Integration Danang Mursita

Integration Danang Mursita

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Integration • • • •

The Indefinite Integral The Definite Integral The Fundamental Theorem of Calculus Application of Integration : Area between two curves


The Indefinite Integral • Definition : A function F(x) is called an antiderivative of a function f(x) if the derivative of F(x) is f(X) or F ‘(x) = f(x) • Examples : 1. F(x) = x2 + 3  f(x) = 2x 2. F(x) = x2 – 10  f(x) = 2x 3. F(x) = x2 + 15  f(x) = 2x

• If F ‘(x) = f(x) then the functions of the form F(x) + C are antiderivative of f(x) • The process for finding antiderivative is called antidifferentiation or integration. • Notation : f(x) dx  F(x)  C  http://danangmursita.staff.telkomuniversity.ac.id/

Integration Formula d 1. sin x   cos x   cos x dx  sin x  C dx 2.... 3....etc

d a. x   1  dx  x  C dx d  xr 1  b.  ... dx  r  1 

r1 x r  x dx  r  1  C r  1


Problems 1)  2) 

sin x cos 2 x cos x 2 sin x

dx

6)  sec xtan x  cos x dx

dx

sin 2x 7)  dx cos x 8)  dx

3)  x dx 3 2 4)  x dx

5) 

1 x3

dx

3 2 9)  x  2 x dx

10) 

x2  3 x3

dx


The Properties of Indefinite Integral

1)  cf(x) dx  c f(x) dx

2)  f(x)  g(x)dx   f(x) dx   g(x) dx Examples : 1. x 2 x  1dx 2. x x  12 dx 3. 

x 2 sin x  x x2

dx


Integration by Substitution • Let y = (fog)(x). The derivative of y with respect to x is y ‘ = (fog)’(x) = f ‘(g(x)) g’(x) [ The Chain rule] • If u = g(x) then the formula of integration is du  f(u)u' dx   f(u) dx dx   f(u) du  F(u)  C  Fg(x)  C


Problems 6)  1  2x 5 dx

1)  2x x 2  1 dx





2)  cos 1  x 2 x dx 3) 

x 1 x 2  2x  1

7) 

dx

4)  sin x cos 2 x dx 5)  cos x sin2 x dx

8) 

sin 2x

5  cos 2x 3 x2 x3 1

dx

dx

 

sin 1x 9)  dx 2 x

 

10)  x 2 sec 2 x 3 dx http://danangmursita.staff.telkomuniversity.ac.id/

Problem : The Definite Integral • How to find the area of a region D bounded below by the x-axis, on the sides by the lines x = a and x = b , and above by a curve y = f(x), where f(x) is continuous on [a,b] and f(x) ≥ 0 for all x in (a,b) • D = { (x,y) | a ≤ x ≤ b, 0 ≤ y ≤ f(x) }


Steps : The Definite Integral 1. Divide [a,b] into n subintervals by choosing points x1, x2, …,xn such that a < x1 < x2 < …< xn < b. These points are said to form a partition of [a,b]. Let x1, x2, …, xn are the length of the partition. 2. Choose the any point x1,x2,…,xn in subinterval and we have a Riemann sum : n f(x1) x1 + f(x2) x2 + … + f(xn) xn =  f xk xk k 1 3. Increase n ( n  ) in such a way the length of the partition approaches zero (xk  0 ) and form the n n limit     lim

 f xk xk 

xk0 k1

lim

 f xk xk

n k1


Illustration : The Definite Integral

y = f(x)

Y

xk

a

X b

Xk-1

xk


Movie : The Definite Integral


Definition : The Definite Integral • If a function f(x) is defined on [a,b], then the Definite Integral of f(x) from a to b is defined to be (provided the limit exits) b

 f(x)dx 

a

n

lim

 f xk xk 

xk0 k 1

n

lim

 f xk xk

n k 1

• If the limit exits then the function f(x) is said integrable on [a,b] • The values a and b are called respectively lower and upper limits of integration and f(x) is called the Integrand. http://danangmursita.staff.telkomuniversity.ac.id/

The First Fundamental Theorem • If f(x) is continuous on [a,b] and F(x) is antiderivative of f(x) on [a,b] then b

b  f(x) dx  F(x) a  F(b)  F(a)

a

Examples : 2

(1)  x dx 1 

2

(2)  cos x dx 0


The Properties of Definite Integral b

b

b

a a

a

a

(1)  p f(x)  qg(x)dx  p  f(x) dx  q g(x) dx (2)  f(x) dx  0 a b

a

a b

b c

a b

a

(3)  f(x) dx    f(x) dx (4)

