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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!
11. Gradien garis singgung di setiap titik pada kurva \(f (x)\) adalah \(3x^2 – 6x +5\). Jika kurva tersebut melalui titik \((1,3)\) maka persamaan kurva tersebut adalah
a) \(x^3 – 3x^2 + 5x – 6\)
b) \(x^3 – 3x^2 + 5x +6\)
c) \(6x^3 – 6x^2 + 5x – 8\)
d) \(6x^3 – 6x^2 + 5x – 2\)
e) \(6^3 – 6x^2 + 5x +2\)
Pembahasan:
\(y=x^3-3x^2+5x+c\)
\(-3=1^3-3*1^2+5*1+c\)
\(-3=3+c\)
\(c=6\)
\(f(x)=x^3-3x^2+5x+6\)
12. Jika \(f(x)=ax+b\),\(\int_{0}^{1}\limits f(x)dx=1\) dan \(\int_{1}^{2}\limits f(x)dx=5\),maka \(a+b\)=
a) \(3\)
b) \(4\)
c) \(5\)
d) \(-3\)
e) \(-4\)
Pembahasan:
\(\int_{0}^{2}f(x)dx=\int_{0}^{1}f(x)dx+\int_{1}^{2}f(x)dx\)
\(\int_{0}^{2}(ax+b)dx=1+5\)
\(\frac{1}{2}ax^2+bx]_0^2=6 \)
\(\frac{1}{2} a.2^2+b.2)=6\)
\(2a+2b=6\)
\(a+b=3\)
13. Diketahui \(f(x)=\frac{1}{3} x^3+\frac{1}{4}\), maka \(\int_{1}^{2}\limits \sqrt{4+(f'(x))}^2 dx\)
a) \(\frac{13}{6}\)
b) \(\frac{14}{6}\)
c) \(\frac{16}{6}\)
d) \(\frac{17}{6}\)
e) \(\frac{18}{6}\)
Pembahasan:
\(\int_{1}^{2}\sqrt{4+(f'(x))}^2 dx\)
=\(\int_{1}^{2}\sqrt{4+(x^4-2+\frac{1}{x^4})}dx \)
=\(\int_{1}^{2}\sqrt{x^4+2+\frac{1}{x^4}}dx \)
=\(\int_{1}^{2}(x^2+\frac{1}{x^2} )dx \)
=\(\frac{1}{3} x^3-\frac{1}{x} ]_1^2 \)
=\(\frac{1}{3} 2^3-\frac{1}{2})-(\frac{1}{3} 1^3-\frac{1}{2}) \)
=\(\frac{13}{6}\)
14. \(\int\limits \sin x sin?3x dx\)
a) \(\frac{1}{2}\cos 2x-\frac{1}{4}\cos 4x+c\)
b) \(\frac{1}{2}\sin 2x-\frac{1}{4}\sin 4x+c\)
c) \(\frac{1}{4}\sin 2x-\frac{1}{8}\sin 4x+c\)
d) \(-2\sin 2x-4\sin 4x+c\)
e) \(2\sin 2x-4\sin 4x+c\)
Pembahasan:
=\(\int \sin 3x\sin x dx\)
=\(\int \frac{-1}{2}{\cos 4x-\cos 2x} dx\)
=\(\frac{1}{2}\int\cos 4x dx+\frac{1}{2} \int\cos 2x dx\)
=\(\frac{-1}{8}\sin 4x+\frac{1}{4}\sin 2x+c \)
=\(\frac{1}{4}\sin 2x-\frac{1}{8}\sin 4x+c\)
15. \(\int_{2}^{5}\limits \frac{x+1}{\sqrt{x-1}}dx\)
a) \(\frac{35}{4}\)
b) \(\frac{26}{3}\)
c) \(\frac{22}{3}\)
d) \(\frac{35}{6}\)
e) \(\frac{23}{7}\)
Pembahasan:
\(\int_{2}^{5}\frac{x+1}{\sqrt{x-1}}dx\)
=\(\int_{1}^{2}\frac{p^2+2}{p}*2p dp \)
=\(2\int_{1}^{2}(p^2+2)dp \)
=\(2{\frac{1}{3} p^3+2p]_1^2 } \)
=\(2{(\frac{1}{3}*8+2*2)-(\frac{1}{3}+2)}\)
=\(2\frac{20}{3}-\frac{7}{3}\)
=\(\frac{26}{3}\)
