Saturday, November 29, 2014

Latihan Soal Integral

11/29/2014 05:05:00 PM MATEMATIKA Soal latihan No comments

Pembahasan Latihan Soal Integral (1) UN SMA

21 Votes

Diketahui $\int_a^3$ (3x² + 2x + 1) dx = 25 Nilai $\frac{1}{2}$ a = …

A. – 4

B. – 2

C. – 1

D. 1

E. 2

PEMBAHASAN :

$\int_a^3$ (3x² + 2x + 1) dx = x³ + x² + x $\mid_a^3$

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 $\frac{1}{2}$ a = 1

JAWABAN : D
Nilai $\int_0^\pi$ sin 2x cos x dx = …

A. -4/3

B. -1/3

C. 1/3

D. 2/3

E. 4/3

PEMBAHASAN :

$\int_0^\pi$ sin 2x cos x dx = $\int_0^\pi$ 2 sin x cos x cos x dx

= $\int_0^\pi$ 2 sin x cos² x dx

misal u = cos x $\Rightarrow$ du = -sin x dx

= $\int_0^\pi$ 2 u² (-du)

= - $\frac{1}{3}$ u³ $\mid_0^\pi$

Substitusi u = cos x

= - $\frac{1}{3}$ cos³ x $\mid_0^\pi$

= - $\frac{1}{3}$ cos³ $(\pi)$ + $\frac{1}{3}$ cos³ 0

= - $\frac{1}{3}$ (-1)³ + $\frac{1}{3}$ .1³

= $\frac{1}{3}$ + $\frac{1}{3}$

= $\frac{2}{3}$

JAWABAN : D
Hasil dari $\int_0^1$ 3x $\sqrt{3x^2+1}$ dx = …

A. 7/2

B. 8/3

C. 7/3

D. 4/3

E. 2/3

PEMBAHASAN :

$\int_0^1$ 3x $\sqrt{3x^2+1}$ dx = …

misal u = 3x² + 1 $\Rightarrow$ du = 6x dx

= $\int_0^1$ $\sqrt{u}$ $\frac{du}{2}$

= $\int_0^1$ $\frac{1}{2}$ u^1/2 du

= $\frac{1}{2}$ . $\frac{2}{3}$ u^3/2 $\mid_0^1$

substitusi u = 3x² + 1, sehingga diperoleh

= $\frac{1}{3}$ (3x² + 1)^3/2 $\mid_0^1$

= $\frac{1}{3}$ (3.1² + 1)^3/2 – $\frac{1}{3}$ (3.0² + 1)^3/2

= $\frac{1}{3}$ 8 – $\frac{1}{3}$ .1

= $\frac{7}{3}$

JAWABAN : C
Hasil dari $\int$ cos⁵ x dx = …

A. - $\frac{1}{6}$ cos⁶ x sin x + C

B. $\frac{1}{6}$ cos⁶ x sin x + C

C. –sin x + $\frac{2}{3}$ sin³ x + $\frac{1}{5}$ sin⁵ x + C

D. sin x – $\frac{2}{3}$ sin³ x + $\frac{1}{5}$ sin⁵ x + C

E. sin x + $\frac{2}{3}$ sin³ x + $\frac{1}{5}$ sin⁵ x + C

PEMBAHASAN :

$\int$ cos⁵ x dx = $\int$ cos x (cos² x)² dx

= $\int$ cos x (1 – sin² x)² dx

= $\int$ cos x (1 – 2 sin² x + sin⁴ x) dx

misal u = sin x $\Rightarrow$ du = cos x

= $\int$ (1 – 2u² + u⁴) du

= u – $\frac{2}{3}$ u³ + $\frac{1}{5}$ u⁵ + C

substitusi u = sin x,

= sin x – $\frac{2}{3}$ sin³ x + $\frac{1}{5}$ sin⁵ x + C

JAWABAN : D
Hasil dari $\int$ 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 $\Rightarrow$ du = 2x dx

dv = cos x dx $\Rightarrow$ v = sin x

$\int$ u dv = uv – $\int$ v du

= sin x (x² + 1) – $\int$ sin x 2x dx

parsial lagi

m = 2x $\Rightarrow$ dm = 2 dx

dn = sin x dx $\Rightarrow$ n = -cos x

= sin x (x² + 1) – (2x (-cos x) – $\int$ -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
Diketahui $\int_p^3$ (3x² – 2x + 2) dx = 40. Nilai $\frac{1}{2}$ p = …

