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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) = 3x 3 + x 2 – 2x – 7 , maka F’(x) = 9x 2 + 2x – 2 Jika f adalah fungsi yang didefinisikan sebagai f(x) = 9x 2 + 2x – 2, maka f turunan dari F and F adalah anti turunan dari f Jika G(x) = 3x 3 + x 2 – 2x + 5, maka G juga anti turunan dari f karena G’(x) = 9x 2 + 2x – 2 Secara umum, fungsi yang didefinisikan sebagai 3x 3 + x 2 – 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 A x (9x 2 + 2x – 2) = 3x 3 + x 2 – 2x + C 1

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Page 1: staffnew.uny.ac.idstaffnew.uny.ac.id/.../Bahan+Ajar+Kalkulus+Integral.docx · Web viewBahan Ajar KALKULUS INTEGRAL Oleh: ENDANG LISTYANI ANTI TURUNAN (hal 299) Anda tentu sudah mengenal

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.

DefinisiSuatu 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 TurunanAx(9x2 + 2x – 2) = 3x3 + x2 – 2x + C

Bagaimana dengan

Ax(

12 x3 + 2x2 – 2√ x ) = ?

Tentu bukan pekerjaan mudah menentukan suatu fungsi yang jika diturunkan

berbentuk

12 x3 + 2x2 – 2√ xUntuk itu diperlukan notasi dan aturan-aturan yang dapat mempermudah

menentukan anti turunan

1

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Anti turunan dinotasikan sebagai ∫ .... dx

∫ f ( x ) dx = F(x) + C

Teorema A (hal 301)

∫ xr dx =

xr+1

r+1+C

untuk r bilangan rasional dan r ≠ - 1Sifat-sifat

1. ∫ dx = x + C

2. ∫ af ( x )dx = a∫ f ( x )dx

3. ∫ [ f ( x )+g( x ) ]dx = ∫ f ( x )dx + ∫ g (x )dxDengan demikian

Ax(

12 x3 + 2x2 – 2√ x ) = ∫ (

12 x3 + 2x2 – 2√ x ) dx

=

12∫ x

3dx +

2∫ x2 dx - 2∫ x

12dx

=

12

. 14x4

+ C1 + 2 . 1

3x3

+ C2 – 2(

23x

32

+ C3 )

=

18x4

+ 2 . 1

3x3

-

43x√ x+C

Soal-soal 5.1 halaman 307. Tugas individuUntuk dikerjakan hari kamis tanggal 14 Feb 2013No 2 – 26 no genap sajaNo 27 – 50 semua

FUNGSI-FUNGSI APA YANG DAPAT DIINTEGRALKAN?

2

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Sebarang fungsi yang terintegralkan pada [a,b] harus terbatas di [a , b]. Yaitu: terdapat konstanta M sedemikian sehingga |f ( x )|≤MContoh

Lihat soal no 21 hal 347

3

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 ]

Dengan demikian f ( x )= 1

x2 tidak terintegralkan pada

[-2 , 2]

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4

f(x) = x3 + sin x

Ada M = 9 sehingga |f ( x )|≤M untuk setiap x pada [-2 , 2]

Jadi f(x) = x3 + sin x 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]

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]

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f(x) = x3 + sin x

Ada M = sehingga |f ( x )|≤M untuk setiap x pada [-2 , 2]

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]

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)

5

f(x) = tan x

pada [-2 , 2] maksimum dan minimum di ∞ dan di -∞

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(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]

FUNGSI LOGARITMA ASLI

6

f(x) = sin(1/x)

Ada M =..... sehingga |f ( x )|≤M untuk setiap x pada [-2 , 2]

Jadi f terintegralkan pada [-2 , 2]

f(x) = {x2 jika −2≤x≤0

1 jika 0<x≤2

Ada M = 4 sehingga |f ( x )|≤M untuk setiap x pada [-2 , 2]

Jadi f terintegralkan pada [-2 , 2]

