Bahan Ajar Kalkulus Integral

(1)

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_{+ x}2_{– 2x – 7 , maka F’(x) = 9x}2_{+ 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_{+ x}2_{– 2x + 5, maka G juga anti turunan dari f karena G’(x) = 9x}2

+ 2x – 2

Secara umum, fungsi yang didefinisikan sebagai 3x3_{+ 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

Ax(9x2 + 2x – 2) = 3x3 + x2 – 2x + C

Bagaimana dengan

Ax(

1

2

_x3_{+ 2x}2_{– 2}

√

x

_{) = ?}

Tentu bukan pekerjaan mudah menentukan suatu fungsi yang jika diturunkan berbentuk

1

2

_x3_{+ 2x}2_{– 2}

√

x

(2)

Anti turunan dinotasikan sebagai

∫

.... dx

∫

f

(

x

)

dx

_{= F(x) + C}

Teorema A (hal 301)

∫

x

r

dx

=

x

r

+

1

r

+

1

+

C

_{untuk r bilangan rasional dan r ≠ - 1}

Sifat-sifat

1.

∫

dx

= x + C

2.

∫

af

(

x

)

dx

= a

∫

f

(

x

)

dx

3.

∫

[

f

(

x

)+

g

(

x

) ]

dx

=

∫

f

(

x

)

dx

+

∫

g

(

x

)

dx

Dengan demikian

Ax(

1

2

_x3_{+ 2x}2_{– 2}

√

x

_{) =}

∫

₍

1

2

_x3_{+ 2x}2_{– 2}

√

x

₎_dx

=

1

2

∫

x

3

dx

+

2

∫

x

2

dx

-

2

∫

x

1 2

dx

=

1

2

.

1

4

x

4

+ C

1

+

2.

1

3

x

3

+ C

2

– 2(

2

3

x

3 2

+ C

3

)

=

1

8

x

4

+

2.

1

3

x

3

-

4

3

x

√

x

+

C

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

(3)

Sebarang fungsi yang terintegralkan pada [a,b] harus terbatas di

[a , b]. Yaitu: terdapat konstanta M sedemikian sehingga

|

f

(

x

)|≤

M

Contoh

Lihat soal no 21 hal 347

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]}

Perhatikan fungsi

f

(

x

)=

1

x

2 _{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

(4)

f(x) = x3_{+ sin x}

Ada M = sehingga

|

f

(

x

)|≤

M

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) = 1/(x-1)

pada [-2 , 2] maksimum dan minimum di ∞ _{dan di} -∞

|

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) = 1/(x+3)

Ada M = 1 sehingga

|

f

(

x

)|≤

1

(5)

f(x) = tan x

pada [-2 , 2] maksimum dan minimum di ∞ _{dan di -} ∞

|

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

(6)

FUNGSI LOGARITMA ASLI

f(x) =

{

x

2

jika

−

2

≤

x

≤

0

1

jika

0

<

x

≤

2

Ada M = 4 sehingga

|

f

(

x

)|≤

M

(7)

Teorema (hal 453)

ln 1 = 0

19)

∫

2ln

x

dx

Misal ln x = u

1

x

dx=du

∫

2ln

x

dx

=

2

∫

u du=u

2

+C

=(

ln

x

)

2

+C

20)

∫

−

1

(8)

1

x

dx=du

∫

−

1

x

(

ln

x

)

2

dx

₌

−

∫

u

−2

du

= u-1_{+ C =}

1

u

_{+ C =}

1

ln

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

x =

±

√

y

bukan fungsi , jadi y = f(x) = x2_{tidak mempunyai balikan/INVERS.}

Tetapi jika domainnya di batasimisalnya y = f(x) = x2_{didefinisikan pada [0 ,} _∞ _]

maka y = f(x) = x2_{mempunyai invers yaitu x = f}-1_{(y) =}

√

y

1

(9)

[image:9.612.90.187.70.224.2]

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 menggunakan bentuk y = f-1_{(x). Perhatikan bahwa posisi x dan y dipertukarkan}

[image:9.612.90.329.376.585.2]

Dengan 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

(10)

[image:10.612.87.501.68.335.2]

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

y = ln x x = ey

(11)

Turunan dari e

x

Contoh

y

=

e

x

⇒

dy

dx

=

e

(12)

y

=

e

x

⇒

dy

dx

=

e

x

⇒

e

x

dx

=

dy

∫

e

x

dx

=

y

+

C

=

e

x

+

C

Contoh Integralkan

∫

e

2x+1

dx

Misal 2x+1 = u 2 dx = du

∫

e

2x+1

dx

=

1

2

∫

e

u

d u=

1

2

e

u

+C=

1

2

e

2x+1

+

C

Fungsi eksponen umum

y=

a

x

⇒

dy

dx

=?

