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Selamat datang di halaman contoh soal dan pembahasan "Integral". Di halaman ini akan membahas tentang contoh soal dan pembahasan lengkap mengenai Integral.
Untuk lebih jelasnya mari kita lihat contoh soal dan pembahasan dibawah ini!
41. Nilai \(\int_{0}^{\pi }\limits \sin 2x\cos x dx =\)
a) \(-\frac{4}{3}\)
b) \(-\frac{1}{3}\)
c) \(\frac{1}{3}\)
d) \(\frac{2}{3}\)
e) \(\frac{4}{3}\)
Pembahasan:
\(\int_{0}^{\pi }\sin 2x\cos xdx = \int_{0}^{\pi }\frac{1}{2}\left ( \sin 3x+\sin x \right )dx\)
\(=\frac{1}{2}\left [ -\frac{1}{3}\cos 3x-\cos x \right ]_{0}^{\pi }\)
\(=\frac{1}{2}\left [ \left ( -\frac{1}{3}\cos 3\pi -\cos \pi \right )-\left ( -\frac{1}{3}\cos 0-\cos 0 \right ) \right ]\)
\(=\frac{1}{2}\left [ \left \{ -\frac{1}{3}\left ( -1 \right )-\left ( -1 \right ) \right \}-\left ( -\frac{1}{3}. 1-1 \right ) \right ]\)
\(=\frac{1}{2}\left ( \frac{4}{3}+\frac{4}{3} \right )\)
\(=\frac{1}{2}\left ( \frac{8}{3} \right )=\frac{4}{3}\)
42. Volume benda putar yang terjadi, jika daerah antara kurva \(y=x^{2}+1\) dan \(y=x+3\) , di putar mengelilingi sumbu \(X\) adalah
a) \(\frac{67}{5}\pi\) satuan volum
b) \(\frac{107}{5}\pi\) satuan volum
c) \(\frac{117}{5}\pi\) satuan volum
d) \(\frac{133}{5}\pi\) satuan volum
e) \(\frac{183}{5}\pi\) satuan volum
Pembahasan:
\(V=\pi\int_{-1}^{2}\left\{ \left( x+3 \right)^{2}-\left( x^{2}+1 \right)^{2} \right\} dx \)
\( =\pi\int_{-1}^{2}\left( x^{2}+6x+9-x^{4}-2x^{2}-1 \right)dx\)
\(=\pi\int_{-1}^{2}\left( -x^{4}-x^{2}+6x+8 \right)dx\)
\(=\pi \left [ -\frac{1}{5}x^{5}-\frac{1}{3}x^{3}+3x^{2}+8x \right ]_{-1}^{2}\)
\(=\pi \left [ \left \{ -\frac{1}{5}\times 2^{5}-\frac{1}{3}\times 2^{3}+3\times 2^{2}+8\times 2 \right \}-\left \{ -\frac{1}{5}\left ( -1 \right )^{5} -\frac{1}{2}\left ( -1 \right )3+3\left ( -1 \right )^{2}+8\left ( -1 \right )\right \} \right ]\)
\(=\pi \left [ \left ( -\frac{32}{5}-\frac{8}{3}+12+16 \right )-\left ( \frac{1}{5}+\frac{1}{3}+3-8 \right ) \right ]\)
