Bahan Ajar Kalkulus Integral

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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 dansebaliknya.

    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+ x22x 7 , maka F(x) = 9x2+ 2x2

    Jika f adalah fungsi yang didefinisikan sebagai

    f(x) = 9x2+ 2x2, maka f turunan dari F and F adalah anti turunan dari f

    Jika G(x) = 3x3+ x22x + 5, maka G juga anti turunan dari f karena G(x) = 9x2

    + 2x2

    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+ 2x2) = 3x3+ x22x + C

    Bagaimana dengan

    Ax(

    2

    1x3+ 2x22 x ) = ?

    Tentu bukan pekerjaan mudah menentukan suatu fungsi yang jika diturunkan

    berbentuk

    2

    1x

    3+ 2x

    22 x

    Untuk itu diperlukan notasi dan aturan-aturan yang dapat mempermudah

    menentukan anti turunan

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    Anti turunan dinotasikan sebagai .... dx

    dxxf )( = F(x) + C

    Teorema A (hal 301)

    dxxr

    = Cr

    xr

    1

    1

    untuk r bilangan rasional dan r - 1

    Sifat-sifat

    1. dx = x + C

    2. dxxaf )( = a dxxf )(

    3. dxxgxf ])()([ = dxxf )( + dxxg )( Dengan demikian

    Ax( 21 x3+ 2x22 x ) = ( 2

    1 x3+ 2x22 x ) dx

    = dxx 3

    2

    1+ dxx

    22 - dxx 21

    2

    =4

    4

    1.

    2

    1x + C1+

    3

    3

    1.2 x + C22( 2

    3

    3

    2x + C3)

    =

    4

    8

    1x

    +

    3

    3

    2x

    - C3

    4xx

    Soal-soal 5.1 halaman 307. Tugas individu

    Untuk dikerjakan hari kamis tanggal 14 Feb 2013

    No 226 no genap saja

    No 2750 semua

    FUNGSI-FUNGSI APA YANG DAPAT DIINTEGRALKAN?

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    Sebarang fungsi yang terintegralkan pada [a,b] harus terbatas di[a , b]. Yaitu: terdapat konstanta M sedemikian sehingga

    Mxf )(

    Contoh

    Lihat soal no 21 hal 347

    f(x) = x + sin x

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

    Jadi f(x) = x3+ sin x terintegralkan pada [-2 , 2]

    Perhatikan fungsi2

    1)(x

    xf pada selang [-2 , 2]

    merupakan fungsi tak terbatas.

    Tidak terdapat suatu bilangan M sedemikian sehingga

    2],[-2)( padaxsemuauntukMxf

    Dengan demikian2

    1)(

    xxf tidak terintegralkan pada

    -2 , 2

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    f(x) = x3+ sin x

    Ada M = sehingga Mxf )( 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 1)( xf 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 Msehinngga Mxf )(

    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 -

    tidak ada nilai Msehinngga Mxf )(

    jadi f(x) = 1/(x-1) tidak terintegralkan pada [-2 , 2]

    f(x) = tan x

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

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    f(x) = tan x

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

    tidak ada nilai Msehinngga Mxf )(

    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 Mxf )( untuk

    setiap x pada [-2 , 2]

    Jadi f terintegralkan pada [-2 , 2]

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    FUNGSI LOGARITMA ASLI

    f(x) =

    201

    022

    xjika

    xjikax

    Ada M = 4 sehingga Mxf )(

    untuk setiap x pada [-2 , 2]

    Jadi f terintegralkan pada [-2 , 2]

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    Teorema (hal 453)

    ln 1 = 0

    19) dxx

    xln2

    Misal ln x = u

    dudxx

    1

    dxx

    xln2= CxCuduu

    22 )(ln2

    20)

    dxxx 2)(ln

    1

    Misal ln x = u

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    dudxx

    1

    dx

    xx 2)(ln

    1= udu

    2

    = u-1+ C =u

    1+ C =

    xln

    1+ C

    Hitunglah

    FUNGSI BALIKAN/INVERS

    Misalkan y = f(x) = x3+ 1

    x = f-1

    (y) = 3 1y

    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) = x

    2 didefinisikan pada [0 , ]

    maka y = f(x) = x

    2

    mempunyai invers yaitu x = f

    -1

    (y) = y

    1

    2

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    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 dipertukarkan

    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

    y = f(x)

    y = f-1

    (x)

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    Grafik y = exdiperoleh dengan mencerminkan grafik y = ln x terhadap garis y = x