(5)

b

 f(x) dx   f(x) dx   f(x) dx c g(b)

 fog (x) g' (x) dx   f(u) du

a

g(a)


Problems 3

(1) 

1 x2

2

dx

1 9

(2)  2x x dx 4 4

3   3 (3)    5 x  x 2  dx  1 x 

2

 (4)    t  dt 2    sin t 2

6

2

1 2

(7)  x 8  x 2 dx 1 3

x2

(8) 

x 2  4x  7

1 

4

dx

tan t sec 2 t dt

(9) 

0 

(5)  2x  3 dx 0

(6)  2  x dx

4

(10) 

0

cos 2t dt 7  3 sin 2t


The Second Fundamental Theorem • Let f(x) is continuous on open interval I and a  I. If G(x) is defined by x

G(x)   f(t) dt a

Then G’(x) = f(x) at each point x in interval I G(x) 

v(x)

 f(t) dt



G' (x)  f v(x) v' (x)  f u(x)u' (x)

u(x)


Problems x

(1) G(x)   3t 2  1 dt  G' (2) and G"(2) 2 x

cos t

(2)H(x)  

0 x

(3)F(x)  

0

t2  3 t3

t2  7

dt  H' (0) and H" (0) dt    x  

(a) find x where F(x) attains its minimum value (b) find open int erval which F(x) is only increa sin g x3

(4)F(x)   sin2 t dt  F' (x) and F" (x) x2


Area of Region (1) • Let D is region that bounded below by the x-axis, on the sides by the lines x = a and x = b , and above by a curve y = f(x), where f(x) is continuous on [a,b] and f(x) ≥ 0 for all x in (a,b) or D = { (x,y) | a ≤ x ≤ b, 0 ≤ y ≤ f(x) } • Then the area of D bis A   f(x) dx a


Area of Region (2) • Let D is region that bounded below by the x-axis, on the sides by the lines x = a and x = b , and above by a curve y = f(x), where f(x) is continuous on [a,b] and f(x) ≤ 0 for all x in (a,b) or D = { (x,y) | a ≤ x ≤ b, f(x) ≤ y ≤ 0 } • Then the area of D is b

A    f(x) dx a http://danangmursita.staff.telkomuniversity.ac.id/

Example # 1 • Find the area of region that bounded by the curve y = sin x, x-axis, and on the sides by the lines x = 0 and x = 3π/2


Area of Region (3) • Let D is region that bounded below by the line y = c , on the sides by the y-axis and a curve x = g(y), and above by the line y = d, where g(y) is continuous on [c,d] and g(y) ≥ 0 for all y in (c,d) or D = { (x,y) | 0 ≤ x ≤ g(y), c ≤ y ≤ d } • Then the area of D is d

A   g(y) dy c


Area of Region (4) • Let D is region that bounded below by the line y = c , on the sides by the y-axis and a curve x = g(y), and above by the line y = d, where g(y) is continuous on [c,d] and g(y) ≤ 0 for all y in (c,d) or D = { (x,y) | g(y) ≤ x ≤ 0, c ≤ y ≤ d } • Then the area of D is d

A    g(y) dy c


Example # 2 • Find the area of region that bounded by g(y) = y3 – 2y2 – 3y , y-axis and on the sides by the lines y = -1 and y = 4.

-1

- 1/3

3

4


Area between two curves (1) • Let D is region that bounded below by a curve y = g(x), on the sides by the lines x = a and x = b , and above by a curve y = f(x), where g(x) and f(x) are continuous on [a,b] or D = { (x,y) | a ≤ x ≤ b, g(x) ≤ y ≤ f(x) } • Then the area of D is b

A   f(x)  g(x)dx a


Area between two curves (2) • Let D is region that bounded below by the line y = c , on the sides by a curve x = g(y) and a curve x = f(y), and above by the line y = d, where g(y) and f(y) are continuous on [c,d] and f(y) ≥ g(y) for all y in (c,d) or D = { (x,y) | g(y) ≤ x ≤ f(y), c ≤ y ≤ d } • Then the area of D is d

A   f(y)  g(y)dy c


Examples 1. Find the area of the region enclosed by the curves y = x2 and y = 4x by integrating a) With respect to x b) With respect to y

