Contoh Soal 7 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 31.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{2}}]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 3}\\ \text{b}.{\displaystyle 3\frac{1}{2}}\\ \text{c}.{\displaystyle 4}\\ \text{d}.{\displaystyle 4\frac{1}{2}}\\ \text{e}.{\displaystyle 5}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{2}}]dx}\\ & ={\displaystyle {\int }_{-1}^{-\frac{1}{2}}-1dx+{\int }_{-\frac{1}{2}}^{\frac{1}{2}}0dx+{\int }_{\frac{1}{2}}^{\frac{3}{2}}1dx+{\int }_{\frac{3}{2}}^{\frac{5}{2}}2dx+{\int }_{\frac{5}{2}}^{3}3dx}\\ & =-1{|}_{-1}^{-\frac{1}{2}}+0{|}_{-\frac{1}{2}}^{\frac{1}{2}}+x{|}_{\frac{1}{2}}^{\frac{3}{2}}+2x{|}_{\frac{3}{2}}^{\frac{5}{2}}+3x{|}_{\frac{5}{2}}^3\\ & =-(-{\displaystyle \frac{1}{2}+1})+0+\left({\displaystyle \frac{3}{2}-\frac{1}{2}}\right)+2(\frac{5}{2}-\frac{3}{2})+3(3-{\displaystyle \frac{5}{2}})\\ & =-{\displaystyle \frac{1}{2}+0+1+2+\frac{3}{2}}\\ & ={\displaystyle 4}\\ & \text{Perhatikan tabel berikut}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{3}}]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 3}\\ \text{b}.{\displaystyle 3\frac{1}{3}}\\ \text{c}.{\displaystyle 4\frac{2}{3}}\\ \text{d}.{\displaystyle 5\frac{1}{3}}\\ \text{e}.{\displaystyle 6\frac{1}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{3}}]dx}\\ & ={\displaystyle {\int }_{-1}^{-\frac{1}{3}}-1dx+{\int }_{-\frac{1}{3}}^{\frac{2}{3}}0dx+{\int }_{\frac{2}{3}}^{\frac{5}{3}}1dx+{\int }_{\frac{5}{3}}^{\frac{8}{3}}2dx+{\int }_{\frac{8}{3}}^{3}3dx}\\ & =-1{|}_{-1}^{-\frac{1}{3}}+0{|}_{-\frac{1}{3}}^{\frac{2}{3}}+x{|}_{\frac{2}{3}}^{\frac{5}{3}}+2x{|}_{\frac{5}{3}}^{\frac{8}{3}}+3x{|}_{\frac{8}{3}}^3\\ & =-(-{\displaystyle \frac{1}{3}+1})+0+\left({\displaystyle \frac{5}{3}-\frac{2}{3}}\right)+2(\frac{8}{3}-\frac{5}{3})+3(3-{\displaystyle \frac{8}{3}})\\ & =-{\displaystyle \frac{2}{3}+0+1+2+\frac{3}{3}}\\ & ={\displaystyle 3\frac{1}{3}}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_0^n\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle n^2}\\ \text{b}.{\displaystyle \frac{n(n-1)}{2}}\\ \text{c}.{\displaystyle \frac{n(n+1)}{2}}\\ \text{d}.{\displaystyle \frac{(n+1)(n+2)}{2}}\\ \text{e}.{\displaystyle \frac{n^2+1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_0^{n}f(x)dx={\displaystyle {\int }_0^n\left[x\right]dx}}\\ & ={\displaystyle {\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx+\cdots +{\int }_{n-1}^n(n-1)dx}\\ & =0+x{|}_1^2+2x{|}_2^3+\cdots +(n-1)x{|}_{n-1}^n\\ & =0+(2-1)+2(3-2)+\cdots +(n-1)(n-(n-1))\\ & =0+1+2+\cdots (n-1)\\ & ={\displaystyle \frac{(n-1)((n-1)+1)}{2}}\\ & ={\displaystyle \frac{n(n-1)}{2}}\\ & \text{Perhatikan tabel berikut}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-2022}^{2022}\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left[x\right]dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left[x\right]dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_{-2022}^{2022}\left[x\right]dx}\\ & ={\displaystyle {\int }_{-2022}^{-2021}\left[x\right]dx+{\displaystyle {\int }_{-2021}^{-2020}\left[x\right]dx+\cdots +{\int }_{2021}^{2022}\left[x\right]dx}}\\ & =-2022+(-2021)+\cdots +2020+2021\\ & =-2022\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-2022}^{2022}(x-\left[x\right])dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left[x\right]dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left[x\right]dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_{-2022}^{2022}(x-\left[x\right])dx}\\ & ={\displaystyle {\int }_{-2022}^{2022}xdx-\left({\displaystyle {\int }_{-2022}^{-2021}\left[x\right]dx+\cdots +{\int }_{2021}^{2022}\left[x\right]dx}\right)}\\ & \text{Sebelumnya perlu diingat bahwa}\\ & {\displaystyle {\int }_{-2022}^{2022}xdx\text{adalah}\text{fungsi ganjil, maka}=0}\\ & \text{Sehingga nilai akhirnya adalah}\\ & =0-(-2022)\text{lihat pembahasan sebelumnya}\\ & =2022\end{array}\end{array}$.