Nomor 31
Nilai dari $ \int \limits_{-1}^2 (x-1)(3x+1) dx =...$
$\spadesuit \, $ Rumus dasar : $\int kx^n dx = \frac{k}{n+1} x^{n+1} + c $
$\begin{align*} \int \limits_{-1}^2 (x-1)(3x+1) dx &= \int \limits_{-1}^2 (x-1)(3x+1) dx \\ &= \int \limits_{-1}^2 (3x^2-2x-1) dx \\ &= \left[ \frac{3}{3} x^3 - \frac{2}{2} x^2 - x \right]_{-1}^2 \\ &= \left[ x^3 - x^2 - x \right]_{-1}^2 \\ &= [2^3 - 2^2 - 2] - [(-1)^3 - (-1)^2 - (-1)] \\ &= 2-(-1) \\ &= 3 \end{align*}$
Jadi, $\int \limits_{-1}^2 (x-1)(3x+1) dx = 3 . \heartsuit$
$\begin{align*} \int \limits_{-1}^2 (x-1)(3x+1) dx &= \int \limits_{-1}^2 (x-1)(3x+1) dx \\ &= \int \limits_{-1}^2 (3x^2-2x-1) dx \\ &= \left[ \frac{3}{3} x^3 - \frac{2}{2} x^2 - x \right]_{-1}^2 \\ &= \left[ x^3 - x^2 - x \right]_{-1}^2 \\ &= [2^3 - 2^2 - 2] - [(-1)^3 - (-1)^2 - (-1)] \\ &= 2-(-1) \\ &= 3 \end{align*}$
Jadi, $\int \limits_{-1}^2 (x-1)(3x+1) dx = 3 . \heartsuit$
Nomor 32
Nilai dari $ \int \limits_{0}^{\frac{\pi}{2}} (\sin 2x \cos 2x) dx=...$
$\clubsuit \,$ Rumus dasar :
$\sin 2px = 2 \sin px . \cos px \Rightarrow \sin px . \cos px = \frac{1}{2} \sin 2px $
dan $ \int \sin ax \, dx = -\frac{1}{a} \cos ax + c$
$\begin{align*} \int \limits_{0}^{\frac{\pi}{2}} (\sin 2x \cos 2x) dx &= \int \limits_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin 2.(2x) \, dx \\ &= \int \limits_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin 4x \, dx \\ &= \left[ \frac{1}{2} . -\frac{1}{4} \cos 4x \right]_{0}^{\frac{\pi}{2}} \\ &= -\frac{1}{8} \left[ \cos 4x \right]_{0}^{\frac{\pi}{2}} \\ &= -\frac{1}{8} . ( \left[ \cos 4.\frac{\pi}{2} \right] - \left[ \cos 4.0 \right] ) \\ &= -\frac{1}{8} . ( \left[ \cos 2\pi \right] - \left[ \cos 0 \right] ) \\ &= -\frac{1}{8} . ( \left[ 1 \right] - \left[ 1 \right] ) \\ &= -\frac{1}{8} . 0 \\ &= 0 \end{align*}$
Jadi, $\int \limits_{0}^{\frac{\pi}{2}} (\sin 2x \cos 2x) dx = 0 .\heartsuit $
$\sin 2px = 2 \sin px . \cos px \Rightarrow \sin px . \cos px = \frac{1}{2} \sin 2px $
dan $ \int \sin ax \, dx = -\frac{1}{a} \cos ax + c$
$\begin{align*} \int \limits_{0}^{\frac{\pi}{2}} (\sin 2x \cos 2x) dx &= \int \limits_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin 2.(2x) \, dx \\ &= \int \limits_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin 4x \, dx \\ &= \left[ \frac{1}{2} . -\frac{1}{4} \cos 4x \right]_{0}^{\frac{\pi}{2}} \\ &= -\frac{1}{8} \left[ \cos 4x \right]_{0}^{\frac{\pi}{2}} \\ &= -\frac{1}{8} . ( \left[ \cos 4.\frac{\pi}{2} \right] - \left[ \cos 4.0 \right] ) \\ &= -\frac{1}{8} . ( \left[ \cos 2\pi \right] - \left[ \cos 0 \right] ) \\ &= -\frac{1}{8} . ( \left[ 1 \right] - \left[ 1 \right] ) \\ &= -\frac{1}{8} . 0 \\ &= 0 \end{align*}$
Jadi, $\int \limits_{0}^{\frac{\pi}{2}} (\sin 2x \cos 2x) dx = 0 .\heartsuit $
Nomor 33
Hasil dari $ \int (\cos ^23x \sin 3x)dx = ...$
$\spadesuit \, $ Menentukan integral dengan substitusi :