2. Sketch the region enclosed by the curves and find its area by any method a) y = x2 + 4 and x + y = 6 b) y = x3, y = - x and y = 8


Problems Sketch the region and find its area by any methods 1. y = x3 – 4x, y = 0, x = 0, x = 2 2. x = y2 – 4y, x = 0, y = 0, y = 4 3. x2 = y, x = y – 2 4. y2 = - x, y = x – 6, y = - 1, y = 4 5. y = x3 – 2x2, y = 2x2 – 3x, x = 0, x = 3


Menghitung volume benda putar Metoda Cakram a. Daerah

D  ( x, y) | a  x  b , 0  y  f ( x)

diputar terhadap sumbu x

f(x)

D

a

b

Daerah D

Benda putar 29


Untuk menghitung volume benda putar gunakan pendekatan Iris , hampiri, jumlahkan dan ambil limitnya.

Jika irisan berbentuk persegi panjang dengan tinggi f(x) dan alas xdiputar terhadap sumbu x akan diperoleh suatu cakram lingkaran dengan tebal x dan jari-jari f(x).

f(x)

D

x

a

b

sehingga

V   f 2 ( x) x

f(x) b

V    f 2 ( x) dx x

a

30


Contoh: Tentukan volume benda putar yang terjadi jika daerah D yang dibatasi oleh y  x 2, sumbu x, dan garis x=2 diputar terhadap sumbu x

Jika irisan diputar terhadap sumbu x akan diperoleh cakram dengan jari-jari x 2 dan tebal x

y  x2

x2 x

2

Sehingga

V   ( x 2 ) 2 x   x 4 x Volume benda putar

x2

2



32 V    x dx  x |   5 5 0 4

x

5 2 0

31


b. Daerah

D  ( x, y) | c  y  d , 0  x  g ( y)

diputar terhadap sumbu y

d d x=g(y)

D c

c

Daerah D

Benda putar

32


Untuk menghitung volume benda putar gunakan pendekatan Iris , hampiri, jumlahkan dan ambil limitnya.

Jika irisan berbentuk persegi panjang dengan tinggi g(y) dan alas ydiputar terhadap sumbu y akan diperoleh suatu cakram lingkaran dengan tebal y dan Jari-jari g(y).

d

y

x=g(y)

D

sehingga

c

V   g 2 ( y ) y g ( y)

d

y

V    g 2 ( y ) dy c

33


Contoh : Tentukan volume benda putar yang terjadi jika daerah yang dibatasi oleh y  x 2garis y = 4, dan sumbu y diputar terhadap sumbu y Jika irisan dengan tinggi y dan tebal y diputar terhadap sumbu y akan diperoleh cakram dengan jari-jari y dan tebal y

4

y y

y  x2

x

y

Sehingga

V   ( y ) 2 y   y y Volume benda putar 4

y

y

V    ydy  0


 2

y 2 | 04  8

34

7.2.2 Metoda Cincin a. Daerah

D  ( x, y) | a  x  b , g ( x)  y  h( x)

diputar terhadap sumbu x h(x) D

g(x)

a

b

Daerah D

Benda putar

35 http://danangmursita.staff.telkomuniversity.ac.id/

Untuk menghitung volume benda putar gunakan pendekatan Iris , hampiri, jumlahkan dan ambil limitnya. h(x)

Jika irisan berbentuk persegi panjang dengan tinggi h(x)-g(x) dan alas xdiputar terhadap sumbu x akan diperoleh suatu cincin dengan tebal x dan jari –jari luar h(x) dan jari-jari dalam g(x).

D

g(x) a

h(x)

x

b

sehingga

V   (h 2 ( x)  g 2 ( x))x

x g(x)

b

V    (h 2 ( x)  g 2 ( x)) dx a


Contoh: Tentukan volume benda putar yang terjadi jika daerah D yang dibatasi oleh y  x 2, sumbu x, dan garis x=2 diputar terhadap garis y=-1

Jika irisan diputar terhadap garis y=1 Akan diperoleh suatu cincin dengan Jari-jari dalam 1 dan jari-jari luar 1  x 2 y  x2

Sehingga

V   (( x 2  1) 2  12 )x

1 x2

D x

2

1

  ( x 4  2 x 2  1  1)x y=-1

  ( x 4  2 x 2 )x

Volume benda putar : 2

V    x 4  2 x 2 dx   ( 15 x 5  23 x 3 | 02 )   ( 325  163 )  186 15  0


Metoda Kulit Tabung Diketahui

D  ( x, y) | a  x  b , 0  y  f ( x)

Jika D diputar terhadap sumbu y diperoleh benda putar

f(x) D a

b

Daerah D

Benda putar Volume benda putar ?