16. \(\int\limits e^x\cos x dx\)=
a) \(\cos x e^x+\sin x e^x+c\)
b) \(\frac{1}{2}\cos x e^x+\frac{1}{2}\sin x e^x+c\)
c) \(\frac{1}{2}\cos x e^x+\sin x e^x+c\)
d) \(\cos x e^x+\frac{1}{2}\sin x e^x+c\)
e) \(2\cos x e^x+2\sin x e^x+c\)
Pembahasan:
\(\int e^x\cos x dx=\int \cos xe^x dx\)
=\(\cos x e^x-\int e^x (-\sin x) dx\)
=\(\cos x e^x+\int e^x \sin x dx\)
=\(\cos x e^x+\sin xe^x-\int e^x\cos x dx\)
=\(\cos x e^x+\sin xe^x\)
=\(\frac{1}{2}\cos xe^x+\frac{1}{2}\sin xe^x+c\)
17. Luas daerah yang dibatasi oleh kurva \(f(x)=x^2+2x+3\) dan \(g(x)=3-x\) adalah
a) \(3\)
b) \(4,5\)
c) \(6\)
d) \(7,5\)
e) \(9\)
Pembahasan:
\(L=\int_{-3}^{0}f(x)-g(x)dx \)
=\(\int_{-3}^{0}{(x^2+2x+3)-(3-x)}dx \)
=\(\int_{-3}^{0}(x^2+3x)dx \)
=\(\frac{1}{3}x^3+\frac{3}{2} x^2\)
=\(\frac{1}{3}0^3+\frac{3}{2}.0^2 )-(\frac{1}{3}.(-3)^3+\frac{3}{2}.(-3)^2 ) \)
=\(0-(-9+\frac{27}{2}) \)
=\(4,5\)
18. Volume benda putar yang terjadi jika daerah yang dibatasi oleh kurva \(f(x)= 3x – 2\) , garis \(x = 1\), dan \(x = 3\) kemudian diputar mengelilingi sumbu \(X\) adalah
a) \(34\pi\)
b) \(38\pi\)
c) \(46\pi\)
d) \(50\pi\)
e) \(52\pi\)
Pembahasan:
= \(\pi \int_{1}^{3}(3x-2)^2 dx\)
=\(\pi \int_{1}^{3}(9x^2-12x+4)dx \)
=\(\pi {3x^3-6x^2+4x]_1^3} \)
=\(\pi {(3*3^3-6*3^2+4*3)-(3*1^3-6*1^2+4*1)} \)
=\(38\pi\)
19. Volume benda putar yang terjadi jika daerah yang dibatasi oleh kurva \(y = - x^2\) dan \(y = - x – 2\) kemudian diputar mengelilingi summbu \(X\) adalaah
a) \(13 \frac{2}{3} \pi\)
b) \(14 \frac{2}{5} \pi\)
c) \(15 \frac{2}{3} \pi\)
d) \(17 \frac{2}{5} \pi\)
e) \(18 \frac{2}{3} \pi\)
Pembahasan:
=\(\pi\int_{1}^{2}{(-x^2 )^2-(-x-2)^2 }dx\)
=\(\pi \int_{1}^{2}x^4-x^2-4x-4 dx\)
=\(\pi \frac{1}{5} x^5-\frac{1}{3} x^3-2x^2-4x]_(-1)^2\)
=\(\pi {(\frac{1}{5}*2^5-\frac{1}{3}*2^3-2*2^2-4*2)-(\frac{1}{5}*2^5-\frac{1}{3}*2^3-2*(-1)^2-4*(-1))} \)
=\(\pi \left | -12\frac{4}{15} -2\frac{2}{15}\right | \)
=\(14 \frac{2}{5}\pi\)
20. hasil dari \(\int\limits (2x-1)(x^2-x+3)^3 dx\)
a) \(\frac{1}{3}(x^2-x+3)^3+c\)
b) \(\frac{1}{4}(x^2-x+3)^3+c\)
c) \(\frac{1}{4}(x^2-x+3)^4+c\)
d) \(\frac{1}{2}(x^2-x+3)^4+c\)
e) \((x^2-x+3)^4+c\)
Pembahasan:
\(\int (2x-1)(x^2-x+3)^3 dx\)
\(\int (2x-1)u^3 \frac{du}{2x-1} \)
\(\int u^3 du \)
\(\frac{1}{4}u^4+c \)
\(\frac{1}{4}(x^2-x+3)^4 +c\)