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Teorema (hal 453)

ln 1 = 0

19) ∫ 2 ln x

xdx

Misal ln x = u1xdx=du

7

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∫ 2 ln xxdx

= 2∫ u du=u2+C=( ln x )2+C

20) ∫ −1x ( ln x )2

dx

Misal ln x = u

1xdx=du

∫ −1x ( ln x )2

dx = −∫u−2 du

= u-1 + C =

1u + C =

1ln x + C

Hitunglah

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

8

1

2)

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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

y = f(x) mempunyai invers jika y merupakan fungsi satu-satu atau fungsi monotonJika 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 menggunakan bentuk y = f-1(x). Perhatikan bahwa posisi x dan y dipertukarkanDengan demikian grafik fungsi y = f-1(x) dapat diperoleh dengan mencerminkan grafik y = f(x) terhadap garis y = x y = 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 = f(x)

y = f-1(x)

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Grafik y = ex diperoleh dengan mencerminkan grafik y = ln x terhadap garis y = x (hal 468)

y = f(x) = ex disebut fungsi eksponen asli

Sifat-sifat:

ln e = 1

Contoh

10

y = ln xx = ey

y = ln xx = ey

y = ex

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Turunan dari ex

Contoh

11

y=ex⇒ dydx

=ex

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y=ex⇒dydx

=ex⇒ exdx=dy

∫ exdx= y+C= ex+C

ContohIntegralkan

∫ e2 x+1dx

Misal 2x+1 = u2 dx = du

∫ e2 x+1dx =

12∫ e

udu=12eu+C=1

2e2 x+1+C

Fungsi eksponen umum

y=ax⇒ dydx

=?

y=ax⇔ y=ex ln a⇒dydx

=ex ln a . ln a

dydx

=ax ln a

ecara Umum

12

∫ exdx=e x+C

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Contoh

Tentukanlah

dydx

1) y = 52x+3 (2) y = 7x2−6 x

y=ax⇒ dydx

=ax ln a⇒ax ln a dx=dy

∫ ax ln a dx=∫ dy∫ ax dx=1

ln a ∫dy

=1ln a

y+C

=(1ln a

)ax+C

Contoh

Fungsi logaritma terhadap basis a

13

u=sin xdu = cos x dx

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y= loga x⇔ x=a y⇒ ln x= y ln a⇒ y=ln xln a

y= ln xln a

⇒ loga x=ln xln a

y=loga x⇒dydx

=?

y=loga x⇒ y=ln xln a

⇒ dydx

= 1x ln a

Fungsi Invers Trigonometri (hal 494)

y=sin x⇔ x=arc sin y

Bagaimana grafik fungsi y = arc sin x ?

14

Definisi (hal 479)

y=loga x⇔ x=a y

log a x=ln xln a

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y=arc sin x⇒ dydx

=?

y=arc sin x⇒ x=sin ydxdy

=cos y

1

yy

y=arc sin x⇒dydx

=1√1− x2

1√1−x2

dx=dy

¿

∫1√1−x2

dx=∫ dy ¿ ¿∫ ¿ ¿

15

y

x

dxdy

=cos y=√1−x2

dydx

=1√1−x2

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∫∫ 1√1−x2

dx=∫dy ¿ = y+C ¿ =arcsin x+C ¿¿∫ ¿

Introduction to Differential EquationWe know that the expression F’(x) = f(x) is equivalent with dF(x) = f(x) dx,

so we can write ∫ 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

dydx = 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

dydx = g (x), then y = ∫ g (x ) dx ,

y = ∫2x dx = x2 + C

Method 2

Think

dydx as dy is divided by dx, so we can write

dy=2x dx

Integrate two sides ∫ dy = ∫2x dx

16

∫ 1√1−u2

du=arcsinu+C

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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

dydx = 2x is called differential equation.

