Contoh Soal 3 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 11.& \text{Diketahui}{\displaystyle \int f(x)dx=a{x}^2+bx+c}\\ & \text{dengan}a\ne 0.\text{Jika}a,f(a),2b\text{membentuk}\\ & \text{barisan arimetika dan}f(b)=6,\text{maka}\\ & \text{nilai}{\displaystyle {\int }_0^{1}f(x)dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{17}{4}}\\ \text{b}.{\displaystyle \frac{21}{4}}\\ \text{c}.{\displaystyle \frac{25}{4}}\\ \text{d}.{\displaystyle \frac{13}{4}}\\ \text{e}.{\displaystyle \frac{11}{4}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int f(x)dx=a{x}^2+bx+c}\\ & \Rightarrow f(x)=2ax+b\\ & \Rightarrow f(a)=2a^2+b\\ & \Rightarrow f(b)=2ab+b=6\\ & \text{Karena}a,f(a),2b\text{membentuk barisan}\\ & \text{aritmetika, maka}f(a)={\displaystyle \frac{a+2b}{2}}\\ & \text{Sehingga}\\ & f(a)=2a^2+b={\displaystyle \frac{a+2b}{2}}\\ & \iff 4a^2-a=0\iff a(4a-1)=0\\ & \iff a=0\text{atau}a={\displaystyle \frac{1}{4}}\\ & \text{Dan karena}a\ne 0,\text{maka}a={\displaystyle \frac{1}{4}\Rightarrow b=4}\\ & \text{Selanjutnya}\\ & f(x)=2ax+b={\displaystyle \frac{1}{2}x+4}\\ & \text{Dari fakta di atas, maka}\\ & {\displaystyle {\int }_0^{1}f(x)dx={\int }_0^1{\displaystyle \frac{1}{2}x+4dx}}\\ & ={\displaystyle \frac{1}{4}{x}^2+4x{|}_0^1}\\ & =\left({\displaystyle \frac{1}{4}+4}\right)-0={\displaystyle \frac{17}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Diketahui}f(x)=a+bx\text{dan}F(x)\\ & \text{adalah anti turunan}f(x).\text{Jika}\\ & F(1)-F(0)=3,\text{maka}2a+b\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 10}\\ \text{b}.{\displaystyle 6}\\ \text{c}.{\displaystyle 5}\\ \text{d}.{\displaystyle 4}\\ \text{e}.{\displaystyle 3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & f(x)=a+bx,\text{sedangkan}\\ & F(x)={\displaystyle \int f(x)dx=\int (a+bx)dx}\\ & =ax+{\displaystyle \frac{1}{2}b{x}^2+C}\\ & \text{Dan diketahui pula}\\ & \begin{array}{ll}F(0)=a(1)+{\displaystyle \frac{1}{2}b(1{)}^2+C}& \\ F(1)=a(0)+{\displaystyle \frac{1}{2}b(0{)}^2+C}& -\\ 3=a+{\displaystyle \frac{1}{2}b}\end{array}\\ & \text{Sehingga}\\ & 2a+b=2(a+{\displaystyle \frac{1}{2}b})=2.3=6\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Jika}y={\displaystyle \frac{1}{3}(x^3+\frac{3}{x}),\text{maka}}\\ & {\displaystyle {\int }_1^2\sqrt{4+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{13}{14}}\\ \text{b}.{\displaystyle \frac{13}{6}}\\ \text{c}.{\displaystyle \frac{15}{6}}\\ \text{d}.{\displaystyle \frac{16}{6}}\\ \text{e}.