Bahan Ajar KALKULUS INTEGRAL
Oleh: ENDANG LISTYANI
ANTI TURUNAN
(hal 299)
Anda tentu sudah mengenal invers atau balikan suatu operasi hitung
Invers dari operasi penjumlahan adalah pengurangan; perkalian dengan
pembagian, pemangkatan dengan penarikan akar.
Demikian pula turunan merupakan invers atau balikan dari anti turunan dan
sebaliknya.
Definisi
Suatu fungsi F disebut suatu anti turunan dari suatu fungsi f pada interval I jika
F’(x) = f(x) untuk setiap x pada I
Ilustrasi
Jika F(x) = 3x3 + x2 – 2x – 7 , maka F’(x) = 9x2 + 2x – 2
Jika f adalah fungsi yang didefinisikan sebagai
f(x) = 9x2 + 2x – 2, maka f turunan dari F and F adalah anti turunan dari f
Jika G(x) = 3x3 + x2 – 2x + 5, maka G juga anti turunan dari f karena G’(x) = 9x2
+ 2x – 2
Secara umum, fungsi yang didefinisikan sebagai 3x3 + x2 – 2x + C, dengan C
adalah konstanta, merupakan anti turunan dari f
Secara umum, jika suatu fungsi F adalah suatu anti turunan dari f pada interval
I dan jika G didefinisikan sebagai G(x) = F(x) + C dimana C adalah konstanta
sebarang, maka G’(x) = F’(x) = f(x)
dan G juga merupakan anti turunan dari f pada interval I
Notasi untuk Anti Turunan
Ax(9x2 + 2x – 2) = 3x3 + x2 – 2x + C
Bagaimana dengan
1
Ax( x3 + 2x2 – 2
2
x)=?
Tentu bukan pekerjaan mudah menentukan suatu fungsi yang jika diturunkan
berbentuk
1
3
2
2 x + 2x – 2
x
Untuk itu diperlukan notasi dan aturan-aturan yang dapat mempermudah
menentukan anti turunan
1
Anti turunan dinotasikan sebagai

.... dx
 f ( x) dx = F(x) + C
Teorema A (hal 301)
r
 x dx =
x r 1
C
r 1
untuk r bilangan rasional dan r ≠ - 1
Sifat-sifat
1.
2.
3.
 dx = x + C
 af ( x)dx = a  f ( x)dx
 [ f ( x)  g ( x)]dx =  f ( x)dx +  g ( x)dx
Dengan demikian
1
Ax( x3 + 2x2 – 2
2
=
x)= 
1
( x3 + 2x2 – 2
2
1 3
x dx + 2 x 2dx

