Auo undan plan
uve
.
nas
1
uneb'on S
indeono
ble
tarsinvaun on an indeyol fox), h
1h net
['o.il
on
1h groph of
44 1hs nagon w thnd hy
zigned o uo
nuoan
& Ik indrrvol ta, b]
Sew) d
A
Net signod a roa
A
A
A
A, 4A,)-A
Arwa unden
Ven
ical
a
cunve
inan B
- region
Th
x.
Ck
bounded
b
given funtion
n)
Tha
b
a
Continuou and
the
Then
cloed inter vl [a,b),
bounded
b the gnoph o . tke
nes X=a
on
non negative functon
-anea
of
n-oki
the
rleq
ion
the vertca
X=bh 9iveh b
b
fo)oh.
A tws
Skadc he
.
2
qnaph
Dede rrmine the
3Sed
up 1he
boundarien
a
&b
definite'inkegra
9nteqrate.
eample:
Find fhe aua in
bounded by fu)=
Solwion
-
h i n t quad rand
4x-xk th
x-any
-(x-22n+4)+4
4-(-2)
Here On
.
X=
2
So the
bounderu'es
arre
avn
y=o
0, 4
3
a
O
4 - ) Jx
Henu
bo un ale d
area
3
2
4
3
2
So te
fr)=
anea
in The
4x-»h
first
ouasrand. unden th
eaua to
3
uaru
uni
eunve
n d te
area bounded by he
aaea
cuves
'
y- 4 , Y-
Soution
/-0
ad
Now
So
aKb J=o
-4
Bu
the
X
Hene
Henw2
=
boundes
=
2
b
X
0
=
2, -2
ane a
needs
positive Vdu
accep table
th boundarries
Bounded
a
are
aa
22,
4) dn
anea
2
3
32
maru unib
x-4
Aneo unden a Curve-given funeton, region boun do d by
t e horigontd inen. given funchion
n thi case
Ar ea
Example
anis.
nd
=
olukion prrocoedure:
& he .
As previou one's otepn.
Rind th finot uodrand bounded by fh cun ve
Slurion:
4
tan'/n- tan
rom
qruph,
e
m ples
y
on
an o
boundarien
=
kertigontel Verical
o
are
He nce. te tloned
axis
And x- ton
X
=0,
d
4
tan d
Anea
Jo
Med ,wkiek
-nl Canl
=
.
S,
bounded
he
flrea
n 2
betueen two
Type 1 F f ond
Intenva
T
belos
b
nigh by
In n
2
Curve»
a
are
conAinuous
all
lo, ]
g0),
the
uine
on
x-
the
le 4E
b7.a M=a
u
on
b
f-3]d
fo)dx-
9n) d
[fn-om]dx
b
wppm unahion -
ower
finehon
then
tug,
by
the line
A
A
xin
The
on
gunctons
the ragion bounded above
of
unih
s9uone
if fu)>g0) sor
To.b)
ahe a
aea
The
1ype 2 4
v
w
aru
continuouw undtionn and if
Y) v ) fon al J in fe, J]. Than the anea o
tea ion boundod
tegion
rught by
wY),and below
A
ve)
the efE
on
on
tha
above
by
by « = v(Y),
by
Y=a. &
u
y=d b
o(-v)] d>
AA
- ||
funchion
ult
funchonright
dy
hind tha
Cunves
reqion enetored by the
arre a o1 e
y
,
Y- *4 6
=0,
-
Stion
5
y
|
M
Now
Now
t
x
Velues where
fhoy
insennect.
x= * t 6
So
=
we
tind
Lt'
5
need
So 4on t
3,or -2
the
irot
poaitive
arua
fon fhe Aecond
A
on
one ^p
X=3 b
bounaruen a
from M 3
acceplable
rrom
to
X -0
-5
A
-(x
uare
uare unith.
6
to
n-
3
F i n d the
=
8
anea of region
enclased
by
the
J=x-2
Solution
yM-2
Point
So the intereching
onu ;
=1,9-2
e g tfunehon
u i g h t
f u r n
c h o n
Hente the
bouraled
arrea
n
-[)-]4.
19
Aanre
fut
unve
Cunve
Hins The
aua
o tu segmant eut sA from Paza bolas
Solion:
To fnd hu
interdecHng poinh
4an
-
61a'x
04a,
Area
v
te t
n
a1f segmerd
ha
dea(aa)-
324- 6
Squar Uhis
h Find
the
and y
he
anea. en clord by
to
Curwe»
4aoHa)
4b(b-
Soluhon:
To Rnd the
point 6} indersertionn
4alxa) = 4b(b-
a n t a = b-bx
tb)n
=
(bta) Cb-a)
=b-a
b-a
So the enclaned Area
=
2|4a(xiaj.yt+
b(b-)
2
dx
bb-a
+4
Cb-)
b-a
Hind th arua interiort t y24x-x exterion to
Taax
lying
Conterponalng
in
the
frrat uadnant.
Hene 4hd h
totod arua,
Solution:
Gtiven
-
a
M-a)
y4
So Ean iaeeliag
fort indereneehiny
: A r in
peinkb:
a
2Xx
an-
th tird uadnant=
n0
24x-R-an d
L
20M-dn
=
a-M
a Sing
a-(adn= a usg do
a Coo do
C1t
C14 Canzo) do
a
,
t0
-0-o
2
Area Iin Tke Sirot uadnand- " -
a(
totd
Henca
the
Caleulote
angern
64a &
+
aua
-
atea
y-
2(-5 h
b
enclased
m.uni
. unt.
th eunve
120M.
;
Jnttrectiy poín
64a-= 12an
X+36a= looa
+2:6a
dn-2/tya-Rd
Area t r - 2 a n
-
+
W
=
6a) t lba
=4a,
a
-lCa
l-lßa nat »ad accaptable