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1. 7. APPLICATIONS OF
THE DEFINED INTEGRAL
1

2. 2
7.1 Area Between Two Curves
a. Let
{}
)( 0, |),(xf y bx ayx D≤ ≤ ≤ ≤ =
a
b
f(x)
D
Area of region D = ?
x∆
Steps :
1. Divided D into n pieces, the area of each piece
s
is approximated by area of rectangular
with height f(x) and length of base
x∆
xxf A∆ ≈∆)(
2. The area of D is approximated by sum area of rectangular .
If , The area of D is
∫
b
a
dxxf)(
A =
0 x∆→

3. Example :Find the area of region that is bounded
by parabola
xaxis
, and
x
= 2.
3
8
3
1
2
0
3
2
0
2
=
⎥
⎦
⎤
= =
∫
x dxx A
,
2
x y=
2
x y=
2
Area of pieces
x∆
2
x
x x A∆ ≈∆
2
Area of region

4. 4
b) Let
{}
)( )(, |),(xhy xgbx ayx D≤ ≤ ≤≤ =
x∆
x xg xh A∆ − ≈∆))( )((
h(x)
g(x)
ab
Area of region D = ?
Steps:
1. Divided D into n pieces, the area of each
pieces is approximated by area of rectangular
with height h(x)-f(x) and length of base
x∆
2. The area of D is approximated by sum of area rectangular .
If , The area of D is
A =
∫
−
b
a
dx xg xh))( )((
D
h(x)-g(x)
0 x∆→

5. 5
Example:Find the area of region that is bounded by
y
=
x
+4
and parabola
2
2
− =x y
2 4
2
− = +x x
2
2
− =x y
0 6
2
= −−x x
0)2 )(3 (= + −x x
The straight line and parabola
intersects at
y=x+4
-23
x = -2, x = 3
x∆
)2 ()4 (
2
− − +x x
Area of pieces
x x x A∆ − − + ≈∆))2 ()4 ((
2

6. 6
∫∫
−−
+ + − = − − + =
3
2
3
2
2 2
)6 ( ))2 ()4 ((dx x x dx x x A
6
125
6
2
1
3
1
3
2
2 3
=
⎥
⎦
⎤
+ + −=
−
x x x
Area of region :

7. 7
Example : Find the area of region that is bounded by x axis,
2
x y=
And y = -x + 2
Answer
Intersection points
2
2
+−=x x
2
x y=
0 2
2
= −+x x
x x A∆ ≈ ∆
2
1
2
0)1 )(2 (= − +x x
x = -2, x = 1
y=-x+2
1
If pieces is vertical , then region must be
divided into 2 sub region
x∆
x∆
x x A∆ +−≈ ∆)2 (
2
Area of pieces I
Area of pieces II

8. 8
∫
= = =
1
0
1
0
3
3
1 2
1
3
1
|x dxx A
The area of region I
The area of region II
2
1
2
2
1
2
1
2
|2 2x x dx x A+ −= +−=
∫
2
1
)2 ()42(
2
1
= +−− +−=
The area of region
6
5
2
1
3
1
2 1
= + = + =A A A

9. 9
c). Let
{}
)( )(, |),(yhx ygdy cyx D≤ ≤ ≤ ≤ =
y∆
∫
−
d
c
dy yg yh))( )((
h(y)
g(y)
c
d
D
Area of region D = ?
Langkah :
1. Divided D into n pieces, the area of each
pieces is approximated by area of rectangular
with height h(y)-f(y) and length of base
y∆
y yg yh A∆ − ≈∆))( )((
y∆
A =
h(y)-g(y)
2. The area of D is approximated by sum area of rectangulars .
If , The area of D is
0 y∆→

10. 10
2
3y x−=
2
31y y−=+
02
2
=−+y y
2
3y x−=
Example
: Find the area of region that is bounded by
and
1− =x y
Answer :
The intersection point between
parabola and straight line are
0)1 )(2 (= − +y y
y = -2 dany = 1
1− =x y
-2
1
y∆
)1 () 3(
2
+ − −
y y
Area of pieces
y y y A∆ + − − =∆))1 () 3((
2

11. 11
The area of region is :
∫
−
+ − − =
1
2
2
))1 () 3((dy y y L
∫
−
+ − − =
1
2
2
)2 (dy y y
.
2
9
2
2
1
3
1
1
2
2 3
=
⎥
⎦
⎤
+ − −=
−
y y y

12. 12
7.2NN ilai rata-rata



b
a
dxxf
ab
value Average)(
1

Secara Geometri untuk
0)(xf
, maka
Luas daerah dibawah kurva y = f(x),
bx a


a b
0
y=f(x)
Average value
of f
gambar 6.9

13. 25
7.3 The Length of Curve
A parameter form of curve on
x = f(t)
y = g(t)
bta≤≤,
Point A(f(a),g(a)) is called original point and B(f(b),g(b)) is called
terminal point of curve.
Definition: A curve is called smooth if
(1)
(i)
'f'g
andare continuous on [a,b]
(ii)
'f'g
andis not zero at the same time on (a,b)
2
R