A. 2

B. 1

C. – 1

D. – 2

E. – 4

PEMBAHASAN :

$\int_p^3$ (3x² – 2x + 2) dx = x³ – x² + 2x $\mid_p^3$

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 $\frac{1}{2}$ p = -1

JAWABAN : C
Hasil dari $\int_0^{\frac{\pi}{2}}$ sin 3x cos 5x dx = …

A. -10/6

B. -8/10

C. -5/16

D. -4/16

E. 0

PEMBAHASAN :

$\int_0^{\frac{\pi}{2}}$ sin 3x cos 5x dx = $\int_0^{\frac{\pi}{2}}$ $\frac{1}{2}$ [sin 8x + sin (-2x)] dx

[Sifat Trigonometri]

= $\int_0^{\frac{\pi}{2}}$ $\frac{1}{2}$ sin 8x dx – $\int_0^{\frac{\pi}{2}}$ $\frac{1}{2}$ sin 2x dx

misal u = 8x $\Rightarrow$ du = 8 dx

v = 2x $\Rightarrow$ dv = 2 dx

= $\int_0^{\frac{\pi}{2}}$ $\frac{1}{2}$ sin u $\frac{du}{8}$ – $\int_0^{\frac{\pi}{2}}$ $\frac{1}{2}$ sin v $\frac{dv}{2}$

= - $\frac{1}{16}$ cos u $\mid_0^{\frac{\pi}{2}}$ + $\frac{1}{4}$ cos v $\mid_0^{\frac{\pi}{2}}$

substitusi u = 8x dan v = 2x

= - $\frac{1}{16}$ cos 8x $\mid_0^{\frac{\pi}{2}}$ + $\frac{1}{4}$ cos 2x $\mid_0^{\frac{\pi}{2}}$

= [- $\frac{1}{16}$ (cos 8( $\frac{\pi}{2}$ ) – cos 8(0))] + [ $\frac{1}{4}$ (cos 2( $\frac{\pi}{2}$ ) – cos 2(0))]

= [- $\frac{1}{16}$ (1 – 1)] + [ $\frac{1}{4}$ (-1 – 1)]

= $-\frac{1}{2}$

JAWABAN :
$\int_0^\pi$ x sin x dx = …

A. $\frac{\pi}{4}$

B. $\frac{\pi}{3}$

C. $\frac{\pi}{2}$

D. $\pi$

E. $\frac{3\pi}{2}$

PEMBAHASAN :

dalam penyelesaian soal ini akan menggunakan Integral Parsial

u = x $\Rightarrow$ du = dx

dv = sin x dx $\Rightarrow$ v = -cos x

$\int$ u dv = uv – $\int$ v du

= -x cos x – $\int$ (-cos x) dx

= [-x cos x + sin x] $\mid_0^\pi$

= [- $\pi$ cos ( $\pi$ ) + sin ( $\pi$ )] – [-0 cos 0 + sin 0]

= - $\pi$ (-1)

= $\pi$

JAWABAN : D
Nilai $\int_0^{\frac{\pi}{2}}$ (2x + sin x) dx = …

A. $\frac{1}{4}\pi^2$ – 1

B. $\frac{1}{4}\pi^2$

C. $\frac{1}{4}\pi^2$ + 1

D. $\frac{1}{2}\pi^2$ – 1

E. $\frac{1}{2}\pi^2$ + 1

PEMBAHASAN :

$\int_0^{\frac{\pi}{2}}$ (2x + sin x) dx = x² – cos x $\mid_0^{\frac{\pi}{2}}$

= (( $\frac{\pi}{2}$ )² – cos ( $\frac{\pi}{2}$ )) – (0² – cos 0)

= ( $\frac{\pi^2}{4}$ – 0) – (0² – 1)

= $\frac{\pi^2}{4}$ + 1

JAWABAN : C
Nilai $\int$ 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 $\Rightarrow$ du = 2x dx

$\int$ x sin(x² + 1) dx = $\int$ sin u $\frac{du}{2}$

= $-\frac{1}{2}$ cos u + C

substitusi u = x² + 1

= $-\frac{1}{2}$ cos (x² + 1) + C

JAWABAN : C

https://aimprof08.wordpress.com/2012/12/01/pembahasan-latihan-soal-integral-1-un-sma/

0 comments:

Post a Comment

Subscribe to: Post Comments (Atom)