$\begin{align*} \int (\cos ^23x \sin 3x)dx &= \int (\cos 3x)^2 \sin 3x \, dx \\ &= \int (u)^2 \sin 3x \, \frac{du}{u^\prime} \, \text{(misal : } \, u=\cos 3x ) \\ &= \int (u)^2 \sin 3x \, \frac{du}{-3\sin 3x} \\ &= -\frac{1}{3} \int (u)^2 \, du \\ &= -\frac{1}{3} . \frac{1}{3} . u^3 +c \\ &= -\frac{1}{9}. (\cos 3x)^3 +c \\ &= -\frac{1}{9}.\cos ^3 3x +c \end{align*}$
Jadi, $\int (\cos ^23x \sin 3x)dx = -\frac{1}{9}.\cos ^3 3x + c . \heartsuit $
$\begin{align*} \int (\cos ^23x \sin 3x)dx &= \int (\cos 3x)^2 \sin 3x \, dx \\ &= \int (u)^2 \sin 3x \, \frac{du}{u^\prime} \, \text{(misal : } \, u=\cos 3x ) \\ &= \int (u)^2 \sin 3x \, \frac{du}{-3\sin 3x} \\ &= -\frac{1}{3} \int (u)^2 \, du \\ &= -\frac{1}{3} . \frac{1}{3} . u^3 +c \\ &= -\frac{1}{9}. (\cos 3x)^3 +c \\ &= -\frac{1}{9}.\cos ^3 3x +c \end{align*}$
Jadi, $\int (\cos ^23x \sin 3x)dx = -\frac{1}{9}.\cos ^3 3x + c . \heartsuit $
Nomor 34
Luas daerah yang diarsir pada gambar dapat dinyatakan dengan rumus ...
$\clubsuit \, $ Titik potong kurva $y=x^2+4x+4$ pada sumbu X : $y=0$
$x^2+4x+4=0 \Rightarrow (x+2)^2=0 \Rightarrow x=-2$
$\clubsuit \, $ Gambarnya:
$\clubsuit \, $ Menentukan luas A dan B :
$L_A= \int \limits_{-2}^{1} (x^2+4x+4) dx $
$L_B= \int \limits_{1}^{10} (10-x) dx $
$\clubsuit \, $ Luas total :
Jadi, $L_{\text{(arsir)}}=L_A + L_B =\int \limits_{-2}^{1} (x^2+4x+4) dx + \int \limits_{1}^{10} (10-x) dx . \heartsuit$
$x^2+4x+4=0 \Rightarrow (x+2)^2=0 \Rightarrow x=-2$
$\clubsuit \, $ Gambarnya:
$\clubsuit \, $ Menentukan luas A dan B :
$L_A= \int \limits_{-2}^{1} (x^2+4x+4) dx $
$L_B= \int \limits_{1}^{10} (10-x) dx $
$\clubsuit \, $ Luas total :
Jadi, $L_{\text{(arsir)}}=L_A + L_B =\int \limits_{-2}^{1} (x^2+4x+4) dx + \int \limits_{1}^{10} (10-x) dx . \heartsuit$
Nomor 35
Volume benda putar yang terbentuk dari daerah di kuadran I yang dibatasi oleh kurva $y=\sqrt{3}x^2 $ , lingkaran $x^2+y^2=4$ dan sumbu X,
diputar mengelilingi sumbu X adalah ...
$\spadesuit \, $ Gambarnya :
$\spadesuit \, $ Menentukan volume A dan B :
$V_A = \pi \int \limits_{0}^{1} \left( \sqrt{3}x^2 \right)^2 \, dx = \pi \int \limits_{0}^{1} 3x^4 \, dx = \frac{3}{5} \pi $
$V_B = \pi \int \limits_{1}^{2} (4-x^2) \, dx = \frac{5}{3} \pi $
$\spadesuit \, $ Menentukan volume arsir :
$V_{(arsir)} = V_A + V_B = \frac{3}{5} \pi + \frac{5}{3} \pi = \frac{34}{15} \pi$
Jadi, $V_{(arsir)} = \frac{34}{15} \pi . \heartsuit $
$\spadesuit \, $ Menentukan volume A dan B :
$V_A = \pi \int \limits_{0}^{1} \left( \sqrt{3}x^2 \right)^2 \, dx = \pi \int \limits_{0}^{1} 3x^4 \, dx = \frac{3}{5} \pi $
$V_B = \pi \int \limits_{1}^{2} (4-x^2) \, dx = \frac{5}{3} \pi $
$\spadesuit \, $ Menentukan volume arsir :
$V_{(arsir)} = V_A + V_B = \frac{3}{5} \pi + \frac{5}{3} \pi = \frac{34}{15} \pi$
Jadi, $V_{(arsir)} = \frac{34}{15} \pi . \heartsuit $
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