2Bsection5_6part1

2Bsection5_6part1 - A r e a 
b e tw e e n 
2 
c u r v...

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Unformatted text preview: A r e a 
b e tw e e n 
2 
c u r v e s . S e c t i o n 
5 . 6 
P a rt 
1 PROBLEM
1 Given
2
functions
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
function A=? left‐most value
of
x lower
function
 right‐most value
of
x ANSWER Areas
between
2
functions
as
Riemann
Sum: The
height
of
a
rectangle
is
of
the
form
f(x)‐g(x),
 where
f
is
the
upper
curve,
and
g
is
the
lower
one.
 PROBLEM
2 Given
2
functions
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=? Left

function
 Right
function smallest value
of
y ANSWER We
use
horizontal
rectangles.
The
height
of
a
rectange
is
f(y)‐g(y), where
f
is
the
right
curve
and
f
is
the
left
one. Right
function: x=f(y) left
function: x=g(y) ? ? You
need
the
graphs! ? upper
function ? Look
at
the
graph lower
function
 What
about
a
and
b? They
are
the
x‐coordinates
of
the
points
where the
2
curves
intersect. ‐>
set
the
2
curves
=
to
each
other
(and
solve
for
x). ? x2 x4=x







x(x3‐1)=0







x(x‐1)(x2+x+1)=0







x=0
or
x=1 upper
function lower
function
 a=0 b=1 Comments To
set
up
the
formula
for
the
area
 Don’t
forget Parentheses! ? you
need
to
 ? Easy
 from graph 1. identify
which
is
the
upper
curve
and
which
is
the
lower
curve 2. find
the
extremes
of
integration.
 If
they
are
not
explicitly
given, you
may
need
to
solve
f(x)=g(x) 




Note:
The
final
result
must
be
positive
(it’s
an
area!) Find
the
area
bounded
by

y=2x2+10

and

y=4x+16 1)
graph…
to
see
which
is
the
upper
function
and
which
is
the
lower
one 2)
find
the
extremes
of
integration a=‐1
b=3 Upper
function Lower
function a‐=‐1 b=3 Find
the
area
bounded
by

y=2x2+10

and

y=4x+16 Upper
function Lower
function a‐=‐1 b=3 ) ) Find
the
area
bounded
by

y=2x2+10,
y=4x+16,
x=‐2
and
x=5 Same
curves
as
before Vertical lines The
upper
and 
lower
curve
change
 3
times
in
our 




interval The
extremes
of
 integration
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),
y=
π/2
and
x=0 1)
Graph
to
see
which
is
the
upper
function
and
which
is
the
lower
one 2)
Find
the
intersection
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
function 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) ...
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This note was uploaded on 12/07/2010 for the course MATH MATH 2B taught by Professor Famiglietti,c during the Fall '09 term at UC Irvine.

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