Pembahasan Soal Ujian Nasional Integral Fungsi Aljabar

 

Pembahasan soal-soal Ujian Nasional (UN) SMA bidang studi Matematika IPA untuk materi pembahasan Integral Fungsi Aljabar yang meliputi integral tentu, integral tak tentu dan integral substitusi dari fungsi-fungsi aljabar.

Rumus-rumus integral yang digunakan :
1. ∫ axⁿ dx = $\frac{a}{n+1}$xⁿ⁺¹ + C
2. ∫ (ax + b)ⁿ dx = $\frac{1}{a(n+1)}$(ax + b)ⁿ⁺¹ + C
3. ∫ f(x).g(x)ⁿ dx = $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
    dengan k = $\frac{f\left(x\right)}{g^{\prime }\left(x\right)}$ dan k konstan.


1. UN 2005

Hasil dari ${\int }_0^{1}3x\sqrt{3x^2+1}dx=...$

A.  $\frac{7}{2}$

B.  $\frac{8}{3}$

C.  $\frac{7}{3}$
D.  $\frac{4}{3}$
E.  $\frac{2}{3}$

Pembahasan :
∫ 3x(3x² + 1)${}^{\frac{1}{2}}$ dx
n = $\frac{1}{2}$
k = $\frac{3x}{6x}$ = $\frac{1}{2}$

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{\frac{1}{2}}{\frac{1}{2}+1}$ (3x² + 1)${}^{\frac{1}{2}+1}$ + C
⇒ $\frac{1}{3}$(3x² + 1)${}^{\frac{3}{2}}$ + C

Untuk batas x = 0 sampai x = 1
⇒ $\frac{1}{3}$(3.1² + 1)${}^{\frac{3}{2}}$ − $\frac{1}{3}$(3.0² + 1)${}^{\frac{3}{2}}$
⇒ $\frac{8}{3}$ − $\frac{1}{3}$ = $\frac{7}{3}$

Jawaban : C


2. UN 2007
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 = 25
⇔ [x³ + x² + x${]}_a^3$ = 25
⇔ (3³ + 3² + 3) − (a³ + a² + a) = 25
⇔ a³ + a² + a − 14 = 0
Nilai a yang mungkin adalah faktor dari −14, yaitu :
±1, ±2, ±7, ±14.
Nilai a yang memenuhi adalah a = 2 karena
2³ + 2² + 2 − 14 = 0

Jadi, $\frac{1}{2}$a = $\frac{1}{2}$. 2 = 1

Jawaban : D


3. UN 2008
Hasil dari ${\int }_1^4\frac{2}{x\sqrt{x}}dx=...$
A.  −12
B.  −4
C.  −3
D.  2
E.  $\frac{3}{2}$

Pembahasan :
⇒ ${\int }_1^4$ 2x${}^{-\frac{3}{2}}$ dx
= ${\left[\frac{2}{-\frac{3}{2}+1}{x}^{-\frac{3}{2}+1}\right]}_1^4$
= ${\left[-4x^{-\frac{1}{2}}\right]}_1^4$
= ${\left[\frac{-4}{\sqrt{x}}\right]}_1^4$
= $\frac{-4}{\sqrt{4}}-\frac{-4}{\sqrt{1}}$
= −2 − (−4)
= 2

Jawaban : D


4. UN 2009
Hasil dari ∫ (2x − 1)(x² − x + 3)³ dx = ...
A.  $\frac{1}{3}$(x² − x + 3)³ + C
B.  $\frac{1}{4}$(x² − x + 3)³ + C
C.  $\frac{1}{4}$(x² − x + 3)⁴ + C
D.  $\frac{1}{2}$(x² − x + 3)⁴ + C
E.  (x² − x + 3)⁴ + C

Pembahasan :
∫ (2x − 1)(x² − x + 3)³ dx
n = 3
k = $\frac{2x-1}{2x-1}$ = 1

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{1}{3+1}$(x² − x + 3)³⁺¹ + C
⇒ $\frac{1}{4}$(x² − x + 3)⁴ + C

Jawaban : C


5. UN 2009
Diketahui ${\int }_1^a$ (2x − 3) dx = 12 dan a > 0. Nilai a = ...
A.  2
B.  3
C.  5
D.  7
E.  10

Pembahasan :
${\int }_1^a$ (2x − 3) dx = 12
⇔ [x² − 3x${]}_1^a$ = 12
⇔ (a² − 3a) − (1² − 3.1) = 12
⇔ a² − 3a − 10 = 0
Nilai a yang mungkin adalah faktor dari −10, yaitu :
±1, ±2, ±5, ±10.
Nilai a yang memenuhi adalah a = 5 karena
5² − 3.5 − 10 = 0