\(=\pi \left [- \frac{33}{5}+30 \right ]\)
\(=\frac{117}{5}\pi\)
43. Hasil dari \(\int\limits \cos ^{5}x dx\)
a) \(-\frac{1}{6}\cos ^{6}x\sin x+C\)
b) \(\frac{1}{6}\cos ^{6}x\sin x+C\)
c) \(-\sin x+\frac{2}{3}\sin ^{3}x+\frac{1}{5}\sin ^{5}x+C\)
d) \(\sin x-\frac{2}{3}\sin ^{3}x+\frac{1}{5}\sin ^{5}x+C\)
e) \(\sin x+\frac{2}{3}\sin ^{3}x+\frac{1}{5}\sin ^{5}x+C\)
Pembahasan:
\(\int \cos ^{5}x dx=\int \cos x\left ( \cos ^{4}x \right )dx\)
\( =\int \cos x\left ( \left ( \cos ^{2}x \right )^{2} \right )dx\)
\(=\int \cos x\left ( \left ( 1-\sin ^{2}x \right )^{2} \right )dx\int \cos x\left ( 1-2\sin ^{2}x+\sin ^{4}x \right )dx\)
\(=\int \cos x dx-2\int \sin ^{2}x\cos x dx+\int \sin ^{4}x dx\)
\(=\sin x-\frac{2}{3}\sin ^{3}x+\frac{1}{5}\sin ^{5}x+C\)
44. hasil dari \(\int_{0}^{1}\limits 3x\sqrt{3x^2 + 1} dx =\)
a) \(\frac{ 7}{ 2}\)
b) \(\frac{ 8}{ 3}\)
c) \(\frac{ 7}{ 3}\)
d) \(\frac{ 4}{ 3}\)
e) \(\frac{ 2}{ 3}\)
Pembahasan:
\(\int_{0}^{1}\frac{ 1}{ 2}\sqrt{u}du=\frac{ 1}{ 2}\int_{0}^{1}u^{\frac{ 1}{ 2}}\)
\(=\frac{ 1}{ 2}\left[ \frac{ 2}{ 3}u^{\frac{ 3}{ 2}} \right]_{0}^{1}\)
\(=\frac{ 1}{ 2}\left[ \frac{ 2}{ 3}\left( 3x^{2}+1 \right) ^{\frac{ 3}{ 2}}\right]_{0}^{1}\)
\(=\frac{ 1}{ 2}\left[ \frac{ 2}{ 3}\left( 3+1 \right)^\frac{ 3}{ 2}-\frac{ 2}{ 3}\left( 0+1 \right)^\frac{ 3}{ 2} \right]\frac{ 1}{ 2}\left[ \frac{ 2}{ 3}\sqrt{4^{3}}-\frac{ 2}{ 3}\left( 1 \right) \right]\frac{ 1}{ 2}\left[ \frac{ 16}{ 3}-\frac{ 2}{ 3} \right]\)
\(=\frac{ 1}{ 2}\left[ \frac{ 14}{ 3} \right]=\frac{ 7}{ 3}\)
45. Hasil dari \(\int\limits \cos^{4}2x\sin2x dx\)
a) \(-\frac{1}{10}\sin^{5}2x+C\)
b) \(-\frac{1}{10}\cos^{5}2x+C\)
c) \(-\frac{1}{5}\cos^{5}2x+C\)
d) \(\frac{1}{5}\cos^{5}2x+C\)
e) \(\frac{1}{10}\sin^{5}2x+C\)
Pembahasan:
\(\int\cos^{4}2x\sin2x dx= \int u^{4} \times -\frac{du}{2}\)
\(=-\frac{1}{2}\int u^{4} du=-\frac{1}{2}\left( \frac{1}{5}u^{5} \right)+c=-\frac{1}{10}\cos^{5}2x+c\)
