Agustus | 2011 | Belajar Matematika dan Sains SMP-SMA

Soal integral tertentu XII ipa

09 Agu 2011 4 Komentar

by Raharja in Integral

11. a.
$\int_{0}^{2\pi}\(\sin \theta +\cos \theta \)d\theta \\ = \[-\cos \theta+\sin \theta]_{0}^{2\pi} \\= \[-\cos 2\pi+\sin 2\pi]-\[-\cos 0+\sin 0]\\=\[-1+0]-\[-1+0]\\=0$

11.b.

$\int_{\frac{5\pi}{6}}^{\frac{3\pi}{2}}\(\cos t-\sin t)dt \\=\[\sin t+\cos t]_{\frac{5\pi}{6}}^{\frac{3\pi}{2}}\\=\[\sin \frac{3\pi}{2} +\cos \frac{3\pi}{2}]-\[\sin \frac{5\pi}{6}+\cos \frac{5\pi}{6}]\\=\[-1+0]-\[\frac{1}{2}+\(-\frac{1}{2}\sqrt3)]\\=\frac{1}{2}\sqrt3-1\frac{1}{2}$

11.c.

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\(4\cos \theta-1)d\theta \\=\[4\sin \theta-\theta]_{\frac{\pi}{3}}^{\frac{\pi}{2}}\\=\[4\sin \frac{\pi}{2}-\frac{\pi}{2}]-\[4\sin \frac{\pi}{3}-\frac{\pi}{3}]\\=\[4(1)-\frac{\pi}{2}]-\[4(\frac{1}{2}\sqrt3)-\frac{\pi}{3}]\\=\[4-\frac{\pi}{2}]-\[2\sqrt3-\frac{\pi}{3}]\\=\4-2\sqrt3-\frac{\pi}{6}$

13.a. Diketahui $y=\frac{x}{2x-4}$ dan $\frac{dy}{dx}=4f(x)$ dimana f(x) ialah fungsi dalam x . Hitung nilai dari $\int_{0}^{1}f(x)dx$ .

Jawab : $y=\frac{x}{2x-4}$ maka :

$\frac{dy}{dx}=4f(x)\Rightarrow dy=4f(x)dx\\\Rightarrow \int dy=\int 4f(x)dx\\\Rightarrow y=4\int f(x)dx\\\Rightarrow \int f(x)dx=\frac{1}{4}y+C\\\Rightarrow \int f(x)dx=\frac{1}{4}\(\frac{x}{2x-4})+C=\frac{x}{8x-16}+C$

Sehingga

$\int_{0}^{1} f(x)dx=\[\frac{x}{8x-16}]_{0}^{1}=\frac{1}{8-16}-\frac{0}{0-16}=-\frac{1}{8}$