Jawaban : C


6. UN 2009
Hasil dari ∫ (6x² − 4x)$\sqrt{x^3-x^2-1}$ dx = ...
A.  $\frac{2}{3}\sqrt[3]{3{\left(x^3-x^2-1\right)}^2}+C$
B.  $\frac{2}{3}\sqrt{{\left(x^3-x^2-1\right)}^3}+C$
C.  $\frac{4}{3}\sqrt{{\left(x^3-x^2-1\right)}^3}+C$
D.  $\frac{4}{3}\sqrt[3]{3{\left(x^3-x^2-1\right)}^2}+C$
E.  $\frac{2}{3}\sqrt{{\left(x^3-x^2-1\right)}^2}+C$

Pembahasan :
∫ (6x² − 4x)(x³ − x² − 1)${}^{\frac{1}{2}}$ dx
n = $\frac{1}{2}$
k = $\frac{6x^2-4x}{3x^2-2x}$ = $\frac{2(3x^2-2x)}{3x^2-2x}$ = 2

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{2}{\frac{1}{2}+1}$(x³ − x² − 1)${}^{\frac{1}{2}+1}$ + C
⇒ $\frac{4}{3}$(x³ − x² − 1)${}^{\frac{3}{2}}$ + C
⇒ $\frac{4}{3}\sqrt{{\left(x^3-x^2-1\right)}^3}+C$

Jawaban : C


7. UN 2009
Diketahui ${\int }_1^p$ (x − 1)² dx = 2$\frac{2}{3}$. Nilai p yang memenuhi adalah...
A.  1
B.  1$\frac{1}{3}$
C.  3
D.  6
E.  9

Pembahasan :
${\int }_1^p$ (x − 1)² dx = $\frac{8}{3}$
⇔ ${\left[\frac{1}{1(2+1)}(x-1{)}^{2+1}\right]}_1^p$  = $\frac{8}{3}$
⇔ ${\left[\frac{1}{3}(x-1{)}^3\right]}_1^p$  = $\frac{8}{3}$
⇔ $\frac{1}{3}$(p − 1)³ − $\frac{1}{3}$(1 − 1)³  = $\frac{8}{3}$
⇔ $\frac{1}{3}$(p − 1)³ = $\frac{8}{3}$
⇔ (p − 1)³ = 8
⇔ (p − 1)³ = 2³
⇔ p − 1 = 2
⇔ p = 3

Jawaban : C


8. UN 2010
Nilai dari ${\int }_{-1}^3$2x(3x + 4) dx = ...
A.  88
B.  84
C.  56
D.  48
E.  46

Pembahasan :
${\int }_{-1}^3$2x(3x + 4) dx
⇒ ${\int }_{-1}^3$(6x² + 8x) dx
= ${\left[2x^3+4x^2\right]}_{-1}^3$
= (2.3³ + 4.3²) − (2(−1)³ + 4(−1)²)
= 90 − 2
= 88

Jawaban : A


9. UN 2011
Hasil ${\int }_2^4$(−x² + 6x − 8) dx = ...
A.  $\frac{38}{3}$
B.  $\frac{26}{3}$
C.  $\frac{20}{3}$
D.  $\frac{16}{3}$
E.  $\frac{4}{3}$

Pembahasan :
${\int }_2^4$(−x² + 6x − 8) dx
= ${\left[-\frac{1}{3}{x}^3+3x^2-8x\right]}_2^4$
= (−$\frac{1}{3}$.4³ + 3.4² − 8.4) − (−$\frac{1}{3}$.2³ + 3.2² − 8.2)
= $\frac{4}{3}$

Jawaban : E


10. UN 2011
Hasil $\int \frac{2x+3}{\sqrt{3x^2+9x-1}}dx=...$
A.  $2\sqrt{3x^2+9x-1}+C$
B.  $\frac{1}{3}\sqrt{3x^2+9x-1}+C$
C.  $\frac{2}{3}\sqrt{3x^2+9x-1}+C$
D.  $\frac{1}{2}\sqrt{3x^2+9x-1}+C$
E.  $\frac{3}{2}\sqrt{3x^2+9x-1}+C$

Pembahasan :
∫ (2x + 3)(3x² + 9x − 1)${}^{-\frac{1}{2}}$ dx
n = $-\frac{1}{2}$
k = $\frac{2x+3}{6x+9}$ = $\frac{2x+3}{3(2x+3)}$ = $\frac{1}{3}$

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{\frac{1}{3}}{-\frac{1}{2}+1}$(3x² + 9x − 1)${}^{-\frac{1}{2}+1}$ + C
⇒ $\frac{2}{3}$(3x² + 9x − 1)${}^{\frac{1}{2}}$ + C
⇒ $\frac{2}{3}\sqrt{\left(3x^2+9x-1\right)}+C$

