$\begin{array}{ll}\\ 1.&\textbf{(UAN 2014)}\\ &\textrm{Hasil}\: \: \displaystyle \int_{-1}^{2}\left ( x^{3}+3x^{2}+4x+5 \right )\: dx=\: ....\\ &\begin{array}{lll} \textrm{a}.&33\displaystyle \frac{1}{4}\\ \textrm{b}.&\color{red}33\displaystyle \frac{3}{4}\\ \textrm{c}.&32\displaystyle \frac{1}{4}\\ \textrm{d}.&31\displaystyle \frac{3}{4}\\ \textrm{e}.&23\displaystyle \frac{3}{4} \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\int_{-1}^{2}\left ( x^{3}+3x^{2}+4x+5 \right )\: dx\\ &=\displaystyle \frac{x^{4}}{4}+x^{3}+2x^{2}+5x\: |_{-1}^{2}\\ &=\left (\displaystyle \frac{2^{4}}{4}+2^{3}+2.2^{2}+5.2 \right )\\ &\quad -\left ( \frac{\left ( -1 \right )^{4}}{4}+\left ( -1 \right )^{3}+2.\left ( -1 \right )^{2}+5\left ( -1 \right ) \right )\\ &=\color{red}33\frac{3}{4} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textbf{(UAN 2003)}\\ &\textrm{Jika}\: f(x)=\left ( x-2 \right )^{2}-4\: \: \textrm{dan}\: \: g(x)=-f(x),\\ &\textrm{maka luas daerah yang di batasi kurva }\\ &f\: \: \textrm{dan}\: \: g\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll} \textrm{a}.&10\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{b}.&\color{red}21\displaystyle \frac{1}{3}\: \: \textrm{satuan luas}\\ \textrm{c}.&22\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{d}.&42\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{e}.&45\displaystyle \frac{1}{3}\: \: \textrm{satuan luas} \end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.
$.\qquad\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\displaystyle \int_{0}^{4}\left ( g(x)-f(x) \right )\: \: dx\\ &=\displaystyle \int_{0}^{4}\left ( 4x-x^{2} \right )-\left ( x^{2}-4x \right )\: \: dx\\ &=\displaystyle \int_{0}^{4}\left ( 8x-2x^{2} \right )\: \: dx\\ &=\displaystyle \left [4x^{2}-\frac{2}{3}x^{3} \right ]_{0}^{4}\\ &=\displaystyle \left ( 4.4^{2}-\frac{2}{3}.4^{3} \right )-\left ( 4.0^{2}-\frac{2}{3}.0^{3} \right )\\ &=\displaystyle \left ( 64-\frac{2}{3}.64 \right )-0\\ &=\displaystyle \frac{64}{3}=\color{red}21\frac{1}{3}\: \: \color{black}\textrm{satuan luas} \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan menggunakan rumus}\: \: L=\displaystyle \frac{D\sqrt{D}}{6a^{2}}\\ &\begin{array}{|l|}\hline \begin{aligned}&f(x)=g(x)\\ &f(x)=-f(x),\quad\textrm{ingat}\: \: g(x)= -f(x)\\ &2f(x)=0,\quad\color{blue}\textrm{tidak boleh disederhanakan},\\ &2\times \left (\left ( x-2 \right )^{2}-4 \right )=0,\quad \color{blue}\textrm{karena akan }\\ &2\times \left ( x^{2}-4x \right )=0\quad \color{blue}\textrm{mempengaruhi hasil }\\ &2x^{2}-8x=0,\qquad\quad \color{blue}\textrm{akhir}\\ &\begin{cases} a=2,\: b=-8 & c=0 \\ D=b^{2}-4ac, & D=\left ( -8 \right )^{2}-4(2)(0)=64 \end{cases}\\ &L_{\: \textbf{daerah}}=\displaystyle \frac{\textbf{D}\sqrt{\textbf{D}}}{6\textbf{a}^{2}}\\ &\quad\qquad=\displaystyle \frac{64\sqrt{64}}{6(2)^{2}}\\ &\quad\qquad=\displaystyle \frac{64\times 8}{6\times 4}\\ &\quad\qquad=\displaystyle \frac{64}{3}\\ &\quad\qquad=\color{red}21\displaystyle \frac{1}{3} \end{aligned}\\\hline\end{array} \end{aligned}$.
$\begin{array}{ll}\\ 3.&\textbf{(UAN 2014)}\\ &\textrm{Luas daerah yang diarsir pada gambar }\\ &\textrm{dapat dinyatakan dengan rumus} \end{array}$.
