$\begin{array}{ll}\\ 26.&(\textbf{SBMPTN Mat IPA 2013})\\&\textrm{Diketahui sebuah lingkaran berjari-jari}\: \: 2\\ &\textrm{sebagaimana ilustrasi dibawah. Jika tali}\\ &\textrm{busur pada gambar berjarak 1 dari garis}\\ &\textrm{tengah, maka luas daerah di atas tali busur}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 2\int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{b}.\quad \color{red}\displaystyle 2\int_{0}^{\sqrt{3}}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{1}\sqrt{4-x^{2}}\: dx\\ \textrm{e}.\quad \displaystyle 2\int_{0}^{\sqrt{3}}\sqrt{4-x^{2}}\: dx\end{array} \end{array}$.
$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Diketahui bahwa sebuah lingkaran}\\ &\textrm{dengan}\: \: r=2,\: \textrm{artinya}\: \: x^{2}+y^{2}=r^{2}=2^{2}=4\\ &\textrm{Diketahui pula persamaan garis}\: \: y=1,\: \textrm{maka}\\&x^{2}+1^{2}=4\Leftrightarrow x^{2}=4-1=3\Leftrightarrow x=\pm \sqrt{3} \end{aligned}$
$.\: \qquad\begin{aligned}&\textrm{Sehingga luas daerah di atas tali busur}\\ &\textrm{yaitu berupa}\: \: \textbf{temberang}\: \textrm{adalah}\\ &=\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}\left ( f_{2}(x)-f_{1}(x) \right )dx\\ &=\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ &\textrm{ingat fungsi genap pada pembahasan}\\ &\textrm{sebelumnya, karena}\: f(x)=\sqrt{4-x^{2}}=f(-x)\\ &=\color{red}2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx \end{aligned}$.
$\begin{array}{ll}\\ 27.&(\textbf{SBMPTN Mat IPA 2013})\\&\textrm{Diketahui sebuah lingkaran berjari-jari}\: \: 2\\ &\textrm{sebagaimana ilustrasi dibawah. Jika tali}\\ &\textrm{busur pada gambar berjarak 1 dari garis}\\ &\textrm{tengah, maka luas daerah di bawah tali busur}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 2\pi -2\int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{b}.\quad \displaystyle 2\pi - 2\int_{0}^{\sqrt{3}}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{c}.\quad \displaystyle 4\pi -\int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{d}.\quad \color{red}\displaystyle 4\pi -2\int_{0}^{\sqrt{3}}\left (\sqrt{4-x^{2}}-1 \right )\: dx\\ \textrm{e}.\quad \displaystyle 4\pi -2\int_{0}^{\sqrt{3}}\sqrt{4-x^{2}}\: dx\end{array}\\\\&\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dengan ilustrasi sama dengan no.24 di atas}\\ &\textrm{Kita dapat tentukan luas wilayah di bawah}\\ &\textbf{tembereng},\: \textrm{yaitu}:\\&=\textrm{Luas lingkaran penuh itu}-\textbf{tembereng}\\ &=\pi .r^{2}-2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx\\ &=\pi .(2)^{2}-2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx\\&=\color{red}4\pi -2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 28.&(\textbf{SBMPTN Mat IPA 2013})\\&\textrm{Luas daerah yang dibatasi oleh kurva}\\ &y=2-x^{2}\: \: \textrm{dan}\: \: y=\left | x \right |\: \: \textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle 2\int_{-1}^{0}\left (-x^{2}+x+2 \right )dx\\ \textrm{b}.\quad \displaystyle 2\int_{-1}^{0}\left ( -x^{2}-x+2 \right )dx\\ \textrm{c}.\quad \displaystyle 2\int_{-1}^{0}\left ( -x^{2}+x-2 \right )dx\\ \textrm{d}.\quad \displaystyle \int_{-1}^{1}\left (-x^{2}-x+2 \right )\: dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{1}\left ( -x^{2}+x+2 \right )\: dx\end{array} \end{array}$.
