Integral

Integral---

Integral sebenernya adalah kebalikan dari turunan..
Jika f(x) = ax^n, maka int f(x) dx = a/(n+1) . x^(n+1)
Biar lbh mudah inget, ingat ini aja.."Pangkat ditambah 1, terus koefisien dibagi pangkat"



Perhatikan ilustrasi ini.. :)
d(2x"+5)/dx = 4xd(2x"-1)/dx = 4xd(2x"+10)/dx = 4x
Karena integral adlh kebalikan dr turunan, maka..
int 4x dx = 2x"+C
C adalah konstanta..
Kenapa harus +C ? Krn konstanta berapapun jika diturunkan akan nol..

Latihan! :D
1. 2x + C
2. 2x"+5x + C
3. x"'-(1/2)x" + C
4. (4/3)x"' - 4x + C
5. x""/4 - x"' + (3/2)x" - x + C
 Jangan lupa, tiap akhir harus dikasih +C, karena dia integral tak tentu.. :)

Integral fungsi trigonometri..
Jika..f(x) = sinx
Maka..
-> f'(x) = cosx
-> f"(x) = -sinx
Nah, integral adalah kebalikan turunan..
int cosx dx = sinx + C
int sinx dx = -cosx + C

Nih, coba lagi yuk.. :D
1. int -sinx dx = ?
2. int cos2x dx = ?
3. int sin(5x+3) dx = ?
4. int 2x.cos(5x"-2) = ?
5. int sinx.cosx dx = ?

Lanjut, integral subtitusi.. :D
Contoh soal no.5
int sinx.cosx dx = ?
Penyelesaian :
Misal : sinx = u
Maka :
du/dx = cosx
du = cosx dx
Subtitusi semuanya..
int sinx.cosx dx= int u du= 1/2(u^2) + C
Subtitusi u = sinx= 1/2(sin"x) + C
Gimana ? :D

Semangat.. :D
1. int x.sin(x") dx = ?
2. int 2x.V(x"-5x) dx = ?
3. int 3x/(5x"-2) dx = ?
4. int (4x+3)cos(2x"+3x+4) = ?
5. int cos^4(x) = ?

Lanjut integral parsial.. :)
Buktikan bahwa :
int u dv = uv - int v du

d (uv) = u dv + v du
int d (uv) = int (u dv + v du)
uv = int u dv + int v du
int u dv = uv - int v du
(terbukti)

Contoh soal:
Tentukan nilai dari :
Integral [( x + 2 ) sin(x^2 + 4x -6)] dx
u=x"+4x-6
du=2x+4
int sin u.du/2
=1/2 int sin u du
=-1/2 cos u + c
=-1/2 cos (x"+4x-6) + c

1. Integral dx/(1 + sinx - cosx) = . . .limitnya 2 dh...
2. lim(x->0) (4/(x^2) - 2/(1 - cosx)) = . . .
3. lim(x->0) x^x = . . .

no. 1 integral dx/(1 + sinx - cosx)
integral (2/(1 + y^2))/(1 + (2y/(1+y^2)) - ((1-y^2)/(1+y^2))) dy
integral 2dy/((1+y^2) + 2y - (1-y^2))
integral 2dy/(2y^2 + 2y)
integral dy/(y^2 + y)
y^2 + y = y(y+1)
1/(y^2 + y) = (A/y) + B/(y+1)
<=> 1 = A(y+1) + By = (A+B)y + A diperoleh...
Persamaan A + B = 0
A=1
B=-1
maka 1/(y^2 + y) = (1/y) - (1/(y+1))
integral dy/(y^2 + y) = integral dy/y - integral dy/(y+1)
= lnly+1l - lnly+1l + C = lnl y/(y+1) l
= lnl(tan(x/2))/(tan(x/2) + 1)l + C

2)
u = tan(x/2) 
2arctan(u) = x 
dx= (2/(1 + u²)) du 

Perhatikan bahwa sinx = 2tan(x/2)/(1 + tan²(x/2)) 
sinx = 2u/(1 + u²) 
cosx = (1 - tan²(x/2))/(1 + tan²(x/2)) 
cosx = (1 - u²)/(1 + u²) 

Sehingga diperoleh,∫ (1 /(1 - sinx + cosx))dx 
= ∫ (1/(1 - (2u/(1 + u²)) + ((1 -u²)/(1 + u²))))(2/(1 + u²))du 
= ∫ (1/((1 + u² - 2u + 1 - u²)/(1 + u²))) (2/(1 + u²))du 
= ∫ (1/((-2u + 2)/(1 + u²)))(2/(1 + u²)) du 
= ∫ ((1 + u²)/(-2(u - 1)))(2/(1 + u²))du 
= -∫ (1/(u - 1))du 
= -ln|u - 1| + C
= -ln|tan(x/2) - 1| + C

3)
lim (x → 0) x^x

= lim (x → 0) e^(ln(x^x))
= lim (x → 0) e^(x ln(x))
= e^(lim (x → 0) (xln(x)))
= e^(lim (x → 0) (ln(x)/(1/x)))

Gunakan dalil L'Hopital
= e^(lim (x → 0) ((1/x)/(-1/x²)))
= e^(lim (x → 0) ((-x²)/x)
= e^(lim (x → 0) (-x ))
= e^0 
= 1

created by : admin #XX , admin #XI dan admin #XV