2
x ) dx
1
-
2 x 2 dx
3
1 3
1 1 4
2 2
2
.
x
.
x
=
+ C1 +
+ C2 – 2( x + C3 )
3
2 4
3
=
1 4
x
8
+
2 3 4
x - x x C
3
3
Soal-soal 5.1 halaman 307. Tugas individu
Untuk dikerjakan hari kamis tanggal 14 Feb 2013
No 2 – 26 no genap saja
No 27 – 50 semua
FUNGSI-FUNGSI APA YANG DAPAT DIINTEGRALKAN?
2
Sebarang fungsi yang terintegralkan pada [a,b] harus terbatas di
[a , b]. Yaitu: terdapat konstanta M sedemikian sehingga
f ( x)  M
Contoh
Perhatikan fungsi
f ( x) 
1
x2
pada selang [-2 , 2]
merupakan fungsi tak terbatas.
Tidak terdapat suatu bilangan M sedemikian sehingga
f ( x)  M
untuk semua x pada [-2 , 2]
1
Dengan demikian f ( x ) 
tidak terintegralkan pada
2
x
[-2 , 2]
Lihat soal no 21 hal 347
f(x) = x3 + sin x
Ada M = 9
[-2 , 2]
sehingga
f ( x)  M untuk setiap x pada
Jadi f(x) = x3 + sin x terintegralkan pada [-2 , 2]
3
f(x) = 1/(x-1)
pada [-2 , 2] maksimum dan minimum di  dan di - 
tidak ada nilai M sehinngga f ( x)  M
jadi f(x) = 1/(x-1) tidak terintegralkan pada [-2 , 2]
f(x) = tan x
pada [-2 , 2] maksimum dan minimum di  dan di 
f(x) = x3 + sin x
Ada M =
[-2 , 2]
sehingga
f ( x)  M untuk setiap x pada
Jadi f(x) = x3 + sin x terintegralkan pada [-2 , 2]
f(x) = 1/(x+3)
Ada M = 1 sehingga
f ( x)  1 untuk setiap x pada [-2 , 2]
Jadi f(x) = 1/(x+3) terintegralkan pada [-2 , 2]
f(x) = 1/(x-1)
pada [-2 , 2] maksimum dan minimum di  dan di - 
tidak ada nilai M sehinngga f ( x)  M
jadi f(x) = 1/(x-1) tidak terintegralkan pada [-2 , 2]
4
f(x) = tan x
pada [-2 , 2] maksimum dan minimum di  dan di - 
tidak ada nilai M sehinngga f ( x)  M
jadi f(x) = tan x tidak terintegralkan pada [-2 , 2]
TEOREMA A (hal 342)
(Teorema keintegralan). Jika f terbatas pada [a , b] dan kontinu
di [a , b] kecuali pada sejumlah terhingga titik, maka f
terintegralkan pada [a , b].
Khususnya jika f kontinu pada seluruh selang [a , b], maka f
terintegralkan pada [a , b]
f(x) = sin(1/x)
Ada M =..... sehingga f ( x)  M untuk
setiap x pada [-2 , 2]
Jadi f terintegralkan pada [-2 , 2]
5
 x 2 jika  2  x  0
f(x) = 
 1 jika 0  x  2
Ada M = 4 sehingga f ( x)  M
untuk setiap x pada [-2 , 2]
Jadi f terintegralkan pada [-2 , 2]
FUNGSI LOGARITMA ASLI
6
Teorema (hal 453)
ln 1 = 0
19)
2 ln x
 x dx
Misal ln x = u
1
dx  du
x
2 ln x
2
2
 x dx = 2 u du  u  C  (ln x)  C
20)

1
dx
x(ln x) 2
Misal ln x = u
7
1
dx  du
x
1
dx

x(ln x) 2
=
  u 2 du
= u-1 + C =
1
1
+C=
+C
u
ln x
Hitunglah
1
2
)
0
))
)
FUNGSI BALIKAN/INVERS
Misalkan y = f(x) = x3 + 1
x = f-1(y) = 3
y 1
Apakah setiap fungsi mempunyai invers?
Perhatikan fungsi y = f(x) = x2
x=
 y
bukan fungsi , jadi y = f(x) = x2 tidak mempunyai balikan/INVERS.
Tetapi jika domainnya di batasi misalnya y = f(x) = x2 didefinisikan pada [0 ,
maka y = f(x) = x2 mempunyai invers yaitu x = f-1(y) =
]
y
8
y = f(x) mempunyai invers jika y merupakan fungsi satu-satu atau fungsi monoton
Jika f memiliki invers, maka y = f(x)
x = f-1(y)
Grafik y = f(x) sama/identik dengan grafik x = f-1(y)
Pembahasan lebih lanjut
yang terkait dengan fungsi invers adalah
-1
menggunakan bentuk y = f (x). Perhatikan bahwa posisi x dan y dipertukarkan
Dengan demikian grafik fungsi y = f -1(x) dapat diperoleh dengan mencerminkan
grafik y = f(x) terhadap garis y = x
y=x
y = f-1(x)
y = f(x)
FUNGSI EKSPONEN ASLI
Invers dari fungsi logaritma asli adalah fungsi eksponen asli
y = ln x
x = ey
Grafik y = ln x identik dengan grafik x = ey
9
y = ln x
x = ey
Grafik y = ex diperoleh dengan mencerminkan grafik y = ln x terhadap garis y = x
(hal 468)
y = ln x
x = ey
y = ex
y = f(x) = ex disebut fungsi eksponen asli
Sifat-sifat:
ln e = 1
Contoh
10
Turunan dari ex
y  ex 
dy
 ex
dx
Contoh
11
y  ex 
dy
 e x  e x dx  dy
dx
x
x
 e dx  y  C  e  C
x
x
 e dx  e  C
Contoh
Integralkan
e
2 x 1
dx
Misal 2x+1 = u
2 dx = du
e
2 x 1
dx =
1 u
1
1
e du  e u  C  e 2 x1  C

