Practice Answers

Practice 1: \mathrm{B}(1)=2.5, \mathrm{~B}(2)=5, \mathrm{~B}(3)=8.5, \mathrm{~B}(4)=12, \mathrm{~B}(5)=14.5

\mathrm{B}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt} so B^{\prime}(x)=\frac{d}{d x}\left(\int_{0}^{x} f(t) d t\right)=f(x) (by The Area Function is an Antiderivative theorem): then B^{\prime}(1)=f(1)=2, B^{\prime}(2)=f(2)=3, B^{\prime}(3)=4, B^{\prime}(4)=3 \text {, } and B^{\prime}(5)=2.


Practice 2:

(a) As an area, \int_{1}^{3} x-1 d x is the area of the triangular region between y = x – 1 and the x– axis for 1 \leq x \leq 3: \text { area }=\frac{1}{2}(\text { base)(height })=\frac{1}{2}(2)(2)=2.

(b) F(x)=\frac{x^{2}}{2}-x is an antiderivative of f(x)=x-1 so area = \begin{aligned} &\int_{1}^{3} \mathrm{x}-1 \mathrm{dx}=\mathrm{F}(3)-\mathrm{F}(1)=\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)=2 \end{aligned}.


Practice 3: Area = \int_{1}^{2} 3 x^{2} \mathrm{dx}=\left.\mathrm{x}^{3}\right|_{1} ^{2}=8-1=7.


Practice 4:

(a) distance = \int_{1}^{5} 3 \mathrm{t}^{2} \mathrm{dt}=\left.\mathrm{t}^{3}\right|_{1} ^{5}=125-1 = feet.

(b) In this problem we know the starting point is x = 0, and the total distance ("area") is 343 feet. Our problem is to find the time T (Fig. 16) so 343 \text { feet }=\int_{0}^{T} 3 t^{2} \mathrm{dt}.

343=\int_{0}^{T} 3 \mathrm{t}^{2} \mathrm{dt}=\left.\mathrm{t}^{3}\right|_{0} ^{\mathrm{T}}=\mathrm{T}^{3}-0=\mathrm{T}^{3} so \mathrm{T}=\mathbf{7} \text { seconds. }


Practice 5:

(a) number of new bacteria = \int_{4}^{8} 6 t d t = \left.3 t^{2}\right|_{4} ^{8}=3 \cdot 64-3 \cdot 16=144.

(b) We know the total new population ("area" in Fig. 17) is 2875 – 1000 = 1875 so 1875=\int_{0}^{\mathrm{T}} 6 \mathrm{t} \mathrm{dt}=\left.3 \mathrm{t}^{2}\right|_{0} ^{\mathrm{T}}=3 \mathrm{~T}^{2}-0=3 \mathrm{~T}^{2} so  \mathrm{T}=\mathbf{2 5} \text { minutes. }