Jawaban : C


11. UN 2012
Hasil dari $\int 3x\sqrt{3x^2+1}dx=...$
A.  $-\frac{2}{3}(3x^2+1)\sqrt{3x^2+1}+C$
B.  $-\frac{1}{2}(3x^2+1)\sqrt{3x^2+1}+C$
C.  $\frac{1}{3}(3x^2+1)\sqrt{3x^2+1}+C$
D.  $\frac{1}{2}(3x^2+1)\sqrt{3x^2+1}+C$
E.  $\frac{2}{3}(3x^2+1)\sqrt{3x^2+1}+C$

Pembahasan :
∫ 3x(3x² + 1)${}^{\frac{1}{2}}$ dx
n = $\frac{1}{2}$
k = $\frac{3x}{6x}$ = $\frac{1}{2}$

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{\frac{1}{2}}{\frac{1}{2}+1}$ (3x² + 1)${}^{\frac{1}{2}+1}$ + C
⇒ $\frac{1}{3}$(3x² + 1)${}^{\frac{3}{2}}$ + C
⇒ $\frac{1}{3}$(3x² + 1)$\sqrt{3x^2+1}$ + C

Jawaban : C


12. UN 2012
Nilai dari ${\int }_1^4$(x² − 2x + 2) dx = ...
A.  12
B.  14
C.  16
D.  18
E.  20

Pembahasan :
${\int }_1^4$(x² − 2x + 2) dx
= ${\left[\frac{1}{3}{x}^3-x^2+2x\right]}_1^4$
= ($\frac{1}{3}$.4³ − 4² + 2.4) − ($\frac{1}{3}$.1³ − 1² + 2.1)
= 12

Jawaban : A


13. UN 2013
Hasil dari ${\int }_0^2$3(x + 1)( x − 6) dx = ...
A.  −58
B.  −56
C.  −28
D.  −16
E.  −14

Pembahasan :
${\int }_0^2$3(x + 1)( x − 6) dx
3${\int }_0^2$(x² − 5x − 6) dx
= 3${\left[\frac{1}{3}{x}^3-\frac{5}{2}{x}^2-6x\right]}_0^2$
= 3[($\frac{1}{3}$.2³ − $\frac{5}{2}$.2² − 6.2) − 0]
= 3. $\frac{-58}{3}$
= −58

Jawaban : A


14. UN 2013
Hasil dari $\int \frac{2x}{\sqrt{x^2+1}}dx=...$
A.  $\frac{1}{3}\sqrt{x^2+1}+C$
B.  $\frac{1}{2}\sqrt{x^2+1}+C$
C.  $2\sqrt{x^2+1}+C$
D.  $3\sqrt{x^2+1}+C$
E.  $6\sqrt{x^2+1}+C$

Pembahasan :
∫ 2x(x² + 1)${}^{-\frac{1}{2}}$ dx
n = $-\frac{1}{2}$
k = $\frac{2x}{2x}$ = 1

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{1}{-\frac{1}{2}+1}$(x² + 1)${}^{-\frac{1}{2}+1}$ + C
⇒ 2(x² + 1)${}^{\frac{1}{2}}$ + C
⇒ 2$\sqrt{x^2+1}$ + C

Jawaban : C


15. UN 2014
Hasil ∫ (6x − 12)$\sqrt{x^2-4x+8}$ dx = ...
A.  $\frac{1}{3}$(x² − 4x + 8)${}^{\frac{3}{2}}$ + C
B.  $\frac{1}{2}$(x² − 4x + 8)${}^{\frac{3}{2}}$ + C
C.  $\frac{2}{3}$(x² − 4x + 8)${}^{\frac{3}{2}}$ + C
D.  (x² − 4x + 8)${}^{\frac{3}{2}}$ + C
E.  2(x² − 4x + 8)${}^{\frac{3}{2}}$ + C

Pembahasan :
∫ (6x − 12)(x² − 4x + 8)${}^{\frac{1}{2}}$ dx
n = $\frac{1}{2}$
k = $\frac{6x-12}{2x-4}$ = $\frac{3(2x-4)}{2x-4}$ = 3

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{3}{\frac{1}{2}+1}$(x² − 4x + 8)${}^{\frac{1}{2}+1}$ + C
⇒ 2(x² − 4x + 8)${}^{\frac{3}{2}}$ + C

Jawaban : E


16. UN 2014
Hasil ${\int }_0^1$(x³ + 2x − 5) dx = ...
A.  $-\frac{16}{4}$
B.  $-\frac{15}{4}$
C.  0
D.  $\frac{15}{4}$
E.  $\frac{16}{4}$

Pembahasan :
${\int }_0^1$(x³ + 2x − 5) dx
= ${\left[\frac{1}{4}{x}^4+x^2-5x\right]}_0^1$
= ($\frac{1}{4}$.1⁴ + 1² − 5.1) − 0
= $-\frac{15}{4}$

