Contoh Soal 3 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 11.&\textrm{Diketahui}\: \: \displaystyle \int f(x)\: dx=ax^{2}+bx+c\\ &\textrm{dengan}\: \: a\neq 0.\: \textrm{Jika}\: a,\: f(a),\: 2b\: \: \textrm{membentuk}\\ &\textrm{barisan arimetika dan}\: \: f(b)=6,\: \: \textrm{maka}\\ &\textrm{nilai}\: \: \displaystyle \int_{0}^{1}f(x)\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \color{red}\displaystyle \frac{17}{4}\\ \textrm{b}.\quad \displaystyle \frac{21}{4}\\ \textrm{c}.\quad \displaystyle \frac{25}{4}\\ \textrm{d}.\quad \displaystyle \frac{13}{4}\\ \textrm{e}.\quad \displaystyle \frac{11}{4}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int f(x)\: dx=ax^{2}+bx+c\\ &\Rightarrow f(x)=2ax+b\\ &\Rightarrow f(a)=2a^{2}+b\\ &\Rightarrow f(b)=2ab+b=6\\ &\textrm{Karena}\: \: a,f(a),2b\: \: \textrm{membentuk barisan}\\ &\textrm{aritmetika, maka}\: \:  f(a)=\displaystyle \frac{a+2b}{2}\\ &\textrm{Sehingga}\\ &f(a)=2a^{2}+b=\displaystyle \frac{a+2b}{2}\\ &\Leftrightarrow 4a^{2}-a=0\Leftrightarrow a(4a-1)=0\\ &\Leftrightarrow a=0\: \: \textrm{atau}\: \: a=\displaystyle \frac{1}{4}\\ &\textrm{Dan karena}\: \: a\neq 0,\: \: \textrm{maka}\: \: a=\displaystyle \frac{1}{4}\Rightarrow b=4\\ &\textrm{Selanjutnya}\\ &f(x)=2ax+b=\displaystyle \frac{1}{2}x+4\\ &\textrm{Dari fakta di atas, maka}\\ &\displaystyle \int_{0}^{1} f(x)\: dx=\int_{0}^{1} \displaystyle \frac{1}{2}x+4\: dx\\ &=\displaystyle \frac{1}{4}x^{2}+4x|_{0}^{1}\\ &=\left ( \displaystyle \frac{1}{4}+4 \right )-0=\color{red}\displaystyle \frac{17}{4} \end{aligned} \end{array}$.

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

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: y=\displaystyle \frac{1}{3}\left ( x^{3}+\frac{3}{x} \right ),\: \textrm{maka}\\ &\displaystyle \int_{1}^{2}\sqrt{4+\left (\displaystyle \frac{dy}{dx}  \right )^{2}}\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \frac{13}{14}\\ \textrm{b}.\quad \displaystyle \frac{13}{6}\\ \textrm{c}.\quad \color{red}\displaystyle \frac{15}{6}\\ \textrm{d}.\quad \displaystyle \frac{16}{6}\\ \textrm{e}.\quad \displaystyle \frac{17}{6}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&y=\displaystyle \frac{1}{3}\left ( x^{3}+\frac{3}{x} \right )\\ &\Leftrightarrow dy=\left (x^{2}-\displaystyle \frac{1}{x^{2}}  \right )\: dx\\ &\Leftrightarrow \displaystyle \frac{dy}{dx}=\left (x^{2}-\displaystyle \frac{1}{x^{2}}  \right )\\ &\Leftrightarrow \left ( \displaystyle \frac{dy}{dx} \right )^{2}=\left (x^{2}-\displaystyle \frac{1}{x^{2}}  \right )^{2}=\displaystyle x^{4}-2+\displaystyle \frac{1}{x^{4}}\\ &\textrm{maka}\\ &\displaystyle \int_{1}^{2}\sqrt{4+\left (\displaystyle \frac{dy}{dx}  \right )^{2}}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{4+\left ( \displaystyle x^{4}-2+\displaystyle \frac{1}{x^{4}} \right )}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{\left ( \displaystyle x^{4}+2+\displaystyle \frac{1}{x^{4}} \right )}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{\left ( \displaystyle x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}}\: dx\\ &=\displaystyle \int_{1}^{2}x^{2}+\displaystyle \frac{1}{x^{2}}\: dx\\&=\displaystyle \frac{1}{3}x^{3}-\displaystyle \frac{1}{x}|_{1}^{2}\\ &=\left ( \displaystyle \frac{8}{3}-\frac{1}{2} \right )-\left ( \displaystyle \frac{1}{3}-1 \right )\\ &=\displaystyle \frac{16-3-2+6}{6}\\&=\color{red}\displaystyle \frac{17}{6} \end{aligned}  \end{array}$.