y=

a

x

⇔

y

=e

xlna

⇒

dy

dx

=e

xlna

. ln

a

dy

dx

=a

x

ln

a

ecara Umum

(13)

Contoh

Tentukanlah

dy

dx

1) y = 52x+3 _{(2) y =}

7

x

2₋₆_x

y

=

a

x

⇒

dy

dx

=

a

x

ln

a

⇒

a

x

ln

a dx

=

dy

∫

a

x

ln

a dx

=

∫

dy

∫

a

x

dx

=

1

ln

a

∫

dy

=

1

ln

a

y

+

C

=(

1

ln

a

)

a

x

+

C

Contoh

Fungsi logaritma terhadap basis a

(14)

y=

log

_a

x

⇔

x=

a

y

⇒

ln

x=

y

ln

a

⇒

y=

ln

x

ln

a

y=

ln

x

ln

a

⇒

log

a

x=

ln

x

ln

a

y

=

log

_a

x

⇒

dy

dx

=?

y

=

log

_a

x

⇒

y=

ln

x

ln

a

⇒

dy

dx

=

1

x

ln

a

Fungsi Invers Trigonometri (hal 494)

y

=

sin

x

⇔

x

=

arc

sin

y

Bagaimana grafik fungsi y = arc sin x ? Definisi (hal 479)

y=

log

_a

x

⇔

x

=

a

y

log

_a

x

=

ln

x

(15)

y=

arc

sin

x

⇒

dy

dx

=?

y

=

arc

sin

x

⇒

x

=

sin

y

dx

dy

=

cos

y

1

yy

¿

1

√

1

−x

2

dx=

∫

dy

y=arc

sin

x⇒

dy

dx

=

1

√

1

−x

2

1

√

1

−x

2

dx

=dy

∫

¿ ¿ ¿

∫

¿ ¿

y x

dx

dy

=

cos

y

=

√

1

−

x

2

dy

dx

=

1

(16)

∫

√

1

1−x2 dx=

∫

dy

=y+C =arcsin x+C

∫

¿ ¿ ¿ ¿

∫

¿

Introduction to Differential Equation

We 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

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

dx

₌_g₍_x_{), then y =}

∫

g

(

x

)

dx

_,

y =

∫

2x dx

= x2_{+ C}

Method 2

Think

dy

dx

_{as dy is divided by}_{dx, so we can write}

dy

=

2x dx

Integrate two sides

∫

dy

=

∫

2x dx

∫

1

(17)

y + C1 = x2 + C2

y = x2_{+ C}

2 - 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

dx

_{= 2x}_{is called differential equation.}

Other examples of differential equation are

dy

dx

_{= 2xy}

y dy = (x2_{+ 1) dx}

d

2

y

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

dx

₌

x

+

3x

2

y

2

can be written as

y2_{dy = (x + 3x}2_{) dx}

Here, x and y term are separated. To solve this equation, we use method 2

∫

y

2

dy

=

∫

(

x

+

3x

2

)

dx

y

3

_{+ C}₁₌

x

2

_{+ x}3_{+ C} 2

y3₌

3x

2

_{+ 3x}3_{+ 3C}

2 – 3C1

y3₌

3x

2

(18)

y =

3

√

3

x

2

+

3

x

3

+

C

Supposed we have y = 6 for x = 0, then we can find C:

6 =

3

√

C

C = 216 Hence,

y =

3

√

3

x

2

+

3

x

3

+

216

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

dy

dx

₌

1

3

(

3

x

2

+

3

x

3

+

216

)

−

2

3

(3x + 9x2₎

=

x

+

3

x

2

(

3

2

x

2

+

3

x

3

+

216

)

2

3

and the right side of the differential equation becomes

x

+

3

x

2

y

2

₌

x

+

3

x

2

(

2

3

x

2

+

3

x

3

+

216

)

2

3

. 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

(19)

Solution

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

dy

dx

_{= 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

Solution

dy

dx

₌

∫

_{(2 – 4x) dx = 2x – 2x}2_{+ C} 1

y =

∫

(2x – 2x2_{+ C}

1) dx = x2 - x3 + C1x + C2

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

3 = (-1)2_- _(-1)3_{+ C}

1(-1) + C2 …………..(1)

2 = (0)2_- ₍₀₎3_{+ C}

1(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, Dx2y = 6x, find an

equation of the curve

Solution

dy

dx

_{= 3x}2_{+ C} 1

3

2

3

2

3

2

3

2

3

2

(20)

y = x3_{+ C}

1x + C2

The slope of the tangent line to the curve at the point (1,3) is1, so 3(1)2_{+ C}

1 = 1, C1 = -2

The point (1,3) is on the curve, so 3 = 13_{+ (-2)(1) + C}

2 , 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

Solution

Suppose V = f(h)

dV

dh

₌

π

_(2h+3)2_{, V =}

∫

π

_(2h+3)2_dh

V =

1

2

∫

π

_(2h+3)2_d(2h+3)

=

1

2

_.

