Contoh Soal 4 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 16.& \text{Luas daerah yang batasi oleh parabola}\\ & y=4x-x^2,y=-2x+8,\text{dan sumbu-Y}\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 4\frac{2}{3}\text{satuan luas}}\\ \text{b}.{\displaystyle 6\frac{2}{3}\text{satuan luas}}\\ \text{c}.{\displaystyle 12\frac{2}{3}\text{satuan luas}}\\ \text{d}.{\displaystyle 20\frac{2}{3}\text{satuan luas}}\\ \text{e}.{\displaystyle 30\frac{2}{3}\text{satuan luas}}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Luas arsiran}=\text{Luas trapesium}-\text{luas II}\\ & \text{Proses penyelesaian 1}\\ & \text{Kurva}y=4x-x^2,\text{maka koordinat titik}\\ & \text{puncaknya adalah}:(p,q).\text{Pilih}y=(x)=0\\ & \iff y=f(x)=4x-x^2=x(4-x)=0\\ & \iff x_1=0\text{atau}{x}_2=4\\ & \text{dengan}p={\displaystyle \frac{x_1+x_2}{2}=\frac{0+4}{2}=2,\text{dan}}\\ & q=y=f(2)=4.2-2^2=8-4=4\\ & \text{Sehingga}(p,q)=(2,4)\\ & \text{Proses penyelesaian 2}\\ & \text{Untuk garis}y=-2x+8,\text{maka koordinat}\\ & \text{titik potongnya terhadap sumbu-X dan Y}:\\ & \begin{array}{|ccc|}\hline x& 0& 4\\ y& 8& 0\\ (x,y)& (0,8)& (4,0)\\ \hline\end{array}\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Luas arsiran}=\text{Luas trapesium}-\text{luas II}\\ & ={\displaystyle \frac{(8+4)\times 2}{2}-{\displaystyle {\int }_0^{2}4x-x^{2}dx}}\\ & =12-\left((2x^2-{\displaystyle \frac{1}{3}{x}^3}){|}_0^2\right)\\ & =12-((8-{\displaystyle \frac{8}{3}})-0)\\ & =12+{\displaystyle \frac{8}{3}-8=6\frac{2}{3}\text{satuan luas}}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Luas daerah yang batasi oleh kurva}y=4-x^2\\ & y=-x+2,\text{dan}0\le x\le 2\text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{8}{3}\text{satuan luas}}\\ \text{b}.{\displaystyle \frac{10}{3}\text{satuan luas}}\\ \text{c}.{\displaystyle \frac{14}{3}\text{satuan luas}}\\ \text{d}.{\displaystyle \frac{16}{3}\text{satuan luas}}\\ \text{e}.{\displaystyle \frac{26}{3}\text{satuan luas}}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Proses penyelesaian 1}\\ & \text{Kurva}y=4-x^2,\text{maka koordinat titik}\\ & \text{puncaknya adalah}:(p,q).\text{Pilih}y=(x)=0\\ & \iff y=f(x)=4-x^2=(2+x)(2-x)=0\\ & \iff x_1=-2\text{atau}{x}_2=2\\ & \text{dengan}p={\displaystyle \frac{x_1+x_2}{2}=\frac{-2+2}{2}=0,\text{dan}}\\ & q=y=f(0)=4-0^2=4\\ & \text{Sehingga}(p,q)=(0,4)\\ & \text{Proses penyelesaian 2}\\ & \text{Untuk garis}y=-x+2,\text{maka koordinat}\\ & \text{titik potongnya terhadap sumbu-X dan Y}:\\ & \begin{array}{|ccc|}\hline x& 0& 2\\ y& 2& 0\\ (x,y)& (0,2)& (2,0)\\ \hline\end{array}\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Luas arsiran}=\text{Dibawah kurva}-\text{segitiga}\\ & ={\displaystyle {\int }_0^{2}4-x^{2}dx-{\displaystyle \frac{1}{2}\times \text{alas}\times \text{tinggi}}}\\ & =(4x-{\displaystyle \frac{1}{3}{x}^3}){|}_0^2-{\displaystyle \frac{2\times 2}{2}}\\ & =8-{\displaystyle \frac{8}{3}-2=6-2{\displaystyle \frac{2}{3}=3\frac{1}{3}}}\\ & =\frac{10}{3}\text{satuan luas}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Luas daerah di bawah kurva}y=-x^2+8x\\ & \text{di atas}y=6x-24\text{dan terletak di }\\ & \text{kuadran I adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(x^2-2x-24)dx}\\ \text{b}.