Pembahasan Soal UN 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 = \(\mathrm{\frac{a}{n+1}}\)xⁿ⁺¹ + C
2. ∫ (ax + b)ⁿ dx = \(\mathrm{\frac{1}{a(n+1)}}\)(ax + b)ⁿ⁺¹ + C
3. ∫ f(x).g(x)ⁿ dx = \(\mathrm{\frac{k}{n+1}}\) g(x)ⁿ⁺¹ + C
    dengan k = \(\mathrm{\frac{f(x)}{g'(x)}}\) dan k konstan.

1. UN 2005

Hasil dari \(\mathrm{\int_{0}^{1}3x\sqrt{3x^{2}+1}\:dx=…}\)

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 = \(\mathrm{\frac{3x}{6x}}\) = \(\frac{1}{2}\)

⇒ \(\mathrm{\frac{k}{n+1}}\) g(x)ⁿ⁺¹ + C
⇒ \(\frac{\frac{1}{2}}{\frac{1}{2}+1}\) (3x² + 1)\(\mathrm{^{\frac{1}{2}+1}}\) + C
⇒ \(\frac{1}{3}\)(3x² + 1)\(\mathrm{^{\frac{3}{2}}}\) + 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 \(\mathrm{\int_{a}^{3}}\) (3x² + 2x + 1) dx = 25. Nilai \(\mathrm{\frac{1}{2}a=…}\)
A.  −4
B.  −2
C.  −1
D.  1
E.  2

Pembahasan :
\(\mathrm{\int_{a}^{3}}\) (3x² + 2x + 1) dx = 25
⇔ [x³ + x² + x\(\mathrm{]_{a}^{3}}\) = 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 \(\mathrm{\int_{1}^{4}\frac{2}{x\sqrt{x}}\:dx=…}\)
A.  −12
B.  −4
C.  −3
D.  2
E.  \(\frac{3}{2}\)

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

⇒ \(\mathrm{\frac{k}{n+1}}\) 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 \(\mathrm{\int_{1}^{a}}\) (2x − 3) dx = 12 dan a > 0. Nilai a = …
A.  2
B.  3
C.  5
D.  7
E.  10

Pembahasan :
\(\mathrm{\int_{1}^{a}}\) (2x − 3) dx = 12
⇔ [x² − 3x\(\mathrm{]_{1}^{a}}\) = 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)\(\mathrm{\sqrt{x^{3}-x^{2}-1}}\) dx = …
A.  \(\mathrm{\frac{2}{3}\sqrt[3]{\left (x^{3}-x^{2}-1  \right )^{2}}+C}\)
B.  \(\mathrm{\frac{2}{3}\sqrt{\left (x^{3}-x^{2}-1  \right )^{3}}+C}\)
C.  \(\mathrm{\frac{4}{3}\sqrt{\left (x^{3}-x^{2}-1  \right )^{3}}+C}\)
D.  \(\mathrm{\frac{4}{3}\sqrt[3]{\left (x^{3}-x^{2}-1  \right )^{2}}+C}\)
E.  \(\mathrm{\frac{2}{3}\sqrt{\left (x^{3}-x^{2}-1  \right )^{2}}+C}\)

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

⇒ \(\mathrm{\frac{k}{n+1}}\) 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
⇒ \(\mathrm{\frac{4}{3}\sqrt{\left (x^{3}-x^{2}-1  \right )^{3}}+C}\)

Jawaban : C

7. UN 2009
Diketahui \(\mathrm{\int_{1}^{p}}\) (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 :
\(\mathrm{\int_{1}^{p}}\) (x − 1)² dx = \(\frac{8}{3}\)
⇔ \(\mathrm{\left [ \frac{1}{1(2+1)}(x-1)^{2+1} \right ]_{1}^{p}}\)  = \(\frac{8}{3}\)
⇔ \(\mathrm{\left [ \frac{1}{3}(x-1)^{3} \right ]_{1}^{p}}\)  = \(\frac{8}{3}\)
⇔ \(\frac{1}{3}\)(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 \(\mathrm{\int_{-1}^{3}}\)2x(3x + 4) dx = …
A.  88
B.  84
C.  56
D.  48
E.  46

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

Jawaban : A

9. UN 2011
Hasil \(\mathrm{\int_{2}^{4}}\)(−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 :
\(\mathrm{\int_{2}^{4}}\)(−x² + 6x − 8) dx
= \(\mathrm{\left [ -\frac{1}{3}x^{3}+3x^{2}-8x \right ]_{2}^{4}}\)
= (−\(\frac{1}{3}\).4³ + 3.4² − 8.4) − (−\(\frac{1}{3}\).2³ + 3.2² − 8.2)
= \(\frac{4}{3}\)

Jawaban : E

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

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

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

Jawaban : C

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

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

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

Jawaban : C

12. UN 2012
Nilai dari \(\mathrm{\int_{1}^{4}}\)(x² − 2x + 2) dx = …
A.  12
B.  14
C.  16
D.  18
E.  20

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

Jawaban : A

13. UN 2013
Hasil dari \(\mathrm{\int_{0}^{2}}\)3(x + 1)( x − 6) dx = …
A.  −58
B.  −56
C.  −28
D.  −16
E.  −14

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

Jawaban : A

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

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

⇒ \(\mathrm{\frac{k}{n+1}}\) 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\(\mathrm{\sqrt{x^{2}+1}}\) + C

Jawaban : C

15. UN 2014
Hasil ∫ (6x − 12)\(\mathrm{\sqrt{x^{2}-4x+8}}\) 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 = \(\mathrm{\frac{6x-12}{2x-4}}\) = \(\mathrm{\frac{3(2x-4)}{2x-4}}\) = 3

⇒ \(\mathrm{\frac{k}{n+1}}\) 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 \(\mathrm{\int_{0}^{1}}\)(x³ + 2x − 5) dx = …
A.  \(-\frac{16}{4}\)
B.  \(-\frac{15}{4}\)
C.  0
D.  \(\frac{15}{4}\)
E.  \(\frac{16}{4}\)

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

Jawaban : B