
computed
by
 means
of
a
single
 integral
in
dx?
 A. YES B. No C. I am not sure B. No Note that, if you use vertical integration, both curves should be expressed as functions of x. x=(1/2)y2-3 y2=2x+6 y= ±(2x+6)1/2 We need 2 functions! -3<x<-1 Upper curve: y= +(2x+6)1/2 -1<x<5 Upper curve: y= +(2x+6)1/2 Lower curve: y= -(2x+6)1/2 Lower curve: y= x-1 Ques$on
 Can
the
area

 be
computed
by
 means
of
a
single
 integral
in
dy?
 A. YES B. No C. I am not sure A. YES Left curve: x= (1/2)y2-3 Right curve: x= y+1 [Both the right and the left curve are expresses as functions of y] =d Left curve Right curve =c a much easier calculation! Ques)on
 Consider
the
region
enclosed
by
the
curves:
 1)
x=y2‐1

(from
leD)
 2)
y=x2


(from
right)
 3)
y=0


(from
below)
 4)
y=1

(from
above).
 It
is
faster
to
compute
the
area
by
means
of

 A. Horizontal Integration (in dy) B. Vertical Integration (in dx) ...
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