integral
TRANSCRIPT
1. Diketahui (3x2 + 2x + 1) dx = 25 Nilai a = …
A. – 4
B. – 2
C. – 1
D. 1
E. 2
PEMBAHASAN :
(3x2 + 2x + 1) dx = x3 + x2 + x
25 = (33 + 32 + 3) – (a3 + a2 + a)
a3 + a2 + a = 27 + 9 + 3 – 25
a3 + a2 + a – 14 = 0
(a – 2)(a2 + a + 7) = 0
a = 2 atau a2 + a + 7 = 0
jadi a = 1
JAWABAN : D
2. Nilai sin 2x cos x dx = …
A. -4/3
B. -1/3
C. 1/3
D. 2/3
E. 4/3
PEMBAHASAN :
sin 2x cos x dx = 2 sin x cos x cos x dx
= 2 sin x cos2 x dx
misal u = cos x du = -sin x dx
= 2 u2 (-du)
= - u3
Substitusi u = cos x
= - cos3 x
= - cos3 + cos3 0
= - (-1)3 + .13
= +
=
JAWABAN : D
3. Hasil dari 3x dx = …
A. 7/2
B. 8/3
C. 7/3
D. 4/3
E. 2/3
PEMBAHASAN :
3x dx = …
misal u = 3x2 + 1 du = 6x dx
=
= u1/2 du
= . u3/2
substitusi u = 3x2 + 1, sehingga diperoleh
= (3x2 + 1)3/2
= (3.12 + 1)3/2 – (3.02 + 1)3/2
= 8 – .1
=
JAWABAN : C
4. Hasil dari cos5 x dx = …
A. - cos6 x sin x + C
B. cos6 x sin x + C
C. –sin x + sin3 x + sin5 x + C
D. sin x – sin3 x + sin5 x + C
E. sin x + sin3 x + sin5 x + C
PEMBAHASAN :
cos5 x dx = cos x (cos2 x)2 dx
= cos x (1 – sin2 x)2 dx
= cos x (1 – 2 sin2 x + sin4 x) dx
misal u = sin x du = cos x
= (1 – 2u2 + u4) du
= u – u3 + u5 + C
substitusi u = sin x,
= sin x – sin3 x + sin5 x + C
JAWABAN : D
5. Hasil dari cos x (x2 + 1) dx = …
A. x2 sin x + 2x cos x + C
B. (x2 – 1)sin x + 2x cos x + C
C. (x2 + 3)sin x – 2x cos x + C
D. 2x2 cos x + 2x2 sin x + C
E. 2x sin x – (x2 – 1)cos x + C
PEMBAHASAN :
dalam penyelesaian soal ini akan menggunakan Integral Parsial
u = x2 + 1 du = 2x dx
dv = cos x dx v = sin x
u dv = uv – v du
= sin x (x2 + 1) – sin x 2x dx
parsial lagi
m = 2x dm = 2 dx
dn = sin x dx n = -cos x
= sin x (x2 + 1) – (2x (-cos x) – -cos x 2 dx)
= sin x (x2 + 1) – (-2x cos x + 2 sin x) + C
= sin x (x2 + 1) + 2x cos x – 2 sin x + C
= sin x (x2 – 1) + 2x cos x + C
JAWABAN : B
6. Diketahui (3x2 – 2x + 2) dx = 40. Nilai p = …
A. 2
B. 1
C. – 1
D. – 2
E. – 4
PEMBAHASAN :
(3x2 – 2x + 2) dx = x3 – x2 + 2x
40 = (33 – 32 + 6) – (p3 – p2 + 2p)
p3 – p2 + 2p = 27 – 9 + 6 – 40
p3 – p2 + 2p + 16 = 0
(p + 2)(p2 + p + 7) = 0
p = -2 atau p2 + p + 7 = 0
jadi p = -1
JAWABAN : C
7. Hasil dari sin 3x cos 5x dx = …
A. -10/6
B. -8/10
C. -5/16
D. -4/16
E. 0
PEMBAHASAN :
sin 3x cos 5x dx = [sin 8x + sin (-2x)] dx
[Sifat Trigonometri]
= sin 8x dx – sin 2x dx
misal u = 8x du = 8 dx
v = 2x dv = 2 dx
= sin u – sin v
= - cos u + cos v
substitusi u = 8x dan v = 2x
= - cos 8x + cos 2x
= [- (cos 8( ) - cos 8(0))] + [ (cos 2( ) – cos 2(0))]
= [- (1 – 1)] + [ (-1 – 1)]
=
JAWABAN :
8. x sin x dx = …
A.
B.
C.
D.
E.
PEMBAHASAN :
dalam penyelesaian soal ini akan menggunakan Integral Parsial
u = x du = dx
dv = sin x dx v = -cos x
u dv = uv – v du
= -x cos x – (-cos x) dx
= [-x cos x + sin x]
= [- cos ( ) + sin ( )] – [-0 cos 0 + sin 0]
= - (-1)
=
JAWABAN : D
9. Nilai (2x + sin x) dx = …
A. – 1
B.
C. + 1
D. – 1
E. + 1
PEMBAHASAN :
(2x + sin x) dx = x2 – cos x
= (( )2 – cos ( )) – (02 – cos 0)
= ( – 0) – (02 – 1)
= + 1
JAWABAN : C
10.Nilai x sin(x2 + 1) dx = …
A. –cos (x2 + 1) + C
B. cos (x2 + 1) + C
C. –½ cos (x2 + 1) + C
D. ½ cos (x2 + 1) + C
E. –2cos (x2 + 1) + C
PEMBAHASAN :
misal u = x2 + 1 du = 2x dx
x sin(x2 + 1) dx = sin u
= cos u + C
substitusi u = x2 + 1
= cos (x2 + 1) + C
JAWABAN : C