Contoh Soal 6 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 26.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Diketahui sebuah lingkaran berjari-jari}2\\ & \text{sebagaimana ilustrasi dibawah. Jika tali}\\ & \text{busur pada gambar berjarak 1 dari garis}\\ & \text{tengah, maka luas daerah di atas tali busur}\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{b}.{\displaystyle 2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{c}.{\displaystyle {\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{d}.{\displaystyle 2{\int }_0^1\sqrt{4-x^2}dx}\\ \text{e}.{\displaystyle 2{\int }_0^{\sqrt{3}}\sqrt{4-x^2}dx}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa sebuah lingkaran}\\ & \text{dengan}r=2,\text{artinya}{x}^2+y^2=r^2=2^2=4\\ & \text{Diketahui pula persamaan garis}y=1,\text{maka}\\ & x^2+1^2=4\iff x^2=4-1=3\iff x=\pm \sqrt{3}\end{array}$

$.\begin{array}{rl}& \text{Sehingga luas daerah di atas tali busur}\\ & \text{yaitu berupa}\mathbf{\text{temberang}}\text{adalah}\\ & ={\displaystyle {\int }_{-\sqrt{3}}^{\sqrt{3}}(f_2(x)-f_1(x))dx}\\ & ={\displaystyle {\int }_{-\sqrt{3}}^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ & \text{ingat fungsi genap pada pembahasan}\\ & \text{sebelumnya, karena}f(x)=\sqrt{4-x^2}=f(-x)\\ & =2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\end{array}$.

$\begin{array}{l}\\ 27.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Diketahui sebuah lingkaran berjari-jari}2\\ & \text{sebagaimana ilustrasi dibawah. Jika tali}\\ & \text{busur pada gambar berjarak 1 dari garis}\\ & \text{tengah, maka luas daerah di bawah tali busur}\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2\pi -2{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{b}.{\displaystyle 2\pi -2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{c}.{\displaystyle 4\pi -{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{d}.{\displaystyle 4\pi -2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{e}.{\displaystyle 4\pi -2{\int }_0^{\sqrt{3}}\sqrt{4-x^2}dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dengan ilustrasi sama dengan no.24 di atas}\\ & \text{Kita dapat tentukan luas wilayah di bawah}\\ & \mathbf{\text{tembereng}},\text{yaitu}:\\ & =\text{Luas lingkaran penuh itu}-\mathbf{\text{tembereng}}\\ & =\pi .r^2-2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\\ & =\pi .(2{)}^2-2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\\ & =4\pi -2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Luas daerah yang dibatasi oleh kurva}\\ & y=2-x^2\text{dan}y=\left|x\right|\text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2{\int }_{-1}^0(-x^2+x+2)dx}\\ \text{b}.{\displaystyle 2{\int }_{-1}^0(-x^2-x+2)dx}\\ \text{c}.{\displaystyle 2{\int }_{-1}^0(-x^2+x-2)dx}\\ \text{d}.{\displaystyle {\int }_{-1}^1(-x^2-x+2)dx}\\ \text{e}.{\displaystyle {\int }_0^1(-x^2+x+2)dx}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Kita tentukan poin batas-batasnya, yaitu}:\\ & \text{Titik potong kurva dengan sumbu-X},y=0\\ & y=2-x^2=0\iff (\sqrt{2}+x)(\sqrt{2}-x)=0\\ & \iff x=-\sqrt{2}\text{atau}x=\sqrt{2}\\ & \text{Titik potong kurva dengan garis, yaitu}:\\ & 2-x^2=\left|x\right|\iff x^2+\left|x\right|-2=0\\ & \bullet x\ge 0,\left|x\right|=x\\ & \iff x^2+x-2=(x+2)(x-1)=0\\ & \iff x=-2(\mathbf{\text{TM}})\text{atau}x=1\\ & \bullet x<0,\left|x\right|=-x\\ & \iff x^2-x-2=(x-2)(x+1)=0\\ & \iff x=2(\mathbf{\text{TM}})\text{atau}x=-1\\ & \text{Perhatikanlah ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Misalkan}{f}_1(x)=2-x^2\text{dan}{f}_2(x)=\left|x\right|\\ & \text{maka luas arsirannya adalah}:\\ & \mathbf{\text{Luas arsiran}}=2{\displaystyle {\int }_{-1}^0(f_1(x)-f_2(x))dx}\\ & =2{\displaystyle {\int }_{-1}^0(2-x^2-\left|x\right|)dx}\\ & =2{\displaystyle {\int }_{-1}^0(2-x^2-(-x))dx}\\ & =2{\displaystyle {\int }_{-1}^0(-x^2+x+2)dx}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 2}\\ \text{b}.{\displaystyle 4}\\ \text{c}.{\displaystyle 6}\\ \text{d}.{\displaystyle 8}\\ \text{e}.{\displaystyle 10}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

Jika diperjelas lagi

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

$\begin{array}{l}\\ 30.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_a^b\left[x\right]dx+{\int }_a^b[-x]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle a+b}\\ \text{b}.{\displaystyle 0}\\ \text{c}.{\displaystyle a-b}\\ \text{d}.{\displaystyle 2a}\\ \text{e}.{\displaystyle 2b}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel berikut}\\ & \text{Misalkan}f(x)=\left[x\right],g(x)=[-x],\text{dan}\\ & h(x)=f(x)+g(x)=\left[x\right]+[-x]\end{array}$..

$.\begin{array}{|cccc|}\hline \text{Interval}& c\le f(x)<c+1& c\le g(x)<c+1& [h(x)]=c\\ ⋮& ⋮& ⋮& ⋮\\ -3\le x<-2& -3\le x<-2& 2<x\le 3& -3+2=-1\\ -2\le x<-1& -2\le x<-1& 1\le x<2& -2+1=-1\\ -1\le x<0& -1\le x<0& 0\le x<1& -1+0=-1\\ 0\le x<1& 0\le x<1& -1\le x<0& 0+-1=-1\\ 1\le x<2& 1\le x<2& -2\le x<-1& 1+-2=-1\\ 2\le x<3& 2\le x<3& -3\le x<-2& 2+-3=-1\\ ⋮& ⋮& ⋮& ⋮\\ a\le x<b& a\le x<b& -b\le x<-a& a-b\\ \hline\end{array}$.