{\displaystyle \frac{17}{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& y={\displaystyle \frac{1}{3}(x^3+\frac{3}{x})}\\ & \iff dy=(x^2-{\displaystyle \frac{1}{x^2}})dx\\ & \iff {\displaystyle \frac{dy}{dx}=(x^2-{\displaystyle \frac{1}{x^2}})}\\ & \iff {\left({\displaystyle \frac{dy}{dx}}\right)}^2={(x^2-{\displaystyle \frac{1}{x^2}})}^2={\displaystyle x^4-2+{\displaystyle \frac{1}{x^4}}}\\ & \text{maka}\\ & {\displaystyle {\int }_1^2\sqrt{4+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{4+\left({\displaystyle x^4-2+{\displaystyle \frac{1}{x^4}}}\right)}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{\left({\displaystyle x^4+2+{\displaystyle \frac{1}{x^4}}}\right)}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{{\left({\displaystyle x^2+{\displaystyle \frac{1}{x^2}}}\right)}^2}dx}\\ & ={\displaystyle {\int }_1^2{x}^2+{\displaystyle \frac{1}{x^2}dx}}\\ & ={\displaystyle \frac{1}{3}{x}^3-{\displaystyle \frac{1}{x}{|}_1^2}}\\ & =\left({\displaystyle \frac{8}{3}-\frac{1}{2}}\right)-\left({\displaystyle \frac{1}{3}-1}\right)\\ & ={\displaystyle \frac{16-3-2+6}{6}}\\ & ={\displaystyle \frac{17}{6}}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Luas daerah yang diarsir pada}\\ & \text{gambar berikut adalah 12}\\ & \text{satuan luas, maka nilai}a=....\\ & \begin{array}{l}\text{a}.{\displaystyle 4}\\ \text{b}.{\displaystyle 5}\\ \text{c}.{\displaystyle 6}\\ \text{d}.{\displaystyle 7}\\ \text{e}.{\displaystyle 8}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \text{Luas arsiran}=12\\ & =4.a-{\displaystyle {\int }_1^4{\displaystyle \frac{a}{(4-1)(4-7)}(x-1)(x-7)dx}}\\ & =4a-{\displaystyle {\int }_1^4{\displaystyle -\frac{a}{9}(x^2-8x+7)dx}}\\ & =4a-{\displaystyle {\int }_1^4{\displaystyle \frac{a}{9}(-x^2+8x-7)dx}}\\ & =4a-(-{\displaystyle \frac{a}{27}{x}^3+\frac{4a}{9}{x}^2-\frac{7a}{9}x{|}_1^4})\\ & =4a+{\displaystyle \frac{64a}{27}-\frac{64a}{9}+\frac{28a}{9}-\frac{a}{27}+\frac{4a}{9}-\frac{7a}{9}}\\ & =4a-{\displaystyle \frac{18a}{9}=2a=12\iff a=6}\\ & \text{Berikut ilustrasi gambar lengkapnya}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Luas daerah yang diarsir pada}\\ & \text{gambar berikut dapat dinyatakan}\\ & \text{dengan}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^2(3x-x^2)dx}\\ \text{b}.{\displaystyle {\int }_0^2(x+3)dx-{\int }_0^2{x}^{2}dx}\\ \text{c}.{\displaystyle {\int }_0^1(x+3)dx-{\int }_0^2{x}^{2}dx}\\ \text{d}.{\displaystyle {\int }_0^1(x+3-x^2)dx-{\int }_1^2{x}^{2}dx}\\ \text{e}.{\displaystyle {\int }_0^1(x+3-x^2)dx-{\int }_1^2(4-x^2)dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \text{Daerah yang diarsir di atas dibatasi}\\ & \text{oleh 4 buah batas, yaitu}\\ & \bullet \text{Garis}y=4\\ & \bullet \text{Kurva}y=x^2\\ & \bullet \text{Garis}x=0,\text{dan}\\ & \bullet \text{Garis yang melalui titik}(0,3)\text{dan}(1,4)\\ & \begin{array}{|cc|}\hline \text{Titik}& \text{Garis}\\ \begin{array}{rl}& 0,3\\ & \text{dan}\\ & (1,4)\\ & \end{array}& \begin{array}{rl}{\displaystyle y}& =\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1\\ y& ={\displaystyle \frac{4-3}{1-0}(x-0)+3}\\ y& =x+3\end{array}\\ \hline\end{array}\\ & \text{Sehingga luas arisrannya adalah}:\\ & ={\displaystyle {\int }_0^1(x+3)-x^{2}dx+{\int }_1^{2}4-x^{2}dx}\end{array}$.