    (hal 468)

    y = f(x) = exdisebut fungsi eksponen asli

    Sifat-sifat:

    ln e = 1

    Contoh

    y = ln x

    x = ey

    y = ln x

    x = ey

    y = ex

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    Turunan dari ex

    Contoh

    xx edx

    dyey

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    CeCydxe

    dydxeedx

    dyey

    xx

    xxx

    ContohIntegralkan

    dxe x 12

    Misal 2x+1 = u2 dx = du

    dxe x 12 = CeCeude xuu

    12

    2

    1

    2

    1

    2

    1

    Fungsi eksponen umum

    ?dx

    dyay x

    aadx

    dy

    aedx

    dyeyay

    x

    axaxx

    ln

    ln.lnln

    ecara Umum

    Cedxe xx

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    Contoh

    Tentukanlahdx

    dy

    1) y = 52x+3

    (2) y =xx 62

    7

    Caa

    Cya

    dya

    dxa

    dydxaa

    dydxaaaadx

    dyay

    x

    x

    x

    xxx

    )ln

    1(

    ln1

    ln

    1

    ln

    lnln

    Contoh

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    Fungsi logaritma terhadap basis a

    a

    x

    xa

    x

    y

    a

    xyayxaxxy

    a

    ya

    ln

    ln

    logln

    ln

    ln

    lnlnlnlog

    ?log dx

    dy

    xy a

    axdx

    dy

    a

    xyxy a

    ln

    1

    ln

    lnlog

    u=sin x

    du = cos x dx

    Definisi (hal 479)

    ya axxy log

    a

    xxa

    ln

    lnlog

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    Fungsi Invers Trigonometri (hal 494)

    yarcxxy sinsin

    Bagaimana grafik fungsi y = arc sin x ?

    ?sin dxdyxarcy

    ydy

    dx

    yxxarcy

    cos

    sinsin

    1

    yy

    y

    x

    2

    2

    1

    1

    1cos

    xdx

    dy

    xydy

    dx

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    dydxx

    dydx

    x

    xdx

    dyxarcy

    2

    2

    2

    1

    11

    1

    1

    1sin

    Cx

    Cy

    dydxx

    arcsin

    1

    1

    2

    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 dxf(x)dF(x) CF(x) 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 dx

    dy= 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

    dx

    dy= g (x), then y = dxxg )( ,

    y = dx2x = x2+ C

    Cuduu

    arcsin1

    1

    2

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    Method 2

    Think

    dx

    dyas dy is divided by dx, so we can write

    dx2xdy

    Integrate two sides dy = dx2x y + C1= x

    2+ 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

    dx

    dy= 2x is called differential equation.

    Other examples of differential equation are

    dx

    dy= 2xy

    y dy = (x2+ 1) dx

    2

    2

    dx

    yd+ 2

    dx

    dy- 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

    dx

    dy=

    2

    2

    y

    3xx

    can be written as

    y2dy = (x + 3x

    2) dx

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

    dyy2

    = dx)3x(x 2

    3

    y3+ C1=

    2

    x 2+ x

    3+ C2

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    y3=

    2

    3x 2+ 3x

    3+ 3C23C1

    y3=

    23x

    2

    + 3x3+ C

    y =3 3

    2

    32

    3Cx

    x

    Supposed we have y= 6 for x = 0,

    then we can find C:

    6 =3 C

    C = 216

    Hence,

    y =3 3

    2

    21632

    3 x

    x

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

    differential equation becomes

    dx

    dy=

    3

    1 32

    32

    21632

    3

    x

    x(3x + 9x2)

    =

    3

    2

    32

    2

    21632

    3

    3

    xx

    xx

    and the right side of the differential equation becomes

    2

    23

    y

    xx =

    3

    2

    32

    2

    21633

    2

    3

    xx

    xx

    .

    Those give the same expression.

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    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 2x3. Find an equation ofthe curve

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

    dx

    dy= 2x3

    dy = (2x3)dx

    y = (2x3)dx = x23x + C

    The point (3,2) is on a curve, so 2 = 323.3 + C C = 2

    Hence the equation of the curve is y = = x23x + 2

    2. The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve

    Dx2y = 24x. Find an equation of the curve

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

    dx

    dy= (24x) dx = 2x2x2+ C1

    y = (2x2x2+ C1) dx = x2-3

    2x3+ C1x + C2

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

    3 = (-1)2-

    32 (-1)3+ C1(-1) + C2..(1)

    2 = (0)2-

    3

    2(0)3+ C1(0) + C2..(2)

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

    3

    2

    Hence the equation of the curve is

    y = x2-

    3

    2x3+

    3

    2x + 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

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

    dx

    dy= 3x

    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 meterswhen 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

    Solution

    Suppose V = f(h)

    dhdV = (2h+3)2 , V = (2h+3)

    2dh

    V =

    2

    1 (2h+3)2d(2h+3)

    =

    2

    1.