Jawaban : B

17. UN 2015
Hasil ∫ 4x(4x² − 3)⁴ dx = ...
A.  $\frac{1}{10}$(4x² − 3)⁵ + C
B.  $\frac{1}{5}$(4x² − 3)⁵ + C
C.  $\frac{2}{5}$(4x² − 3)⁵ + C
D.  (4x² − 3)⁵ + C
E.  2(4x² − 3)⁵ + C

Pembahasan :
∫ 4x(4x² − 3)⁴ dx
n = 4
k = $\frac{4x}{8x}$ = $\frac{1}{2}$

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{\frac{1}{2}}{4+1}$(4x² − 3)⁴⁺¹ + C
⇒ $\frac{1}{10}$(4x² − 3)⁵ + C

Jawaban : A


18. UN 2015
Nilai dari ${\int }_1^4\left(3\sqrt{x}-\frac{1}{\sqrt{x}}\right)dx$ adalah...
A.  20
B.  12
C.  8
D.  4
E.  2

Pembahasan :
${\int }_1^4$(3x${}^{\frac{1}{2}}$ − x${}^{-\frac{1}{2}}$) dx
= ${\left[\frac{3}{\frac{1}{2}+1}{x}^{\frac{1}{2}+1}-\frac{1}{-\frac{1}{2}+1}{x}^{-\frac{1}{2}+1}\right]}_1^4$
= ${\left[2x^{\frac{3}{2}}-2x^{\frac{1}{2}}\right]}_1^4$
= ${\left[2x\sqrt{x}-2\sqrt{x}\right]}_1^4$
= (2.4√4 − 2√4) − (2.1√1 − 2√1)
= 12 − 0
= 12

Jawaban : B


19. UN 2016
Hasil dari ∫ x(3x − 5)⁴ dx = ...
A.  $-\frac{1}{54}$ (1 + 3x)(3x − 5)⁵ + C
B.  $-\frac{1}{108}$ (1 − 3x)(3x − 5)⁵ + C
C.  $-\frac{1}{270}$ (1 + 3x)(3x − 5)⁵ + C
D.  $\frac{1}{108}$ (1 − 3x)(3x − 5)⁵ + C
E.  $\frac{1}{54}$ (1 + 3x)(3x − 5)⁵ + C

Pembahasan :
∫ x(3x − 5)⁴ dx

Misalkan :
u = 3x − 5 → x = $\frac{u+5}{3}$
du = 3 dx → dx = $\frac{du}{3}$

Substitusi :
⇒ ∫ $\frac{u+5}{3}$u⁴ $\frac{du}{3}$
⇒ $\frac{1}{9}$∫ (u⁵ +5u⁴) du
= $\frac{1}{9}$ ($\frac{1}{6}$u⁶ + u⁵ ) + C
= $\frac{1}{9}$ ($\frac{1}{6}$u + 1)u⁵ + C
= $\frac{1}{9}$ ($\frac{u+6}{6}$)u⁵ + C
= $\frac{1}{54}$ (u + 6)u⁵ + C
= $\frac{1}{54}$ (3x − 5 + 6)(3x − 5)⁵ + C
= $\frac{1}{54}$ (3x + 1)(3x − 5)⁵ + C

Jawaban : E


20. UN 2016
Nilai dari ${\int }_{-1}^2$(3x² + 6x − 1) dx = ...
A.  3
B.  5
C.  9
D.  15
E.  18

Pembahasan :
${\int }_{-1}^2$(3x² + 6x − 1) dx
= ${\left[x^3+3x^2-x\right]}_{-1}^2$
= (2³ + 3.2² − 2) − (1³ + 3.1² − 1)
= 18 −3
= 15

Jawaban : D


21. UN 2016
Hasil dari $\int \frac{6x-9}{\sqrt{x^2-3x-5}}dx$ adalah...
A.  $2\sqrt{x^2-3x-5}+C$
B.  $3\sqrt{x^2-3x-5}+C$
C.  $6\sqrt{x^2-3x-5}+C$
D.  $9\sqrt{x^2-3x-5}+C$
E.  $18\sqrt{x^2-3x-5}+C$

Pembahasan :
∫ (6x − 9)(x² − 3x − 5)${}^{-\frac{1}{2}}$ dx
n = $-\frac{1}{2}$
k = $\frac{6x-9}{2x-3}$ = $\frac{3(2x-3)}{2x-3}$ = 3

⇒ $\frac{k}{n+1}$ g(x)ⁿ⁺¹ + C
⇒ $\frac{3}{-\frac{1}{2}+1}$(x² − 3x − 5)${}^{-\frac{1}{2}+1}$ + C
⇒ 6(x² − 3x − 5)${}^{\frac{1}{2}}$ + C
⇒ 6$\sqrt{x^2-3x-5}$ + C

sumber: https://smatika.blogspot.com/

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