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Pembahasan soal-soal Ujian Nasional (UN) Sekolah Menengan Atas bidang studi Matematika IPA untuk bahan pembahasan Integral Fungsi Aljabar yang mencakup integral tentu, integral tak tentu dan integral substitusi dari fungsi-fungsi aljabar.
Rumus-rumus integral yang dipakai :
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. \(\frac{8}{3}\)
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 hingga 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 yaitu faktor dari −14, yaitu :
±1, ±2, ±7, ±14.
Nilai a yang memenuhi yaitu 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 yaitu faktor dari −10, yaitu :
±1, ±2, ±5, ±10.
Nilai a yang memenuhi yaitu 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
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 = \(\mathrm{\frac{4x}{8x}}\) = \(\frac{1}{2}\)
⇒ \(\mathrm{\frac{k}{n+1}}\) 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 \(\mathrm{\int_{1}^{4}\left ( 3\sqrt{x}-\frac{1}{\sqrt{x}} \right )\:dx}\) adalah...
A. 20
B. 12
C. 8
D. 4
E. 2
Pembahasan :
\(\mathrm{\int_{1}^{4}}\)(3x\(^{\frac{1}{2}}\) − x\(^{-\frac{1}{2}}\)) dx
= \(\mathrm{\left [ \frac{3}{\frac{1}{2}+1}x^{\frac{1}{2}+1}-\frac{1}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1} \right ]_{1}^{4}}\)
= \(\mathrm{\left [ 2x^{\frac{3}{2}}-2x^{\frac{1}{2}} \right ]_{1}^{4}}\)
= \(\mathrm{\left [ 2x\sqrt{x}-2\sqrt{x} \right ]_{1}^{4}}\)
= (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 = \(\mathrm{\frac{u+5}{3}}\)
du = 3 dx → dx = \(\mathrm{\frac{du}{3}}\)
Substitusi :
⇒ ∫ \(\mathrm{\frac{u+5}{3}}\)u4 \(\mathrm{\frac{du}{3}}\)
⇒ \(\frac{1}{9}\)∫ (u5 +5u4) du
= \(\frac{1}{9}\) (\(\frac{1}{6}\)u6 + u5 ) + C
= \(\frac{1}{9}\) (\(\frac{1}{6}\)u + 1)u5 + C
= \(\frac{1}{9}\) (\(\mathrm{\frac{u+6}{6}}\))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 \(\mathrm{\int_{-1}^{2}}\)(3x2 + 6x − 1) dx = ...
A. 3
B. 5
C. 9
D. 15
E. 18
Pembahasan :
\(\mathrm{\int_{-1}^{2}}\)(3x2 + 6x − 1) dx
= \(\mathrm{\left [ x^{3}+3x^{2}-x \right ]_{-1}^{2}}\)
= (23 + 3.22 − 2) − (13 + 3.12 − 1)
= 18 −3
= 15
Jawaban : D
21. UN 2016
Hasil dari \(\mathrm{\int \frac{6x-9}{\sqrt{x^{2}-3x-5}}\:dx}\) adalah...
A. \(\mathrm{2\sqrt{x^{2}-3x-5}+C}\)
B. \(\mathrm{3\sqrt{x^{2}-3x-5}+C}\)
C. \(\mathrm{6\sqrt{x^{2}-3x-5}+C}\)
D. \(\mathrm{9\sqrt{x^{2}-3x-5}+C}\)
E. \(\mathrm{18\sqrt{x^{2}-3x-5}+C}\)
Pembahasan :
∫ (6x − 9)(x2 − 3x − 5)\(^{-\frac{1}{2}}\) dx
n = \(-\frac{1}{2}\)
k = \(\mathrm{\frac{6x-9}{2x-3}}\) = \(\mathrm{\frac{3(2x-3)}{2x-3}}\) = 3
⇒ \(\mathrm{\frac{k}{n+1}}\) 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\(\mathrm{\sqrt{x^{2}-3x-5}}\) + C
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 = \(\mathrm{\frac{4x}{8x}}\) = \(\frac{1}{2}\)
⇒ \(\mathrm{\frac{k}{n+1}}\) 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 \(\mathrm{\int_{1}^{4}\left ( 3\sqrt{x}-\frac{1}{\sqrt{x}} \right )\:dx}\) adalah...
A. 20
B. 12
C. 8
D. 4
E. 2
Pembahasan :
\(\mathrm{\int_{1}^{4}}\)(3x\(^{\frac{1}{2}}\) − x\(^{-\frac{1}{2}}\)) dx
= \(\mathrm{\left [ \frac{3}{\frac{1}{2}+1}x^{\frac{1}{2}+1}-\frac{1}{-\frac{1}{2}+1}x^{-\frac{1}{2}+1} \right ]_{1}^{4}}\)
= \(\mathrm{\left [ 2x^{\frac{3}{2}}-2x^{\frac{1}{2}} \right ]_{1}^{4}}\)
= \(\mathrm{\left [ 2x\sqrt{x}-2\sqrt{x} \right ]_{1}^{4}}\)
= (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 = \(\mathrm{\frac{u+5}{3}}\)
du = 3 dx → dx = \(\mathrm{\frac{du}{3}}\)
Substitusi :
⇒ ∫ \(\mathrm{\frac{u+5}{3}}\)u4 \(\mathrm{\frac{du}{3}}\)
⇒ \(\frac{1}{9}\)∫ (u5 +5u4) du
= \(\frac{1}{9}\) (\(\frac{1}{6}\)u6 + u5 ) + C
= \(\frac{1}{9}\) (\(\frac{1}{6}\)u + 1)u5 + C
= \(\frac{1}{9}\) (\(\mathrm{\frac{u+6}{6}}\))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 \(\mathrm{\int_{-1}^{2}}\)(3x2 + 6x − 1) dx = ...
A. 3
B. 5
C. 9
D. 15
E. 18
Pembahasan :
\(\mathrm{\int_{-1}^{2}}\)(3x2 + 6x − 1) dx
= \(\mathrm{\left [ x^{3}+3x^{2}-x \right ]_{-1}^{2}}\)
= (23 + 3.22 − 2) − (13 + 3.12 − 1)
= 18 −3
= 15
Jawaban : D
21. UN 2016
Hasil dari \(\mathrm{\int \frac{6x-9}{\sqrt{x^{2}-3x-5}}\:dx}\) adalah...
A. \(\mathrm{2\sqrt{x^{2}-3x-5}+C}\)
B. \(\mathrm{3\sqrt{x^{2}-3x-5}+C}\)
C. \(\mathrm{6\sqrt{x^{2}-3x-5}+C}\)
D. \(\mathrm{9\sqrt{x^{2}-3x-5}+C}\)
E. \(\mathrm{18\sqrt{x^{2}-3x-5}+C}\)
Pembahasan :
∫ (6x − 9)(x2 − 3x − 5)\(^{-\frac{1}{2}}\) dx
n = \(-\frac{1}{2}\)
k = \(\mathrm{\frac{6x-9}{2x-3}}\) = \(\mathrm{\frac{3(2x-3)}{2x-3}}\) = 3
⇒ \(\mathrm{\frac{k}{n+1}}\) 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\(\mathrm{\sqrt{x^{2}-3x-5}}\) + C
Emoticon