46. Luas daerah yang dibatasi kurva \(y=4-x^{2}\), \(y=-x+2\) dan \(0\leq x\leq 2\) adalah...
a) \(\frac{8}{3}\) satuan luas
b) \(\frac{10}{3}\) satuan luas
c) \(\frac{14}{3}\) satuan luas
d) \(\frac{16}{3}\) satuan luas
e) \(\frac{26}{3}\) satuan luas
Pembahasan:
\(L=\int_{0}^{2}\left( 4-x^{2} \right)-\left( -x+2 \right) \)
\(=\int_{0}^{2}-x^{2}+x+2 dx=\left[ -\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+2x \right]{2 \atop 0}\)
\(=\left[ -\frac{1}{3}\left( 2 \right)^{3}+\frac{1}{2}\left( 2 \right)^{2}+2.2 \right]-\left[ -\frac{1}{3}.0^{3}+\frac{1}{2}.0^{2}+2.0 \right]\)
\(=\left[ -\frac{8}{3}+2+4 \right]-\left[ 0 \right]=\left[ 6-\frac{8}{3} \right]\)
\(=\frac{10}{3}\)
47. Volime benda putar yang terjadi jika daerah yang dibatasi oleh kurva \(y=x^{2}\), garis \(y=2x\) di kuadran \(1\) diputar \(360^{\circ}\) terhadap sumbu \(X\) adalah
a) \(\frac{20}{15}\pi\) satuan volume
b) \(\frac{30}{15}\pi\) satuan volume
c) \(\frac{54}{15}\pi\) satuan volume
d) \(\frac{64}{15}\pi\) satuan volume
e) \(\frac{144}{15}\pi\) satuan volume
Pembahasan:
\(V= \pi\int_{0}^{2}\left( 2x \right)^{2}-\left( x^{2} \right)^{2}\)
\(=\pi\int_{0}^{2}\left( 4x^{2}-x^{4} \right)dx\)
\(=\pi\left[ \frac{4}{3}x^{3}-\frac{1}{5}x^{5} \right]_{0}^{2}\)
\(=\pi\left[ \left( \frac{4}{3}\left( 2 \right)^{3}-\frac{1}{5}\left( 2 \right)^{5} \right)-\left( \frac{4}{3}\cdot0^{3}-\frac{1}{5}\cdot0^{5} \right) \right]\)
\(=\pi\left[ \frac{4}{3}\cdot8-\frac{1}{5}\cdot32-0 \right]\)
\(=\pi\left[ \frac{32}{3}-\frac{32}{5} \right]\)
\(=\pi\left( \frac{160-96}{15} \right)=\pi\frac{64}{15}\)
48. Hasil \(\int_{2}^{4}\limits \left( -x^{2}+6x-8 \right)dx=\)
a) \(\frac{38}{3}\)
b) \(\frac{26}{3}\)
c) \(\frac{20}{3}\)
d) \(\frac{16}{3}\)
e) \(\frac{4}{3}\)
Pembahasan:
\(\int_{2}^{4}\left( -x^{2}+6x-8 \right)=\left[ -\frac{1}{3}^{3}+3x^{2}-8x \right]_{2}^{4}\)
\(=\left[ \left( -\frac{1}{3}\left( 4 \right)^{3}+3\left( 4 \right)^{2}-8\cdot4 \right)-\left( -\frac{1}{3}\cdot2^{3}+3\cdot2^{2}-8\cdot2 \right) \right]=\left[ -\frac{64}{3}+48-32+\frac{8}{3}-12+16 \right]\)
\(=\left[ -\frac{64}{3}+\frac{8}{3}+20 \right]=\left[ -\frac{56}{3}+20 \right]=\frac{4}{3}\)
49. Hasil \(\int_{0}^{\pi}\limits \left( \sin 3x+ \cos x \right) dx=\)
a) \(\frac{10}{3}\)
b) \(\frac{8}{3}\)
c) \(\frac{4}{3}\)
d) \(\frac{2}{3}\)
e) \(-\frac{4}{3}\)
Pembahasan:
\(\int_{0}^{\pi} \left( \sin 3x+ \cos x \right) dx\)
\(= \left[ -\frac{1}{3}\cos 3x + \sin x \right]_{0}^{\pi}\)
\(=\left[ \left( -\frac{1}{3} \cos 3\pi + \sin \pi \right)-\left( -\frac{1}{3} \cos 0+ \sin 0 \right) \right]\)
\(=\left[ \left( -\frac{1}{3}\cdot -1 +0 \right)-\left( -\frac{1}{3}\cdot 1 +0 \right) \right]\)
\(=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)
50. Diketahui \(\int_{1}^{a}\limits \left( 2x-3 \right) dx=12 dan a\geq0\). Nilai \(a=\)...
a) \(2\)
b) \(3\)
c) \(5\)
d) \(7\)
e) \(10\)
Pembahasan:
\(\int_{1}^{a}\left( 2x-3 \right) dx=12\)
\(\Rightarrow\left[ x^{2}-3x \right]_{1}^{a}=12\)
\(\Rightarrow\left( a^{2}-3a \right)-\left( 1^{2}-3\cdot1 \right)=12\)
\(\Rightarrow a^{2}-3a-10=12\)
\(\Rightarrow \left( a-5 \right)\left( a+2 \right)=0\) \(a=5\) atau \(a=-2\)