$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Kita tentukan poin batas-batasnya, yaitu}:\\ &\textrm{Titik potong kurva dengan sumbu-X},\: y=0\\ &y=2-x^{2}=0\Leftrightarrow (\sqrt{2}+x)(\sqrt{2}-x)=0\\ &\Leftrightarrow x=-\sqrt{2}\: \: \textrm{atau}\: \: x=\sqrt{2}\\ &\textrm{Titik potong kurva dengan garis, yaitu}:\\ &2-x^{2}=\left | x \right |\Leftrightarrow x^{2}+\left | x \right |-2=0\\ &\bullet \: \: x\geq 0,\: \left | x \right |=x\\ &\: \quad \Leftrightarrow x^{2}+x-2=(x+2)(x-1)=0\\ &\: \quad \Leftrightarrow x=-2\: (\textbf{TM})\: \: \textrm{atau}\: \: x=1\\ &\bullet \: \: x< 0,\: \left | x \right |=-x\\ &\: \quad \Leftrightarrow x^{2}-x-2=(x-2)(x+1)=0\\ &\: \quad \Leftrightarrow x=2\: (\textbf{TM})\: \: \textrm{atau}\: \: x=-1\\ &\textrm{Perhatikanlah ilustrasi berikut} \end{aligned}$.
$.\: \qquad\begin{aligned}&\textrm{Misalkan}\: \: f_{1}(x)=2-x^{2}\: \: \textrm{dan}\: \: f_{2}(x)=\left | x \right |\\ &\textrm{maka luas arsirannya adalah}:\\ &\textbf{Luas arsiran}=\: 2\displaystyle \int_{-1}^{0}\left (f_{1}(x)-f_{2}(x) \right )dx\\ &=2\displaystyle \int_{-1}^{0}\left ( 2-x^{2}-\left | x \right | \right )dx\\ &=2\displaystyle \int_{-1}^{0}\left (2-x^{2}-(-x) \right )dx\\ &=2\displaystyle \int_{-1}^{0}\left ( -x^{2}+x+2 \right )dx \end{aligned}$.
$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-1}^{3}\left [ x \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle 2\\ \textrm{b}.\quad \displaystyle 4\\ \textrm{c}.\quad \displaystyle 6\\ \textrm{d}.\quad \displaystyle 8\\ \textrm{e}.\quad \displaystyle 10\end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.
Jika diperjelas lagi
$.\: \qquad\begin{aligned}&\displaystyle \int_{-1}^{3}f(x)\: dx=\displaystyle \int_{-1}^{3}\left [ x \right ]\: dx\\&=\displaystyle \int_{-1}^{0}-1\: dx+\int_{0}^{1}0\: dx+\int_{1}^{2}1\: dx+\int_{2}^{3}2\: dx\\ &=(-x)|_{-1}^{0}+0+x|_{1}^{2}+2x|_{2}^{3}\\ &=\left ( 0-(-(-1)) \right )+0+(2-1)+2(3-2)\\ &=-1+0+1+2\\ &=\color{red}2 \end{aligned}$.
$\begin{array}{ll}\\ 30.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{a}^{b}\left [ x \right ]\: dx+\int_{a}^{b}\left [ -x \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle a+b\\ \textrm{b}.\quad \displaystyle 0\\ \color{red}\textrm{c}.\quad \displaystyle a-b\\ \textrm{d}.\quad \displaystyle 2a\\ \textrm{e}.\quad \displaystyle 2b\end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\textrm{Misalkan}\: \: f(x)=\left [ x \right ],\: g(x)=\left [ -x \right ],\: \textrm{dan}\\ &h(x)=f(x)+g(x)=\left [ x \right ]+\left [ -x \right ] \end{array}$..
$.\: \qquad\begin{array}{|c|c|c|c|}\hline \textrm{Interval}\: \: & c\leq f(x)< c+1&c\leq g(x)< c+1&\left [ h(x) \right ]=c\\\hline \vdots &\vdots &\vdots &\vdots \\\hline -3\leq x< -2&-3\leq x< -2&2< x\leq 3&-3+2=-1\\\hline -2\leq x< -1&-2\leq x< -1&1\leq x< 2&-2+1=-1\\\hline -1\leq x< 0&-1\leq x< 0&0\leq x< 1&-1+0=-1\\\hline 0\leq x< 1&0\leq x< 1&-1\leq x< 0&0+-1=-1\\\hline 1\leq x< 2&1\leq x< 2&-2\leq x< -1&1+-2=-1\\\hline 2\leq x< 3&2\leq x< 3&-3\leq x< -2&2+-3=-1\\\hline \vdots &\vdots &\vdots &\vdots \\\hline a\leq x< b &a\leq x< b &-b\leq x< -a &\color{red}a-b \\\hline \end{array}$.
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