1

3

π

_(2h+3)3_{+ C}

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

1

2

_.

1

3

π

_(2.0+3)3_{+ C,}

C = -

9

2

π

V =

1

6

π

_(2h+3)3_-

9

2

π

If h = 3, V =

1

6

π

_(2.3+3)3_-

9

2

π

_{= 117}

π

The volume of water in the tank when the depth

is 3 m = 117

π

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

(21)

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 =

π

(2h+3)2_{, find}

the volume of water in the tank when the depth is 3 m

Evaluate

π

(2h+3)2_{, find}

Evaluate

π

(2h+3)2_{, find}

Introduction to Area

(22)

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,

Δ

x. There for

Δ

x =

b

−

a

n

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

where x0 = a , x1 = a +

Δ

x , xi = a + i

Δ

x

xn = 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 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

Δ

x units and an altitude f(ci)

units see Fig 2.

a

Δ

x b Fig. 1

Let 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 length

Δ

x =

b

−

a

n

_{, and denote the ith subinterval by [x}_i-1_,x_i_{]. 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 =

lim

n→∞

∑

i=1 n

f

(

c

_i

)

Δ

_x Example

R

(23)

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

Δ

x; x0 = 0 , x1 =

Δ

x, x2 = 2

Δ

x, . . . , xi = i

Δ

x

xn-1= (n-1)

Δ

x , xn = 3

Δ

_{x =}

3

−

0

n

₌

3

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)

There for A =

lim

n→∞

∑

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

(

i

2

−

2

i

+

1

)

27

n

3

= ………….lanjutkan

Gunakan (pilih) rumus sbb

∑

i=1 n

i

=

n

(

n

+

1

)

2

_;

∑

_i₌₁

n

i

2

=

n

(

n

+

1

)(

2

n

+

1

)

6

∑

i

=

1

n

i

3

=

n

2

(

n

+

1

)

2

4

_;

∑

i

=

1

n

i

4

=

n

(

n

+

1

)(

6

n

3

+

9

n

2

+

n

−

1

30

THE DEFINITE INTEGRAL

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

lim

n→∞

∑

i=1 n

f

(

c

_i

)

(24)

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.

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

Δ

i x be

the length of the ith subinterval so that

Δ

i 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 xi* be the point chosen in

[xi -1 , xi]

Form the sum f(xi*)

Δ

1 x + f(x2*)

Δ

2 x + . . . + f(xi*)

Δ

i x + . . . + f(xn*)

Δ

n

x =

∑

i=1 n

f

(

x

_i¿

)

Δ

_i _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

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

∫

a

b

f

(

x

)

dx

(25)

∫

a

b

f

(

x

)

dx

=

lim

‖P‖→0

∑

i=1 n

f

(

x

_i¿

)

Δ

_i

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

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

f

(

x

)

dx

= 0

THE FUNDAMENTAL THEOREM OF THE CALCULUS

Theorem

If 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

)

]

a

b

(26)

Evaluate

∫

1

2

x

3

dx

Solution

∫

1

2

x

3

dx

=

[

1

4

x

4

]

1

2

=

1

4

(

2

)

4

-

1

4

(

1

)

4

= 4 -

1

4

_{= 3}

3

4

Properties of The Definite Integral

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_{= 8x}2_{, 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

θ

, y = 5 sin

θ

0 ¿

θ

¿ 2 π

5. x = 2 cos

θ

+ cos(2

θ

) + 1 , y = 2 sin

θ

+ sin (2

θ

) , 0 ¿

θ

¿

π

Jawab

1. y3_{= 8x}2_{, 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}

(27)

5. x = 2 cos

θ

+ cos(2

θ

) + 1 , y = 2 sin

θ

+ sin (2

θ

) , 0 ¿

θ

¿ π

1. y3_{= 8x}2_{, 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

θ

, y = 5 sin

θ

0 ¿

θ

¿ 2 π

5. x = 2 cos

θ

+ cos(2

θ

) + 1 , y = 2 sin

θ

+ sin (2

θ

) , 0 ¿

θ

¿