Hanyauntuklingkungansendiri

LEMBAR KERJA MAHASISWAINTEGRAL TAK TENTU FUNGSI ALJABAR

DAN INTEGRAL TAK TENTU FUNGSI TRIGONOMETRI

NAMA :NIM :KELAS :

Jelaskan yang ananda ketahui mengenai Kalkulus (Integral) dengan bahasa ananda

sendiri!

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Issac Newton

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Berikan

Gambar

Berikan

Gambar

Gottfried Wilhelm Leibniz

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Georf Friedrich Bernhard Riemman

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Materi :

Integral taktentuadalah proses untukmenentukan anti turunan yang umumdarisuatufungsi

yang diberikan.

Bentukumum integral

Keterangan : F ( x )=¿ fungsi integral umum

f ( x )=¿fungsi integran

C = konstanta

Misalkanf (x) dan g(x ) masing-masing adalah fungsi integran yang dapat di tentukan fungsi

integran umumnya dan c adalah konstanta real maka :

1. ∫1dx=x+c

2. ∫ k dx=kx+c

3. ∫ {f ( x )± g(x )}dx=∫ f ( x )dx±∫ g ( x )dx

4. ∫ k f ( x )dx=k∫ f ( x )dx

5. Dalamkasus n ≠ -1 maka

∫ xndx= 1n+1

xn+1+c

∫ kxndx= kn+1

xn+1+c

6. Dalamkasus n = -1

∫ 1xdx=ln|x|+c

∫ kx dx=k ln|x|+c

1. Tentukan integral-integral taktentuberikutini :

a. ∫12 x−7dx d. ∫ 5√x4dx

b. ∫ 3√x2−4√ x3dx e. ∫ 27√ x2

dx

c. ∫ 2x3 dx f. ∫−4

5√ x4dx

∫ f ( x )dx=F ( x )+c

Penyelesaian

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2. Tentukan integral-integral taktentuberikutini :

a. ∫ x ( x+4 ) ( x−1 )dx c. ∫u (√u+ 1√u )du

b. ∫ (x2−2 )3

x2 dx d. ∫ ( t+1 ) ( t−3 ) ( t−2 )√t

dt

Penyelesaian

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3. DiberikanF (x)=120 {x} ^ {2} - 30 yang merupakan turunan kedua dari F (x). Untuk

¿−1 , fungsi F (x) bernilai 14 dan untuk x=1, fungsi F (x) bernilai 6. Tentukan fungsi

F (x)!

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4. Turunankeduadarif (x) adalah f (x)=6x- jika grafik y=f (x ) melalui titik A (1, 6)

dan garis singgung y=f (x ) di titik A mempunyai gradient 6, tentukan fungsi f (x)!

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5. Sebuahbendabergerakdenganlajuv m/s. Pada saat t sekon laju benda dinyatakan dengan

persamaan v=8−2t . Pada saat t=3 sekon posisibendaberadapadajarak 45 meter

darititikasal. Carilahposisibendapadasaatt=1sekon!

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Rumus integral untukfungsitrigonometri :

∫cos xdx=sin x+c

∫sin x dx=−cos x+c

∫ sec2 x dx=tan x+c

∫ csc2 x dx=−cot x+c

∫ tan x sec x dx=sec x+c

∫cot x csc xdx=−csc x+c

∫cos ax dx=¿¿

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∫sin axdx=¿¿

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∫ sec2axdx=¿¿

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∫ csc2ax dx=¿¿

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∫ tan ax sec axdx=¿¿ ..........................................................................................................

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RumusKebalikan

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........... =

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RumusIdentitasPhytagoras

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RumusTrigonometriSudutRangkap

1. sin 2αSeperti yang sudah dipelajari sebelumnya bahwa : sin (α + β) = sin α cos β + cos α sin β.Untuk β = α, diperolehsin (α + α) = sin α cos α + cos α sin αsin 2 α = 2 sin α cos αJadi, sin 2α = 2 sin α cos α

2. cos 2αRumus lain yang juga telah dipelajari adalah : cos (α + β) = cos α cos β – sin α sin β.Untuk β = α, diperolehcos (α + α) = cos α cos α – sin α sin αcos 2α = cos2α – sin2αJadi, cos 2α = cos2α – sin2αUntuk rumus cos2α dapat juga dituliscos 2α = cos2α – sin2αcos 2α = (1 – sin2α) – sin2αcos 2α = 1 – 2 sin2αJadi, cos 2α = 1 – 2 sin2α

3. tan 2αRumus penjumlahan untuk tangen adalah : tan(α+β)=(tan α + tan β)/(1-tan α.tan β)

Untuk β = α, maka

tan(α+α)=(tan α + tan α)/(1-tan α.tan α) tan2α=(2 tan α)/(1- (tan)^2 α)

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RumusPerkalian Sinus danKosinus

Untuksetiapsudutαdan β sebarang, berlaku :

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1. Tentukan integral-integral taktentuberikutini :

a. ∫ {cos (3 x−2 )−9 sin (2−3x ) }dx d. ∫ x √x+2 cos x−3dx

b. ∫ 4 tan 23x sec 2

3xdx e. ∫16 cos 1

2x dx

c. ∫6 cot(3 x−π4 )csc(3 x−π4 )dxPenyelesaian :

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2. Tentukanhasilsetiap integral berikutini :

a. ∫ ( csc x−2 ) (csc x+2 )dx c. ∫( sin xcos x

+ 1cos x )

2

dx

b. ∫ (5−tan2 x )dx d. ∫{ sin xcos x

sec x−sec2x }dx................................................................................................................................................

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3. Hitunglahhasilsetiap integral berikutini :

a. ∫2sin 11 xcos5 x dx c. ∫6 cos8 xcos2 x dx

b. ∫cos 4 x sin 3x dx d. ∫ 4 sin 3 x sin(3x+ π3 )dx

Penyelesaian

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4. Carilahhasilsetiap integral berikutini :

a. ∫ (sin x+cos x )2dx c. ∫ 4 sin 2 xcos2 xdx

b. ∫sin 2 x dx d. ∫cos4 x dx

Penyelesaian

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