2
2
2
Fungsi eksponen umum
y  ax 
dy
?
dx
dy
 e x ln a . ln a
dx
dy
 a x ln a
dx
y  a x  y  e x ln a 
ecara Umum
12
Contoh
Tentukanlah
dy
dx
1) y = 52x+3
y  ax 
(2) y =
7x
2 6 x
dy
 a x ln a  a x ln a dx  dy
dx
x
 a ln a dx   dy
1
x
 a dx  ln a  dy
1
yC
ln a
1
(
)a x  C
ln a

Contoh
13
u=sin x
du = cos x dx
Fungsi logaritma terhadap basis a
Definisi (hal 479)
y  log a x  x  a y
y  log a x  x  a y  ln x  y ln a  y 
y
ln x
ln a
ln x
ln x
 log a x 
ln a
ln a
log a x 
ln x
ln a
y  log a x 
dy
?
dx
y  log a x  y 
ln x
dy
1


ln a
dx x ln a
14
Fungsi Invers Trigonometri (hal 494)
y  sin x  x  arc sin y
Bagaimana grafik fungsi y = arc sin x ?
y  arc sin x 
dy
?
dx
y  arc sin x  x  sin y
dx
 cos y
dy
1
x
yy
y
dx
 cos y  1  x 2
dy
dy
1

dx
1  x2
15
y  arc sin x 


1
1 x
dy
1

dx
1 x2
1
dx  dy
1 x2
1
dx   dy
2
 1 x
dx   dy
2
 yC
 arcsin x  C

1
1 u
2
du  arcsin u  C
Introduction to Differential Equation
We know that the expression F’(x) = f(x) is equivalent with
so we can write
dF(x) = f(x) dx,
 dF(x)   f(x) dx  F(x)  C
this formula will help us to solve differential equation.
What is differential equation ?
Let start with an example.
Suppose we want to find out xy-equation of a curve passing through a point (-1,
2) with the gradient on each point of the curve is equal to twice the absis of the
point.
Hence
dy
dx
= 2x on each point of the curve.
Now, we will find a function y = f(x) satisfied that condition.
Method 1. If the equation is on the form
dy
= g (x), then y =  g ( x) dx ,
dx
y =  2x dx = x
2
+C
16
Method 2
dy
dx
Think
as dy is divided by dx, so we can write
dy  2x dx
Integrate two sides
 dy =  2x dx
y + C1 = x2 + C2
y = x2 + C2 - C1
y = x2 + C
If the curve pass the point (1,2), we can find C :
2 = (1)2 + C
So C = 1 and the xy-equation is y = x2 + 1
The expression
dy
= 2x
dx
is called differential equation.
Other examples of differential equation are
dy
dx
= 2xy
y dy = (x2 + 1) dx
d2y
dx
2
+2
dy
dx
- 3xy = 0
An equation that contains an unknown function and some of its derivatives
is called differential equation.
In this lecture, we only consider separable first order differential equation.
Notify that the equation
dy x  3x 2
=
dx
y2
can be written as
y2 dy = (x + 3x2) dx
Here, x and y term are separated. To solve this equation, we use method 2
2
2
 y dy =  (x  3x ) dx
y3
3
+ C1 =
x2
+x
2
3
+ C2
17
y3
3x 2
=
+ 3x
2
+ 3C2 – 3C1
y3
3x 2
=
+ 3x
2
+C
3
y=
3
3
3x 2
 3x 3  C
2
Supposed we have y = 6 for x = 0,
then we can find C:
6=
3
C
C = 216
Hence,
y=
3
3x 2
 3 x 3  216
2
Check this result by substituting it in differential equation. The left side of the
differential equation becomes

dy 1  3x 2
3

=
 3x  216

dx 3  2


2
3
(3x + 9x2)
x  3x 2
=
2
3
3 2
3
 x  3x  216
2

and the right side of the differential equation becomes
x  3x 2
y
2
=
x  3x 2
2
3
2 2
3
 x  3x  216
3

