Contoh Soal 7 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 31.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{2} \right ]\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle 3\\ \textrm{b}.\quad \displaystyle 3\frac{1}{2}\\ \textrm{c}.\quad \color{red}\displaystyle 4\\ \textrm{d}.\quad \displaystyle 4\frac{1}{2}\\ \textrm{e}.\quad \displaystyle 5\end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.

$.\: \qquad\begin{aligned}&\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{2} \right ]\: dx\\&=\displaystyle \int_{-1}^{-\frac{1}{2}}-1\: dx+\int_{-\frac{1}{2}}^{\frac{1}{2}}0\: dx+\int_{\frac{1}{2}}^{\frac{3}{2}}1\: dx+\int_{\frac{3}{2}}^{\frac{5}{2}}2\: dx +\int_{\frac{5}{2}}^{3}3\: dx\\ &=-1|_{-1}^{-\frac{1}{2}}+0|_{-\frac{1}{2}}^{\frac{1}{2}}+x|_{\frac{1}{2}}^{\frac{3}{2}}+2x|_{\frac{3}{2}}^{\frac{5}{2}}+3x|_{\frac{5}{2}}^{3}\\ &=-\left ( -\displaystyle \frac{1}{2}+1 \right )+0+\left ( \displaystyle \frac{3}{2}-\frac{1}{2} \right )+2\left ( \frac{5}{2}-\frac{3}{2} \right ) +3\left ( 3-\displaystyle \frac{5}{2} \right )\\&=-\displaystyle \frac{1}{2}+0+1+2+\frac{3}{2}\\ &=\color{red}\displaystyle 4\\ &\textrm{Perhatikan tabel berikut} \end{aligned}$.

$\begin{array}{ll}\\ 32.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{3} \right ]\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle 3\\ \textrm{b}.\quad \color{red}\displaystyle 3\frac{1}{3}\\ \textrm{c}.\quad \displaystyle 4\frac{2}{3}\\ \textrm{d}.\quad \displaystyle 5\frac{1}{3}\\ \textrm{e}.\quad \displaystyle 6\frac{1}{3}\end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.

$.\: \qquad\begin{aligned}&\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{3} \right ]\: dx\\&=\displaystyle \int_{-1}^{-\frac{1}{3}}-1\: dx+\int_{-\frac{1}{3}}^{\frac{2}{3}}0\: dx+\int_{\frac{2}{3}}^{\frac{5}{3}}1\: dx+\int_{\frac{5}{3}}^{\frac{8}{3}}2\: dx +\int_{\frac{8}{3}}^{3}3\: dx\\ &=-1|_{-1}^{-\frac{1}{3}}+0|_{-\frac{1}{3}}^{\frac{2}{3}}+x|_{\frac{2}{3}}^{\frac{5}{3}}+2x|_{\frac{5}{3}}^{\frac{8}{3}}+3x|_{\frac{8}{3}}^{3}\\ &=-\left ( -\displaystyle \frac{1}{3}+1 \right )+0+\left ( \displaystyle \frac{5}{3}-\frac{2}{3} \right )+2\left ( \frac{8}{3}-\frac{5}{3} \right ) +3\left ( 3-\displaystyle \frac{8}{3} \right )\\&=-\displaystyle \frac{2}{3}+0+1+2+\frac{3}{3}\\ &=\color{red}\displaystyle 3\frac{1}{3} \end{aligned}$.

$\begin{array}{ll}\\ 33.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{0}^{n}\left [ x \right ]\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle n^{2}\\ \textrm{b}.\quad \color{red}\displaystyle \frac{n(n-1)}{2}\\ \textrm{c}.\quad \displaystyle \frac{n(n+1)}{2}\\ \textrm{d}.\quad \displaystyle \frac{(n+1)(n+2)}{2}\\ \textrm{e}.\quad \displaystyle \frac{n^{2}+1}{2}\end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.

$.\: \qquad\begin{aligned}&\displaystyle \int_{0}^{n}f(x)\: dx=\displaystyle \int_{0}^{n}\left [ x \right ]\: dx\\&=\displaystyle \int_{0}^{1}0\: dx+\int_{1}^{2}1\: dx+\int_{2}^{3}2\: dx+\cdots +\int_{n-1}^{n}(n-1)\: dx\\ &=0+x|_{1}^{2}+2x|_{2}^{3}+\cdots +(n-1)x|_{n-1}^{n}\\ &=0+(2-1)+2(3-2)+\cdots +(n-1)(n-(n-1))\\&=0+1+2+\cdots (n-1)\\ &=\displaystyle \frac{(n-1)((n-1)+1)}{2}\\ &=\color{red}\displaystyle \frac{n(n-1)}{2}\\ &\textrm{Perhatikan tabel berikut} \end{aligned}$.

$\begin{array}{ll}\\ 34.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-2022}^{2022}\left [ x \right ]\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \color{red}\displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int_{-2022}^{2022}\left [ x \right ]\: dx\\ &=\displaystyle \int_{-2022}^{-2021}\left [ x \right ]\: dx+\displaystyle \int_{-2021}^{-2020}\left [ x \right ]\: dx+\cdots +\int_{2021}^{2022}\left [ x \right ]\: dx\\ &=-2022+(-2021)+\cdots +2020+2021\\ &=\color{red}-2022 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-2022}^{2022}(x-\left [ x \right ])\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{e}.\quad \color{red}\displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int_{-2022}^{2022}(x-\left [ x \right ])\: dx\\ &=\displaystyle \int_{-2022}^{2022}x\: dx-\left (\displaystyle \int_{-2022}^{-2021}\left [ x \right ]\: dx+\cdots +\int_{2021}^{2022}\left [ x \right ]\: dx  \right )\\ &\textrm{Sebelumnya perlu diingat bahwa}\\ &\quad \displaystyle \int_{-2022}^{2022}x\: dx\: \: \textrm{adalah}\: \textrm{fungsi ganjil, maka}=0\\ &\quad \textrm{Sehingga nilai akhirnya adalah}\\  &=0-(-2022)\qquad \textrm{lihat pembahasan sebelumnya}\\ &=\color{red}2022 \end{aligned} \end{array}$.