Diantara bunga tulip dan bunga sakura

Go to Blogger edit html and find these sentences.Now replace these sentences with your own descriptions.

Diantara bunga tulip dan bunga sakura

Go to Blogger edit html and find these sentences.Now replace these sentences with your own descriptions.

Diantara bunga tulip dan bunga sakura

Go to Blogger edit html and find these sentences.Now replace these sentences with your own descriptions.

Diantara bunga tulip dan bunga sakura

Go to Blogger edit html and find these sentences.Now replace these sentences with your own descriptions.

Diantara bunga tulip dan bunga sakura

Go to Blogger edit html and find these sentences.Now replace these sentences with your own descriptions.

Showing posts with label MATEMATIKA Soal latihan. Show all posts

Showing posts with label MATEMATIKA Soal latihan. Show all posts

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/

Subscribe to: Posts (Atom)