.
Those give the same expression.
18
Evaluate
1. The point (3,2) is on a curve, and at any point (x,y) on the curve the tangent
line has a slope equal to 2x – 3. Find an equation of
the curve
Solution
Suppose y = f(x) is the equation of the curve
dy
= 2x – 3
dx
dy = (2x – 3)dx
y=

(2x – 3)dx = x2 – 3x + C
The point (3,2) is on a curve, so 2 = 32 – 3.3 + C
C=2
Hence the equation of the curve is y = = x2 – 3x + 2
2. The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve
Dx2y = 2 – 4x. Find an equation of the curve
Solution
Suppose y = f(x) is the equation of the curve
dy
= 
dx
y=

(2 – 4x) dx = 2x – 2x2 + C1
(2x – 2x2 + C1) dx = x2 -
2
x
3
3
+ C1x + C2
The points (-1,3) and (0,2) are on a curve so,
2
(-1) + C (-1) + C …………..(1)
3
2
2 = (0) (0) + C (0) + C ……………..(2)
3
2
From (1) and (2) : C = 2 , C =
3
3 = (-1)2 -
3
2
3
1
2
1
2
2
1
Hence the equation of the curve is
y=
x2
-
2
x
3
3
+
2
x+2
3
3. An equation of the tangent line to a curve at the
point (1,3) is y = x +2. If at any point (x,y) on the curve, D x2y = 6x,
find an
equation of the curve
Solution
19
Suppose y = f(x) is the equation of the curve
dy
= 3x
dx
2
+ C1
y = x3 + C1x + C2
The slope of the tangent line to the curve at the
point (1,3) is1, so 3(1)2 + C1 = 1, C1 = -2
The point (1,3) is on the curve, so
3 = 13 + (-2)(1) + C2 , C2 = 4
Hence the equation of the curve is y = x3 - 2x + 4
4. The Volume of water in a tank is V cubic meters
when the depth of the water is h meters . If the rate
of chane ofV with respect to h is given by
DhV =
 (2h+3) , find the volume of water in the tank
2
when the depth is 3 m
Solution
Suppose V = f(h)
dV
=  (2h+3) , V =   (2h+3) dh
dh
1
 (2h+3) d(2h+3)
V=

2
1 1
=
.  (2h+3) + C
2 3
11
When h = 0, V = 0, so 0 =
.  (2.0+3)
23
9

C=2
1
9
 (2h+3) - 
V=
6
2
1
9
 (2.3+3) -  = 117
If h = 3, V =
6
2
2
2
2
3
3
+ C,
3
3
The volume of water in the tank when the depth
is 3 m = 117

m3
Evaluate
20
1. The point (3,2) is on a curve, and at any point (x,y) on the curve the tangent
line has a slope equal to 2x – 3. Find an equation of the curve
2. The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve
Dx2y = 2 – 4x. Find an equation of the curve
3. An equation of the tangent line to a curve at the point (1,3) is y = x +2.
If at any point (x,y) on the curve, Dx2y = 6x, find an equation of the curve
4. The Volume of water in a tank is V cubic meters when the depth of the water
is h
meters . If the rate of chane ofV with respect to h is given by D hV =
(2h+3)2, find
the volume of water in the tank when the depth is 3 m

Evaluate
1. The point (3,2) is on a curve, and at any point (x,y) on the curve the tangent
line has a slope equal to 2x – 3. Find an equation of the curve
2. The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve
Dx2y = 2 – 4x. Find an equation of the curve
3. An equation of the tangent line to a curve at the point (1,3) is y = x +2.
If at any point (x,y) on the curve, Dx2y = 6x, find an equation of the curve
4. The Volume of water in a tank is V cubic meters when the depth of the water
is h
meters . If the rate of chane ofV with respect to h is given by DhV =
(2h+3)2, find
the volume of water in the tank when the depth is 3 m