{\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+2x+24)dx}\\ \text{c}.{\displaystyle {\int }_0^6(-x^2+8x)dx+{\int }_6^8(-x^2+2x+24)dx}\\ \text{d}.{\displaystyle {\int }_4^6(6x-24)dx+{\int }_6^8(-x^2+8x)dx}\\ \text{e}.{\displaystyle {\int }_0^4(6x-24)dx+{\int }_0^6(-x^2+8x)dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Kita tentukan poin batas-batasnya dulu}\\ & \text{Titik potong kurva dengan sumbu-X, yaitu}:\\ & y=-x^2+8x=x(-x+8)=0\\ & \iff x=0\text{atau}x=8\\ & \text{Titik potong garis dengan sumbu-X, yaitu}:\\ & y=6x-24=0\iff x=6\\ & \text{Dan titik potong kurva dengan garis adalah}:\\ & y=y\iff -x^2+8x=6x-24\\ & \iff -x^2+2x+24=0\iff (-x-4)(x-6)=0\\ & \iff x_1=-4\text{atau}{x}_2=6\\ & \text{Sehingga luas arsiran}=\text{Luas I}+\text{Luas II}\\ & ={\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+8x)-(6x-24)dx}\\ & ={\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+2x+24)dx}\\ & \text{Perhatikan ulustrasi berikut}\end{array}$ 

$\begin{array}{l}\\ 19.& \text{Luas daerah yang diarsir pada gambar}\\ & \text{di bawah dapat dirumuskan dengan}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_b^{d}g(x)dx-{\int }_b^{c}f(x)dx}\\ \text{b}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_b^d(g(x)-f(x))dx}\\ \text{c}.{\displaystyle {\int }_a^d(f(x)-g(x))dx}\\ \text{d}.{\displaystyle {\int }_a^d(f(x)-g(x))dx-{\int }_c^{d}g(x)dx}\\ \text{e}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_c^d(g(x)-f(x))dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Cukup jelas bahwa}\\ & \text{pada gambar di atas pada selang}\\ & \bullet a\le x\le b:{\displaystyle {\int }_a^b(f(x)-g(x))dx}\\ & \bullet b\le x\le d:{\displaystyle {\int }_b^d(g(x)-f(x))dx}\\ & \text{dikurangi luasan yang}\\ & \text{berselang}b\le x\le c\end{array}$.

$\begin{array}{l}\\ 20.& \text{Daerah}\text{R}\text{di kuadran II, dibatasi oleh}\\ & \text{grafik}y=x^2,y=x+2\text{dan}y=0\\ & \text{Integral yang sesuai kondisi di atas adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_{-2}^{-1}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ \text{b}.{\displaystyle {\int }_{-1}^{-2}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ \text{c}.{\displaystyle {\int }_{-1}^{-2}{x}^{2}dx+{\int }_{-1}^0(x+2)dx}\\ \text{d}.{\displaystyle {\int }_0^2(x^2+x+2)dx}\\ \text{e}.{\displaystyle {\int }_0^2(-x^2+x+2)dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Dari gukup jelas bahwa}\\ & \text{daerah arsiran akan terformulasikan}\\ & {\displaystyle {\int }_{-2}^{-1}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ & \text{Berikut sedikit uraian prosesnya}\\ & \mathbf{\text{Sebagai langkah awal}},\text{tentukan dulu}\\ & \text{titik potong kura dengan garis, yaitu}:\\ & y=y\iff x^2=x+2\iff x^2-x-2=0\\ & \iff (x+1)(x-2)=0\\ & \iff x=-1\text{atau}x=2\\ & \mathbf{\text{Langkah berikutnya}}\\ & \text{Kita tentukan daerah arsiran, yaitu}\\ & \text{seperti terformulasikan di atas}\end{array}$.