Sabtu, 07 Agustus 2010

INTEGRAL 1 dalam MATEMATIKA

Soal dan pembahasan dari UAN IPA

1. Hasil \int_{}^{}\frac{x^{2}dx}{\sqrt{x^{3}-5}} =... (UAN 01)

Pembahasan :

misal u = x^{3}-5, maka \frac{du}{dx}=3x^{2}, sehingga \frac{du}{3}=x^{2}dx.

\int_{}^{}\frac{x^{2}dx}{\sqrt{x^{3}-5}} =\int_{}^{}\frac{du}{3\sqrt{u}}

= \frac{1}{3}\int_{}^{}\frac{du}{{u}^{\frac{1}{2}} }

= \frac{1}{3}\int_{}^{}{{u}^{-\frac{1}{2}} }du

= \frac{1}{3}\frac{1}{\frac{1}{2}}u^\frac{1}{2}+ C

= \frac{2}{3}\sqrt{u}+C

= \frac{2}{3}\sqrt{x^{3}-5}+ C

Jaadi \int_{}^{}\frac{x^{2}dx}{\sqrt{x^{3}-5}} =\frac{2}{3}\sqrt{x^{3}-5}+ C

2. Hasil dari \int_{}^{}\cos x\cos4xdx=...

Pembahasan:

Ingat rumus perkalian trigonometri yang dapat diubah menjadi bentuk penjumlahan: \cos x \cos y=\frac{1}{2}(\cos(x+y)+\cos(x-y)).

\int_{}^{}\cos x \cos 4x dx=\int_{}^{}\frac{1}{2}(\cos(x+4x)+\cos(x-4x))dx

= \int_{}^{}\frac{1}{2}(\cos(5x)+\cos(-3x))dx

= \int_{}^{}\frac{1}{2}(\cos(5x)+\cos(3x))dx

= \frac{1}{2}.\frac{1}{5}\sin(5x)+\frac{1}{2}.\frac{1}{3}\sin(3x)+C

= \frac{1}{10}\sin(5x)+\frac{1}{6}\sin(3x)+C

Jadi \int_{}^{}cos x \cos 4x dx= \frac{1}{10}\sin(5x)+\frac{1}{6}\sin(3x)+C

3. \int_{}^{}\sin^{5}x\cos x dx adalah …. (EBT 88)

Pembahasan

soal ini dapat diselesaikan dengan cara subtitusi u= \sin x , sehingga \frac{du}{dx}= \cos x.
du = \cos x dx,

\int_{}^{}\sin^{5}x\cos x dx= \int_{}^{}u^{5}du

= \frac{1}{6}u^{6} +C

= \frac{1}{6}\sin ^{6}x +C

Jadi \int_{}^{}\sin^{5}x\cos x dx= \frac{1}{6}\sin ^{6}x +C.

4. \int_{\sqrt{6}}^{3\sqrt{2}}x\sqrt{x^{2}-2}dx = .... (UAN 02)

Pembahasan

Dengan subtitusi u = x^{2}-2 , kita dapatkan \frac{du}{dx}=2x, sehingga \frac{du}{2}= x dx.

\int_{\sqrt{6}}^{3\sqrt{2}}x\sqrt{x^{2}-2}dx

=  \int_{\sqrt{6}}^{3\sqrt{2}}\sqrt{u}\frac{du}{2}

= \int_{\sqrt{6}}^{3\sqrt{2}}\frac{1}{2}u^{\frac{1}{2}}du

= \frac{1}{2}.\frac{2}{3}u^{\frac{3}{2}}

= \frac{1}{3}(x^{2}-2)^{\frac{3}{2}}

= \frac{1}{3}((3\sqrt{2})^{2}-2)^{\frac{3}{2}}-\frac{1}{3}((\sqrt{6})^{2}-2)^{\frac{3}{2}}

= \frac{1}{3}(18-2)^{\frac{3}{2}}-\frac{1}{3}(6-2)^{\frac{3}{2}}

= \frac{1}{3}(16)^{\frac{3}{2}}-\frac{1}{3}(4)^{\frac{3}{2}}

= \frac{1}{3}.64 -\frac{1}{3}.8

= \frac{64}{3} -\frac{8}{3} = \frac{56}{3}

Jadi \int_{\sqrt{6}}^{3\sqrt{2}}x\sqrt{x^{2}-2}dx = \frac{56}{3}

5. \int_{0}^{\frac{\pi}{6}}\sin(x+\frac{\pi}{3})\cos(x+\frac{\pi}{3})dx=....

Pembahasan

Soal ini dapat diselesaikan dengan menggunakan rumus sudut rangkap: \sin 2a = 2 \sin a \cos a
\int_{0}^{\frac{\pi}{6}}\sin(x+\frac{\pi}{3})\cos(x+\frac{\pi}{3})dx=

= \int_{0}^{\frac{\pi}{6}}\frac{1}{2}\sin2(x+\frac{\pi}{3})dx

= -\frac{1}{4}\cos 2(x+\frac{\pi}{3})

= -\frac{1}{4}\cos 2(\frac{\pi}{6}+\frac{\pi}{3})-(-\frac{1}{4}\cos 2(0+\frac{\pi}{3}))

= -\frac{1}{4}\cos 2(\frac{3\pi}{6})-(-\frac{1}{4}\cos 2(\frac{\pi}{3}))

= -\frac{1}{4}\cos (\pi)-(-\frac{1}{4}\cos (\frac{2\pi}{3}))

= -\frac{1}{4}(-1)-(-\frac{1}{4}.-(\frac{1}{2}))

= \frac{1}{4}-\frac{1}{8}= \frac{1}{8}

Jadi nilai dari \int_{0}^{\frac{\pi}{6}}\sin(x+\frac{\pi}{3})\cos(x+\frac{\pi}{3})dx= \frac{1}{8}

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