Evaluate
1. The point (3,2) is on a curve, and at any point (x,y) on the curve the tangent
line has a slope equal to 2x – 3. Find an equation of the curve
2. The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve
Dx2y = 2 – 4x. Find an equation of the curve
3. An equation of the tangent line to a curve at the point (1,3) is y = x +2.
If at any point (x,y) on the curve, Dx2y = 6x, find an equation of the curve
4. The Volume of water in a tank is V cubic meters when the depth of the water
is h
meters . If the rate of chane ofV with respect to h is given by D hV =
(2h+3)2, find
the volume of water in the tank when the depth is 3 m

Introduction to Area
21
Consider a region R in the plane as shown in Fig. 1. The region R is bounded by
the x axis, the lines x = a and x = b, and the curve having the equation y = f(x),
where f is a function continuous on the closed interval [a,b].
R
a
b
Fig. 1
Divide the closed interval [a,b] into n subintervals
For simplicity, now we take each of these subintervals as being of equal length,
for instance,
 x. There for  x =
ba
n
Denote the endpoints of these subintervals by x0 , x1 , x2, . . . , xn-1 , xn
x ,
where x0 = a , x1 = a +
xn = b
xi = a + i
x
Let the ith subinterval be denote by
[xi-1,xi].
Because f is continuous on the closed interval [a,b], it is continuous on each
closed subinterval.
By the extreme-value theorem, there is a number in each bsubinterval for which f
has an absolute minimum value.
In the ith subinterval, let this number be ci , so that f(ci) is the absolute minimum
value of f on the subinterval [xi-1,xi].
Consider n rectangles, each vhaving a width
see Fig 2.
f(ci)
 x units and an altitude f(ci) units
R

a
x
b
Fig. 1
Let the sum of the areas of these n rectangle be given by Sn square units, then
Sn = f(c1)
 x + f(c2)  x + . . . + f(cn)  x
n
=

f(ci)
 x …………….(*)
i 1
The summation on thr right side of (*) gives the sum of measures of vthe areas of
n inscribed rectangles. Thus however we define A, it must be such that A
Sn

DEFINITION
Suppose that the function f is continuous on the closed interval [a,b], with f(x)
0 for all x in [a,b], and that R is the region bounded by the curve y=f(x), x axis,
and the lines x = a and x = b. Divide the closed interval [a,b] into n subintervals

each of length
x =
ba
, and denote the ith subinterval by [x
n
i-1,xi].
Then if
f(ci) is the absolute minimum function value on the ith subinterval, the measure of
the area of region R is given by
22
n
A=
 f (ci )  x
lim
n   i 1
Example
Find the area of the region bounded by the curve y = x2, the x axis, and the line
x = 3 by taking inscribed rectangle.
Solution
Devide the interval [0,3] into n subinterval, each of length
 x, x = 2  x, . . . , x = i  x
x = (n-1)  x , x = 3
30 3
=
x =
n
n
x0 = 0 , x1 =
2
n-1
 x;
i
n
Because f is increasing on [0,3], the absolute minimum value of f on the ith
subinterval [xi-1,xi] is f(xi-1)
n
f ( xi 1 )  x

n 
lim
There for A =
i 1
Because xi-1 = (i-1)
f(xi-1) = [(i-1)
Therefore
 x]
 x and f(x) = x ,
2
2
n
n
 f ( xi 1 )  x =  (i  1) 2 (  x)
3
i 1
i 1
n
=
 (i 2  2i  1)
i 1
27
n3
= ………….lanjutkan
Gunakan (pilih) rumus sbb
n
n(n  1)
n(n  1)(2n  1)
2
; i 
i  2
6
i 1
i 1
n
n 2 (n  1) 2 n 4 n(n  1)(6n 3  9n 2  n  1
3
; i 
i 
4
30
i 1
i 1
n
THE DEFINITE INTEGRAL
In the preciding section the measure of the area of a region was defined as the
following limit:
n
f (ci )  x ……………………..(*)