Untuk menghitung volume benda putar gunakan pendekatan Iris , hampiri, jumlahkan dan ambil limitnya. Jika irisan berbentuk persegi panjang dengan tinggi f(x) dan alas xserta berjarak x dari sumbu y diputar terhadap sumbu y akan diperoleh suatu kulit tabung dengan tinggi f(x), jari-jari x, dan tebal x

f(x)

D a

x

b

sehingga

x

x

V  2 x f ( x) x f(x)

x

b

V  2  xf ( x)dx a


Contoh: Tentukan volume benda putar yang terjadi jika daerah D yang dibatasi oleh y  x 2 , sumbu x, dan garis x=2 diputar terhadap sumbu y 2

Jika irisan dengan tinggi x ,tebal x dan berjarak x dari sumbu y diputar terhadap sumbu y akan diperoleh kulit tabung dengan tinggi x 2, tebal xdan jari jari x

y  x2

x2

D x

x

2

Sehingga

V  2 x x 2 x  2 x 3 x Volume benda putar 2

V  2  x dx  3

0


 2

x 4 | 02  8

Catatan : -Metoda cakram/cincin

Irisan dibuat tegak lurus terhadap sumbu putar - Metoda kulit tabung

Irisan dibuat sejajar dengan sumbu putar Jika daerah dan sumbu putarnya sama maka perhitungan dengan menggunakan metoda cakram/cincin dan metoda kulit tabung akan menghasilkan hasil yang sama Contoh Tentukan benda putar yang terjadi jika daerah D yang dibatasi Oleh parabola y  x 2,garis x = 2, dan sumbu x diputar terhadap a. Garis y = 4 b. Garis x = 3


a. Sumbu putar y = 4 (i) Metoda cincin

y=4

(4  x ) 2

Jari-jari dalam = rd  (4  x 2 ) 4

y  x2

Jika irisan diputar terhadap garis y=4 akan diperoleh cincin dengan

Jari-jari luar = rl  4 Sehingga

D

x 2

V   ((4) 2  (4  x 2 ) 2 )x

  (8 x 2  x 4 )x

Volume benda putar 2

V    (8 x 2  x 4 )dx   ( 83 x 3  15 x 5 ) | 02   ( 643  325 )  0


224 15



(ii) Metoda kulit tabung

y=4 4 y

yx

Jika irisan diputar terhadap garis y=4 akan diperoleh kulit tabung dengan Jari-jari = r =

2

Tinggi = h =

y D 2 y

y

2

Tebal =

4 y 2 y

y

Sehingga

V  2 (4  y )(2  y )y Volume benda putar 4

V  2  (8  4 y  2 y  y y )dy 0

 2 (8  4 y  2 y  y y )y

 2 (8 y  83 y 3 / 2  y 2  52 y 5 / 2 ) |04 


224 15



b. Sumbu putar x=3 (i) Metoda cincin

x=3

Jika irisan diputar terhadap garis x=3 diperoleh cincin dengan Jari-jari dalam =

Jari-jari luar = rl  3  y

y  x2

y

1 y

D

rd  1

Sehingga

3 y

V   ((3  y ) 2  (1) 2 )y

2

  (8  6 y  y )y

3 Volume benda putar 4

V    (8  6 y  y )dy   (8 y  4 y 3 / 2  8 |04 )  8 0


(ii) Metoda kulit tabung x=3 Jika irisan diputar terhadap garis x=3 diperoleh kulit tabung dengan yx

2

x

2

Tinggi = h =

x2

Jari-jari = r = Tebal = x

3-x

D

x 2 3-x

x 3

Sehingga

V  2 (3  x) x 2 x  2 (3x 2  x 3 )x

Volume benda putar 2

V  2  (3x 2  x 3 )dx  2 ( x 3  14 x 4 ) |02  2 (8  4)  8 0


Soal Latihan Hitung volume benda putar dari daerah yang terletak di kuadran pertama yang dibatasi oleh y2 = x3 , garis y = 8 dan sumbu Y, bila diputar mengelilingi 1. Sumbu Y 2. Sumbu X 3. Garis x = 4 4. Garis y = 8



Integration Danang Mursita

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