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

$.\: \qquad\begin{aligned}&\textrm{Luas arsiran}=12\\ &=4.a-\displaystyle \int_{1}^{4}\displaystyle \frac{a}{(4-1)(4-7)}(x-1)(x-7)\: dx\\ &=4a-\displaystyle \int_{1}^{4}\displaystyle -\frac{a}{9}(x^{2}-8x+7)\: dx\\ &=4a- \displaystyle \int_{1}^{4}\displaystyle \frac{a}{9}(-x^{2}+8x-7)\: dx\\ &=4a-\left (-\displaystyle \frac{a}{27}x^{3}+\frac{4a}{9}x^{2}-\frac{7a}{9}x|_{1}^{4}  \right )\\ &=4a+\displaystyle \frac{64a}{27}-\frac{64a}{9}+\frac{28a}{9}-\frac{a}{27}+\frac{4a}{9}-\frac{7a}{9}\\ &=4a-\displaystyle \frac{18a}{9}=2a=12\Leftrightarrow a=\color{red}6\\ &\textrm{Berikut ilustrasi gambar lengkapnya} \end{aligned}$.

$\begin{array}{ll}\\ 15.&\textrm{Luas daerah yang diarsir pada}\\ &\textrm{gambar berikut dapat dinyatakan}\\&\textrm{dengan}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \int_{0}^{2}(3x-x^{2})\: dx\\ \textrm{b}.\quad \displaystyle \int_{0}^{2}(x+3)\: dx-\int_{0}^{2}x^{2}\: dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}(x+3)\: dx-\int_{0}^{2}x^{2}\: dx\\ \textrm{d}.\quad \displaystyle \int_{0}^{1}(x+3-x^{2})\: dx-\int_{1}^{2}x^{2}\: dx\\ \textrm{e}.\quad \color{red}\displaystyle \int_{0}^{1}(x+3-x^{2})\: dx-\int_{1}^{2}\left (4-x^{2}  \right )\: dx\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textrm{Daerah yang diarsir di atas dibatasi}\\ &\textrm{oleh 4 buah batas, yaitu}\\ &\bullet \quad \textrm{Garis}\: \: y=4\\ &\bullet \quad \textrm{Kurva}\: \: y=x^{2}\\ &\bullet \quad \textrm{Garis}\: \: x=0,\: \: \textrm{dan}\\ &\bullet \quad \textrm{Garis yang melalui titik}\: (0,3)\: \textrm{dan}\: (1,4)\\ &\begin{array}{|c|c|}\hline \textrm{Titik}&\textrm{Garis}\\\hline \begin{aligned}&{0,3}\\ &\textrm{dan}\\ &(1,4)\\ & \end{aligned}&\begin{aligned}\displaystyle y&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})+y_{1}\\ y&=\displaystyle \frac{4-3}{1-0}(x-0)+3\\ y&=x+3 \end{aligned}\\\hline \end{array}\\ &\textrm{Sehingga luas arisrannya adalah}:\\ &=\color{red}\displaystyle \int_{0}^{1}(x+3)-x^{2}\: dx+\int_{1}^{2}4-x^{2}\: dx \end{aligned}$.