$latex {\bf 6.\ \ (a)}\\Misalkan\quad u=16-7x\quad \Rightarrow \quad du=-7\ dx\\\\\begin{matrix} & \qquad \qquad \qquad \qquad \end{matrix}\qquad \qquad \ \ dx=-\frac{1}{7}\ du\\\\x=1\quad \Rightarrow \quad u=16-7.1=16-7=9\\x=2\quad \Rightarrow \quad u=16-7.2=16-14=2\\\\\begin{array}{rcl}\int \limits_{1}^{2} (16-7x)^3\ dx&=&\int \limits_{9}^{2} u^3\ .\ \big(-\frac{1}{7}\big)du\\\\&=&-\frac{1}{7}\ \int \limits_{9}^{2} u^3\ du\\\\&=&-\frac{1}{7}\ \left [ \frac{u^{4}}{4} \right ]_{9}^{2}\\\\&=&-\frac{1}{7}\ \left ( \frac{2^{4}}{4}-\frac{9^{4}}{4} \right )\\\\&=&-\frac{1}{28}\ \left ( 16-6561 \right )\\\\&=&-\frac{1}{28}\ \left ( -6545 \right )\\\\&=&233\frac{3}{4}\end{array}$
$latex {\bf 6.\ \ (b)}\\\\\begin{array}{rcl}\frac{dy}{dx}&=&3x-5\\\\dy&=&(3x-5)dx\\\\\int y\ dx&=&\int (3x-5)\ dx\\\\y&=&3\times \frac{x^2}{2}-5x+C\\\\y&=&\frac{3}{2}\ x^2-5x+C\end{array}\\\\\\Pada\ titik\ Q(4,\ 8)\ diperoleh:\\\\\begin{array}{rcl}8&=&\frac{3}{2}\times 4^2-5\times 4+C\\8&=&24-20+C\\8&=&4+C\\C&=&4\end{array}\\\\Jadi\ persamaan\ kurva\ tersebut\ adalah\\\\y=\frac{3}{2}\ x^2-5x+4$
$latex {\bf 6.\ \ (c)}\\\\\begin{array}{rcl}\int \limits_{0}^{\frac{\pi }{3}} (2\ cos\ x+3\ sin\ x)\ dx&=&\left \Big[ 2\ sin\ x-3\ cos\ x \right \Big]_{0}^{\frac{\pi }{3}}\\\\&=&\Bigg[\Big(2\ sin\ \frac{\pi }{3}-3\ cos\ \frac{\pi }{3}\Big)-\Big(2\ sin\ 0-3\ cos\ 0\Big)\Bigg]\\\\&=&\Bigg[\Big(2\times \frac{\sqrt{3}}{2}-3\times \frac{1}{2}\Big)-\Big(2\times 0-3\times 1\Big)\Bigg]\\\\&=&\Bigg[\Big(\sqrt{3}-\frac{3}{2}\Big)-\Big(0-3\Big)\Bigg]\\\\&=&\frac{3}{2}+\sqrt{3}\end{array}\\\\\\Jadi\ luas\ daerah\ tersebut\ adalah\ \bigg(\frac{3}{2}+\sqrt{3}\bigg)\ satuan\ luas.$
$latex {\bf 6.\ \ (d)}\\\\\begin{array}{rcl}V&=&\pi \int \limits_{0}^{3} (x^2+2)^2\ dx\\\\&=&\pi \int \limits_{0}^{3} (x^4+2x^2+4)\ dx\\\\&=&\pi \left \Bigg[\frac{x^5}{5}+2\times \frac{x^3}{3}+4x\Bigg]_{0}^{3}\\\\&=&\pi \Bigg[\Big(\frac{3^5}{5}+2\times \frac{3^3}{3}+4.3\Big)-\Big(\frac{0^5}{5}+2\times \frac{0^3}{3}+4.0\Big)\Bigg]\\\\&=&\pi \Big(\frac{243}{5}+18+12-0\Big)\\\\&=&\Big(48\frac{3}{5}+30\Big)\ \pi \\\\&=&\Big(78\frac{3}{5}\Big)\ \pi \end{array}\\\\\\Jadi\ volume\ benda\ tersebut\ adalah\ \Big(78\frac{3}{5}\ \pi \Big)\ satuan\ volume.$
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