\( \left. \begin{array} { l } { f ( x ) = - x ^ { 4 } - 3 x ^ { 3 } } \\ { h ( x ) = - 2 ( x - 2 ) ^ { 2 } ( x + 2 ) } \end{array} \right. \)

1) \( \int \sqrt{3-2 x} x^{2} d x \)

\( \int _ { 0 } ^ { 45 } \sec ^ { 4 } x d x = \)

\( \begin{aligned} \text { 40. } y=\frac{b}{a} \sqrt{a^{2}-x^{2}} & \frac{d y}{d x} & =-\frac{b^{2} x}{d^{2} y} \\ \text { H. } y=\left(a^{2 / 3}-x^{2 / 3}\right)^{3 / 3} & \frac{d y}{d x} & =-\sqrt[3]{\frac{y}{x}}\end{aligned} \)

\( 16 d ^ { 4 } y / d x ^ { 4 } + 24 d ^ { 2 } y / d x ^ { 2 } + 9 y = 0 \)

Sociologists have found that crime rates are influenced by temperature. In a town of 100,000 people, the crime rate has been approximated as \( \mathrm{C}=\frac{1}{10}(\mathrm{~T}-60)^{2}+100 \), where C is the humber of crimes per month and \( T \) is the average monthly temperature in degrees Fahrenheit. The average temperature for May was \( 72^{\circ} \) and by the end of May the temperature was rising at the rate of \( 7^{\circ} \) per month. How fast is the crime rate rising at the end of May? At the end of May, the crime rate is rising by (Simplify your answer.)