Other examples of differential equation are

dydx = 2xy

y dy = (x2 + 1) dx

d2 ydx2

+ 2

dydx - 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

dydx =

x+3x2

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

∫ y2 dy= ∫( x+3x 2 ) dx

y3

3 + C1 =

x2

2 + x3 + C2

y3 =

3x2

2 + 3x3 + 3C2 – 3C1

y3 =

3x2

2 + 3x3 + C

17

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y =

3√ 3 x2

2+3 x3+C

Supposed we have y = 6 for x = 0,

then we can find C:

6 = 3√C

C = 216

Hence,

y =

3√ 3 x2

2+3 x3+216

Check this result by substituting it in differential equation. The left side of the

differential equation becomes

dydx =

13 ( 3 x2

2+3 x3+216)−2

3

(3x + 9x2)

=

x+3x2

( 32 x

2+3 x3+216)23

and the right side of the differential equation becomes

x+3 x2

y2 =

x+3x2

( 23 x

2+3 x3+216)23

.

Those give the same expression.

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

18

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SolutionSuppose y = f(x) is the equation of the curve

dydx = 2x – 3

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

SolutionSuppose y = f(x) is the equation of the curve

dydx = ∫ (2 – 4x) dx = 2x – 2x2 + C1

y = ∫ (2x – 2x2 + C1) dx = x2 - x3 + C1x + C2

The points (-1,3) and (0,2) are on a curve so,

3 = (-1)2 - (-1)3 + C1(-1) + C2 …………..(1)

2 = (0)2 - (0)3 + C1(0) + C2 ……………..(2)

From (1) and (2) : C2 = 2 , C1 =

Hence the equation of the curve is

y = x2 - x3 + x + 2

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

SolutionSuppose y = f(x) is the equation of the curve

dydx = 3x2 + C1

19

32

32

32

32

32

32

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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 = π (2h+3)2, find the volume of water in the tank

when the depth is 3 m

SolutionSuppose V = f(h)

dVdh = π (2h+3)2 , V = ∫ π (2h+3)2 dh

V =

12∫ π (2h+3)2 d(2h+3)

=

12 .

13 π (2h+3)3 + C

When h = 0, V = 0, so 0 =

12 .

13 π (2.0+3)3 + C,

C = -

92 π

V =

16 π (2h+3)3 -

92 π

If h = 3, V =

16 π (2.3+3)3 -

92 π = 117π

The volume of water in the tank when the depth

is 3 m = 117π m3

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 Dx

2y = 2 – 4x. Find an equation of the curve

20

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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, Dx

2y = 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 = π

(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 Dx

2y = 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, Dx

2y = 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 = π

(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 Dx

2y = 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, Dx

2y = 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 = π

(2h+3)2, find the volume of water in the tank when the depth is 3 m

Introduction to Area

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].

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a b Fig. 1

Divide the closed interval [a,b] into n subintervalsFor simplicity, now we take each of these subintervals as being of equal length,

for instance, Δ x. There for Δ x =

b−an

Denote the endpoints of these subintervals by x0 , x1 , x2, . . . , xn-1 , xn

where x0 = a , x1 = a +Δ x , xi = a + iΔ xxn = b

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 c i , so that f(ci) is the absolute minimum value of f on the subinterval [xi-1,xi].

Consider n rectangles, each vhaving a width Δ x units and an altitude f(c i) units see Fig 2.

a Δ x b Fig. 1Let the sum of the areas of these n rectangle be given by Sn square units, then

Sn = f(c1)Δ x + f(c2)Δ x + . . . + f(cn)Δ x

= ∑i=1

n

f(ci)Δ x …………….(*)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 lengthΔ x =

b−an , and denote the ith subinterval by [xi-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

A = limn→∞

∑i=1

n

f (c i )Δ x

ExampleFind the area of the region bounded by the curve y = x2, the x axis, and the line

22

R

R f(ci)

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x = 3 by taking inscribed rectangle.Solution