n 
lim
i 1
To define the definite integral we need to consider a
new kind of limiting
process, of which the limit given in (*) is a special case.
23
Let f be a function defined on the closed interval [a,b]. Divide thus interval into n
subintervals by choosing any (n-1) intermediate points between a and b.
Let x0 = a , and xn = b , and let
x0 < x1 < x2 < . . . < xn-1 < xn
The points x0 , x1 , x2, . . . , xn-1 , xn are not necessarily equidistant. Let
the length of the ith subinterval so that
 i x be
i x = x – x
i
i -1
A set of all such subintervals of the interval [a,b] is called a partition of the
interval [a,b].
One (or more) of these subintervals is longest. The length of the longest
subinterval of the partition called the norm of the partition, is denoted by
P
Choose a point in each subinterval of the partition P. Let xi* be the point chosen
in [xi -1 , xi]
Form the sum f(xi*)
1 x + f(x2*)  2 x + . . . + f(xi*)  i x + . . . + f(xn*)  n x =
n
 f ( xi* ) i x
i 1
Such a sum is called a Riemann sum, named for the mathematician George
Friedrich Bernhard Riemann (1826 – 1866)
Y
x2*
x3*
x1*
xn*
xi*
X
Definition
If f is a function defined on the closed interval [a,b], then the definite integral of f
from a to b, denote by
b
 f ( x) dx , is given by
a
b
n
*
 f ( x) dx = lim  f ( xi ) i x if the limit exists
a
P 0 i 1
Note
That the statement “the function f is integrable on the closed interval [a,b]” is
synonymous with the statement : the definite integral of f from a to b exists
24
b
In the notation for the definite integral
 f ( x) dx ,
a
f (x)
is called the integrand, a is called the lower limit, and b is called the
upper limit.
The symbol

is called an integral sign
Definition

If a > b, then
b
a
a
b
 f ( x) dx = -  f ( x) dx
a

 f ( x) dx = 0
a
THE FUNDAMENTAL THEOREM OF THE CALCULUS
Theorem
If f is a continuous function on the closed interval [a,b], and F
antiderivative of f on [a,b], then
is an any
b
 f ( x) dx = F(b) – F(a)
a
We should write F(b) – F(a) =
F ( x)ba
Example
2
Evaluate
3
x
 dx
1
Solution
2
2
1
1 4  1 4 1 4
x
dx
(
1
)
(
2
)
x
=
=
=
4

 4  4
4
4
1
1
3
=3
3
4
Properties of The Definite Integral
Theorem
25
o
o
o
b
b
a
b
a
 kf ( x) dx = k  f ( x) dx
b
b
a
b
a
b
a
b
a
a
a
 [ f ( x)  g ( x)]dx =  f ( x) dx +  g ( x) dx
 [ f ( x)  g ( x)]dx =  f ( x) dx -  g ( x) dx
Kerjakan secara kelompok, 2 – 3 orang
Tentukan panjang busur
1. y3 = 8x2, dari x = 1 ke x = 8
2. 6xy = x4 + 3 , dari x = 1 ke x = 2
3. 27y2 = 4(x-2)3 , dari x = 2 ke x = 11
4. x = 5 cos  , y = 5 sin 
0 
 2
5. x = 2 cos  + cos(2  ) + 1 , y = 2 sin  + sin (2  ) , 0  

Jawab
Kerjakan secara kelompok, 2 – 3 orang
Tentukan panjang busur
1. y3 = 8x2, dari x = 1 ke x = 8
2. 6xy = x4 + 3 , dari x = 1 ke x = 2
3. 27y2 = 4(x-2)3 , dari x = 2 ke x = 11
4. x = 5 cos  , y = 5 sin 
0 
 2
5. x = 2 cos  + cos(2  ) + 1 , y = 2 sin  + sin (2  ) , 0  

Kerjakan secara kelompok, 2 – 3 orang
26
Tentukan panjang busur
1. y3 = 8x2, dari x = 1 ke x = 8
2. 6xy = x4 + 3 , dari x = 1 ke x = 2
3. 27y2 = 4(x-2)3 , dari x = 2 ke x = 11
4. x = 5 cos  , y = 5 sin 
0 
 2
5. x = 2 cos  + cos(2  ) + 1 , y = 2 sin  + sin (2  ) , 0  

27