14. 26
Let a curve on parameter form (1), we will find length of that curve
Steps
1. Divided interval [a,b] into n subintervals
b t t t t a
n o
= < < < < =...
2 1
ab
●●●●
1
t
i
t
1−i
t
1−n
t
A partition on [a,b]
A partition on curve
●
1
Q
●
●●
●
o
Q
1−i
Q
i
Q
n
Q

15. 27
2. Approximate length of curve
1−i
Q
i
Q
i
s∆
i
w∆
i
s∆
length of arc
i i
Q Q
1−
i
w∆
length of line segment
i i
Q Q
1−
i i
w s∆≈ ∆
The length of arc is approximated by length of line
segment
i
x∆
i
y∆
2 2
) ( ) (
i i
y x∆+ ∆ =
2
1
2
1
)] ( )([ )] ( )([
− −
− + − =
i i i i
tg tg tf tf
We apply mean value theorem to function f and g.
There are , such that
), ( ,ˆ
1i i i i
t t tt
−
∈
t tf tf tf
i i i
∆ = −
−
)(' ) ( )(
1
t tg tg tg
i i i
∆ = −
−
)ˆ(' ) ( )(
1

16. Calculus I (MA 1114) - Faculty of Science
Telkom Institut of Technology28
where
1−
− = ∆
i i i
t t t
so
2 2
] )ˆ('[ ] )('[
i i i i i
t tg t tf w∆ + ∆ = ∆
i i i
t tg tf∆ + =
2 2
)]ˆ('[ )]('[
The length of curve is approximated by length of polygonal arc
∑
=
∆ + ≈
n
i
i i i
t tg tf L
1
2 2
)]ˆ('[ )]('[
If ||P|| Æ0, the length of curve is
dt tg tf L
b
a
∫
+ =
2 2
)]('[ )]('[

17. 29
Remark:
For curve y=f(x),
bx a≤ ≤
dt tg tf L
b
a
∫
+ =
2 2
)]('[ )]('[
dt
dt
dy
dt
dx
b
a
∫
+ =
2 2
] [ ][
dt tg tf L
d
c
∫
+ =
2 2
)]('[ )]('[
dx
dx
dy
dt
dx
dy
dt
dx
b
a
b
a
∫ ∫
⎟
⎠
⎞
⎜
⎝
⎛
+ = ⎟
⎠
⎞
⎜
⎝
⎛
+ =
2 2
2
1 ) 1()(
For curve x=g(y),
d y c≤ ≤
dt
dt
dy
dt
dx
d
c
∫
+ =
2 2
] [ ][
dy
dy
dx
dt
dy
dx
dt
dy
d
c
d
c
∫ ∫
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ =
2 2
2
1 1) (

18. 30
Example : Find length of curve
4 0; ,
2 3
≤≤ = =t t ytx
1.
2
3)('t tx=
t ty2)(',=
dt t t L
∫
+ =
4
0
2 22
)2( )3(
The length of curve
dtt t
∫
+ =
4
0
2 4
4 9
∫
+ =
4
0
2 2
)4 9(dt t t
∫
+ =
4
0
2
4 9dt tt
∫
+
+ =
4
0
2
2/1 2
18
)4 9(
)4 9(
t
td
tt
4
0
2/3 2
3
2
18
1
| )4 9(+ =t
)8 10 80( )8 40 40(
27
1
27
1
− = − =

19. 31
2.
3/2
1
2, 7
3
yx x=≤≤
Answer :
2/1
3x
dx
dy
=
()
∫∫
+ = + =
7
3/1
7
3/1
2
2/1
91 3 1dxx dx x L
)91( )91(
7
3/1
2/1
9
1
x d x+ + =
∫
3
1
27
2 7
3/1
2/3
27
2
37 )8 512( | )91(= − = + =x

20. 32
Problems
A. Sketch and find area of region that is bounded by
2
2 and yx yx==+
1.
3
,, 8and yxy x y ==− =
2.
3.
y
=
x
,
y
= 4
x
,
y
= -
x
+2
4.
y
= sin
x
,
y
= cos
x
,
x
= 0 ,
x
= 2π.
2
270 6 0 xy andyyx −+= −−=
5.

21. 35
E. Find length of curves
1 0,)2 (
3
1
2/3 2
≤ ≤ + =x x y
4 2,
4
ln
2
2
≤ ≤ − =x
x x
y
2/1 0), 1ln(
2
≤ ≤ − =x x y
9 0),3 (
3
1
≤ ≤ − =y yy x
4 1;2/1 2 ,2 3
3 2
≤≤ − = + =t t y t x
π
≤≤ − = =t t yt x0;5 cos 4 , sin 4
1.
2.
3.
4.
5.
6.