MA005: Calculus I

Practice 1: $\int_{1}^{4} f(x)-g(x) d x=7-3=4$.

Practice 2: $\text { (2) }(\mathrm{min} \text {. of f on }[3,5])=4 \leq \int_{3}^{5} \mathrm{f}(x) \mathrm{dx} \leq 2(\max \text {. of } \mathrm{f} \text { on }[3,5])=12 \text {. }$

Practice 3:

(a) $\mathrm{A}(1)=0, \mathrm{~A}(2)=3, \mathrm{~A}(3)=6, \mathrm{~A}(4)=9$

(b) $\mathrm{A}(\mathrm{x})=(\mathrm{x}-1)(3)=3 \mathrm{x}-3$

(c) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{A}(\mathrm{x})=3$

(d) $\mathrm{A}(\mathrm{x})=\int_{\mathrm{1}}^{\mathrm{x}} 3 \mathrm{dx}$

Practice 4:

(a) $\mathrm{B}(0)=0, \mathrm{~B}(1)=1, \mathrm{~B}(2)=4, \mathrm{~B}(3)=9$

(b) $\mathrm{B}(\mathrm{x})=\frac{1}{2} \text { (base) (height) }=\frac{1}{2}(\mathrm{x})(2 \mathrm{x})=\mathrm{x}^{2}$

(c) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{B}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{2}=2 \mathrm{x}$

(d) $\mathrm{B}(\mathrm{x})=\int_{0}^{\mathrm{x}} 2 \mathrm{t} \mathrm{dt}$

Practice 5: The integral = the shaded area in Fig. 10 = $(0.5)(1)+(1)(2)+(0.2)(3)=3.1$