$.\: \quad\begin{array}{ll}\\ &\begin{array}{lll} \textrm{a}.&\displaystyle \int_{0}^{4}4x\: dx-\int_{2}^{4}\left ( 2x-4 \right )\: dx\\ \textrm{b}.&\displaystyle \int_{0}^{4}4x\: dx - \int_{2}^{4}\left ( 2x+4 \right )\: dx\\ \textrm{c}.&\color{red}\displaystyle \int_{0}^{4}2\sqrt{2}\: dx - \int_{2}^{4}\left ( 2x-4 \right )\: dx\\ \textrm{d}.&\displaystyle \int_{0}^{4}2\sqrt{2}\: dx - \int_{2}^{4}\left ( 4-2x \right )\: dx\\ \textrm{e}.&\displaystyle \int_{0}^{4}2\sqrt{x}\: dx - \int_{2}^{4}\left ( 4+2x \right )\: dx \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Luas}_{\color{blue}\textrm{arsiran}}=\displaystyle \int_{0}^{4}y_{1}\: dx-\int_{2}^{4}y_{2}\: dx\\ &=\displaystyle \int_{0}^{4}\sqrt{4x}\: dx-\int_{2}^{4}\left ( 2x-4 \right )\: dx\\ & =\displaystyle \int_{0}^{4}2\sqrt{x}\: dx-\int_{2}^{4}\left ( 2x-4 \right )\: dx. \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textbf{(UAN 2014)}\\ &\textrm{Volume benda putar yang terbentuk dari daerah}\\ &\textrm{yang dikuadran I yang dibatasi oleh kurva}\\ &x=2\sqrt{3}y^{2}\: ,\: \textrm{sumbu Y , dan lingkaran}\: x^{2}+y^{2}=1,\\ &\textrm{diputar mengelilingi sumbu Y adalah}\: ....\\ &\begin{array}{lll} \textrm{a}.&\displaystyle \frac{4}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{b}.&\color{red}\displaystyle \frac{17}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{c}.&\displaystyle \frac{23}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{d}.&\displaystyle \frac{44}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{e}.&\displaystyle \frac{112}{60}\pi \: \: \textrm{satuan volum} \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.
$.\: \quad\begin{aligned}&\textrm{Volumenya jika diputar mengelilingi Sumbu }\\ &\textrm{Y adalah}:\\ & V=\pi \displaystyle \int_{0}^{\frac{1}{2}}x_{1}^{2}\: dy+\pi \int_{\frac{1}{2}}^{1}x_{2}^{2}\: dy\\ &\Leftrightarrow \: \: V=\pi \int_{0}^{\frac{1}{2}}\left ( 2\sqrt{3}y^{2} \right )^{2}\: dy+\pi \int_{\frac{1}{2}}^{1}\left ( 1-y^{2} \right )\: dy\\ &\Leftrightarrow \: \: V=\frac{12}{5}y^{5}\pi |_{0}^{\frac{1}{2}}+\left ( y-\frac{1}{3}y^{3} \right )\pi |_{\frac{1}{2}}^{1}\\ &\Leftrightarrow \: \: V=\pi \left ( \frac{12}{5}\times \frac{1}{32}+\left ( 1-\frac{1}{3} \right )-\left ( \frac{1}{2}-\frac{1}{3}\times \frac{1}{8} \right ) \right )\\ &\Leftrightarrow \: \: V=\color{red}\frac{17}{60}\pi.\end{aligned}$.
$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \displaystyle \frac{d}{dx}g(x)=f(x)\: \: \textrm{di mana} \: f(x)\\ &\textrm{kontinu dari a sampai b, }\\ &\textrm{maka}\: \displaystyle \int_{a}^{b}f(x).g(x)\: dx\\ &\begin{array}{ll}\\ \textrm{a}.\quad 0\\ \textrm{b}.\quad f(b)-f(a)\\ \textrm{c}.\quad g(b)-g(a)\\ \textrm{d}.\quad \displaystyle \frac{\left [ f(b) \right ]^{2}-\left [ f(a) \right ]^{2}}{2}\\ \textrm{e}.\quad \color{red}\displaystyle \frac{\left [ g(b) \right ]^{2}-\left [ g(a) \right ]^{2}}{2}\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int_{a}^{b} f(x).g(x)\: dx\\ &=\displaystyle \int_{a}^{b} g(x).f(x)\: dx\\ &=\displaystyle \int_{a}^{b} g(x).\displaystyle \frac{d\left ( g(x) \right )}{dx}\: dx,\\ &\quad \textrm{ingat bahwa}\quad \displaystyle \frac{d}{dx}g(x)=f(x),\\ &\quad\textrm{f(x) kontinu dari a sampai b}\\ &=\displaystyle \int_{a}^{b} g(x)\: d\left ( g(x) \right )\\ &=\left [\displaystyle \frac{\left ( g(x) \right )^{2}}{2} \right ]_{a}^{b}\\ &=\displaystyle \frac{\left [ g(b) \right ]^{2}-\left [ g(a) \right ]^{2}}{2} \end{aligned} \end{array}$.
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