Pertmuan 10 (Integral Tak Tentu)
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Transcript of Pertmuan 10 (Integral Tak Tentu)
INTEGRAL TAK TENTU(lanjutan)
OLEH:DEDEH HODIYAH
1. Integral Eksponensial
1. exdx = ex + c
2. ekxdx = 1/k ekx + c, k0
Contoh:
1. 9exdx = 9ex + c2. e9xdx = 1/9 e9x + c3. 3e5/4xdx = 3e5/4xdx
= 3[4/5 e5/4x] + c= 12/5 e5/4x + c
2. Integral Logaritma
1. 1/x dx = ln|x|+c
2. lnx dx = x lnx-x + c
Contoh:
1. 4/x dx = 41/x dx = 4 ln |x| + c
2. (-5/x dx + e-2x) = -5 ln |x|- ½ e-2x + c
3. x e-3x2 dx = -1/6e-3x2 + c
3. Integral Trigonometri
1. cos x dx = sin x + c2. sin x dx = -cos x + c3. sec2 x dx = tan x + c4. cosec2x dx = -cot x + c
Contoh:
1. 2 cosx dx = 2cos x dx = 2sinx + c2. (x-3sinx)dx = x - 3sinx dx
= 1/2x2 – 3(-cosx) + c= 1/2x2 + 3cosx + c
3. (2cosx + 3x2)dx = 2sinx +x3 + c
Integral Substitusi
Teorema:Andaikan g suatu fungsi yang terdeferensial dan andaikan F adalah suatu anti turunan dari f maka jika u = g(x), berlakuf(g(x))g’(x)dx = f(u)du
= F(u) + c = F(g(x)) + c
Contoh :
1. (x2-1)4 10x dx = Mis u = x2-1du/dx = 2x
dx = du/2x
(x2-1)4 10x dx = u410x du/2x= u4. 5. du=5 u4 du= 5.[1/5u5]+c= u5 + c= (x2 – 1)5 + c
2. (x4-x2)3(8x3- 4x)dx=Mis: u = x4-x2
du/dx = 4x3 – 2x
dx = du/4x3- 2x
(x4-x2)3(8x3- 4x)dx
= u32(4x3 – 2x)du/ 4x3 – 2x= 2u3du= 2[1/4u4] + c= ½[x4-x2]4 + c
3. sin ax dxMis: u = axdu/dx = a
dx = du/a
sin ax dx =sin u du/a= 1/asin u du= 1/a[-cos u] + c= - 1/acos ax + c
4. sin (ax+b) dx =
Mis: u = ax+b
du/dx = adx = du/a
sin (ax+b) dx =sin u du/a= 1/asin u du= 1/a[-cos u] + c= - 1/acos (ax+b) + c
Rumus-rumus :
1. sin ax dx = -1/a cos ax + c2. cos ax dx = 1/a sin ax + c3. sin (ax+b) dx = -1/a cos (ax+b) + c4. cos (ax+b) dx = 1/a sin (ax+b) + c
Contoh:
5. sin2x dx = 1/2(1-cos 2x)dx= 1/2(1-cos 2x) dx= ½[x – 1/2sin 2x]+c= 1/2x-1/4sin2x + c
6. dx/x-5 =
mis : u = x-5du/dx = 1dx = du
dx/x-5 = du/u= 1/u du= ln |u| + c= ln(x-5)+ c
7. x sin3 (x2)cos(x2) dxmis: u = sin(x2)
du/dx = 2x cos x2
dx = du/2x cos x2
x sin3 (x2)cos(x2) dx = x u3 cos x2du/2x cos x2
= ½ u3du= ½ [1/4 u4] + c= 1/8 u4 + c= 1/8 (sin(x2))4 + c
9. 2
2
9 x
x
INTEGRAL PARSIAL
RUMUS :
u dv = uv- v du
Contoh:
12x(2x – 3)3 dxMis: u = 12x
du = 12 dxdv = (2x-3)3 dxdv = (2x-3)3 dx v = 1/8 (2x – 3)4
+ c
12x(2x – 3)3 dx= uv- v du
= 12x(1/8(2x-3)4) - 1/8(2x-3)4.12 dx= 3/2x(2x-3)4 – 3/2(2x -3)4dx= 3/2x(2x-3)4 – 3/2(1/10(2x-3)5) +c= 3/2x(2x-3)4 – 3/20(2x -3)5 + c
2. 12x sin 2xdxMis: u = 12x
du/dx = 12du = 12 dx
dv = sin 2x dxdv = sin 2x dx v = -1/2 cos 2x
12x sin 2xdx = uv - v du
= 12x(-1/2 cos 2x) +1/2cos 2x 12 dx= - 6x cos 2x + 6 cos 2x dx= - 6 cos 2x + 6 (1/2 sin 2x) + c= - 6 cos 2x + 3 sin 2x + c
Soal Latihan:
1. 6x(x+2)5dx
2.
3. x cos 2x dx
4. 6xsin3xdx
dxxx 42
SEKIAN
TERIMA KASIH