Devide the interval [0,3] into n subinterval, each of length Δ x;

x0 = 0 , x1 = Δ x, x2 = 2Δ x, . . . , xi = iΔ x

xn-1= (n-1) Δ x , xn = 3

Δ x =

3−0n =

3n

Because f is increasing on [0,3], the absolute minimum value of f on the ith subinterval [xi-1,xi] is f(xi-1)

There for A = limn→∞

∑i=1

n

f ( x i−1 )Δ x

Because xi-1 = (i-1)Δ x and f(x) = x2,

f(xi-1) = [(i-1)Δ x]2

Therefore

∑i=1

n

f ( x i−1 )Δ x =

∑i=1

n

( i−1)2

(Δ x)3

=

∑i=1

n

( i2−2 i+1)27n3

= ………….lanjutkanGunakan (pilih) rumus sbb

∑i=1

n

i=n (n+1)2

; ∑i=1

n

i2= n(n+1 )(2n+1 )6

∑i=1

n

i3=n2( n+1)2

4 ;

∑i=1

n

i4=n (n+1)(6n3+9n2+n−130

THE DEFINITE INTEGRAL

In the preciding section the measure of the area of a region was defined as the

following limit:

limn→∞

∑i=1

n

f (c i )Δ x ……………………..(*)

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.

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.

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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 Δi x be

the length of the ith subinterval so that Δi x = xi – xi -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 x i

* be the point chosen

in [xi -1 , xi]

Form the sum f(xi*)Δ1 x + f(x2

*)Δ2 x + . . . + f(xi

*)Δi x + . . . + f(xn

*)Δn x =

∑i=1

n

f ( x i¿ )Δi x

Such a sum is called a Riemann sum, named for the mathematician George

Friedrich Bernhard Riemann (1826 – 1866)

Y

x2* x3

* xn*

x1* xi

* X

DefinitionIf f is a function defined on the closed interval [a,b], then the definite integral of f

from a to b, denote by

∫a

b

f ( x ) dx, is given by

∫a

b

f ( x ) dx = lim‖P‖→ 0

∑i=1

n

f ( x i¿ )Δi x if the limit exists

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

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In the notation for the definite integral

∫a

b

f ( x ) dx,

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

∫a

b

f ( x ) dx = -

∫b

a

f ( x ) dx

∫a

a

f ( x ) dx = 0

THE FUNDAMENTAL THEOREM OF THE CALCULUS

TheoremIf f is a continuous function on the closed interval [a,b], and F is an any antiderivative of f on [a,b], then

∫a

b

f ( x ) dx = F(b) – F(a)

We should write F(b) – F(a) = [F (x )]ab

Example

Evaluate

∫1

2

x3dx

Solution

∫1

2

x3dx = [ 1

4x 4]1

2

=

14(2 )4

-

14(1 )4

= 4 -

14 = 3

34

Properties of The Definite Integral

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Theorem

o

∫a

b

kf ( x ) dx =

k∫a

b

f ( x ) dx

o

∫a

b

[ f ( x ) +g( x ) ]dx =

∫a

b

f ( x ) dx +

∫a

b

g (x ) dx

o

∫a

b

[ f ( x ) −g ( x ) ]dx =

∫a

b

f ( x ) dx -

∫a

b

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 cosθ , y = 5 sinθ 0¿ θ ¿ 2π

5. x = 2 cosθ + cos(2θ ) + 1 , y = 2 sinθ + sin (2θ ) , 0¿ θ ¿ π

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 cosθ , y = 5 sinθ 0¿ θ ¿ 2π

5. x = 2 cosθ + cos(2θ ) + 1 , y = 2 sinθ + sin (2θ ) , 0¿ θ ¿ π

26

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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 cosθ , y = 5 sinθ 0¿ θ ¿ 2π

5. x = 2 cosθ + cos(2θ ) + 1 , y = 2 sinθ + sin (2θ ) , 0¿ θ ¿ π

27