    3

    1(2h+3)

    3+ C

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

    2

    1.

    3

    1(2.0+3)

    3+ C,

    C = -

    2

    9

    V =

    6

    1(2h+3)3-

    2

    9

    If h = 3, V =

    6

    1(2.3+3)3-

    2

    9= 117

    The volume of water in the tank when the depth

    is 3 m = 117m3

    Evaluate

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    1. The point (3,2) is on a curve, and at any point (x,y) on the curve the tangentline has a slope equal to 2x3. 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

    Dx2

    y = 24x. 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 wateris 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 tangentline has a slope equal to 2x3. 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 curveDx

    2y = 24x. 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, Dx2

    y = 6x, find an equation of the curve

    4. The Volume of water in a tank is V cubic meters when the depth of the wateris 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 tangentline has a slope equal to 2x3. 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 curveDx

    2y = 24x. 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 hmeters . 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

    Introduction to Area

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    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].

    a bFig. 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 =

    n

    ab

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

    where x0= a , x1= a + x , xi= a + i 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 eachclosed subinterval.By the extreme-value theorem, there is a number in each bsubinterval for which fhas an absolute minimum value.

    In the ith subinterval, let this number be c i, so that f(ci) is the absolute minimumvalue of f on the subinterval [x i-1,xi].

    Consider n rectangles, each vhaving a width x units and an altitude f(ci) unitssee Fig 2.

    a x bFig. 1Let the sum of the areas of these n rectangle be given by Snsquare units, then

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

    =

    n

    i 1

    f(ci) 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 x =n

    ab , 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 ofthe area of region R is given by

    R

    Rf(ci)

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    A =

    nlim

    n

    iicf

    1

    )( x

    Example

    Find the area of the region bounded by the curve y = x

    2

    , the x axis, and the linex = 3 by taking inscribed rectangle.Solution

    Devide the interval [0,3] into n subinterval, each of length x;x0= 0 , x1= x, x2= 2 x, . . . , xi= i xxn-1= (n-1) x , xn= 3

    x =n

    03=

    n

    3

    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 =

    nlim

    n

    iixf

    11)( x

    Because xi-1= (i-1) x and f(x) = x2,f(xi-1) = [(i-1) x]2Therefore

    n

    iixf1 1)(

    x =

    n

    i i1

    2

    )1( ( x)3

    =

    n

    i nii

    13

    2 27)12(

    = .lanjutkanGunakan (pilih) rumus sbb

    2

    )1(

    1

    nni

    n

    i

    ;

    6

    )12)(1(

    1

    2

    nnni

    n

    i

    4

    )1( 22

    1

    3

    nni

    n

    i

    ;

    30

    196)(1( 23

    1

    4

    nnnnni

    n

    i

    THE DEFINITE INTEGRAL

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

    following limit:

    nlim

    n

    iicf

    1

    )( 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.

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    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, xnare not necessarily equidistant. Let i x be

    the length of the ith subinterval so that i x = xixi -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*) 1 x + f(x2

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

    *) n x =

    n

    iixf

    1

    *)( i x

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

    Friedrich Bernhard Riemann (18261866)

    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

    b

    a

    dxxf )( , is given by

    b

    a

    dxxf )( = lim0P

    n

    iixf

    1

    *)( 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

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    In the notation for the definite integral b

    a

    dxxf )( ,

    )(xf is called the integrand, a is called the lower limit, and b is called theupper limit.

    The symbol is called an integral sign

    Definition

    If a >b, then b

    a

    dxxf )( = - a

    b

    dxxf )(

    a

    a

    dxxf )( = 0

    THE FUNDAMENTAL THEOREM OF THE CALCULUS

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

    b

    a

    dxxf )( = F(b)F(a)

    We should write F(b)F(a) = baxF )(

    Example

    Evaluate 2

    1

    3dxx

    Solution

    2

    1

    3dxx =

    2

    1

    4

    4

    1

    x =

    4)2(4

    1-

    4)1(4

    1= 4 -

    4

    1= 3

    4

    3

    Properties of The Definite Integral

    Theorem

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    o b

    a

    dxxkf )( = b

    a

    dxxfk )(

    o b

    a

    dxxgxf )]()([ = b

    a

    dxxf )( + b

    a

    dxxg )(

    o b

    a

    dxxgxf )]()([ = b

    a

    dxxf )( - b

    a

    dxxg )(

    Kerjakan secara kelompok, 23 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, 23 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, 23 orang

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    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 , y = 5 sin 0 2

    5. x = 2 cos + cos(2 ) + 1 , y = 2 sin + sin (2 ) , 0