8.3.4 Exercises

The function $f$ with $f(x)=3\sin \left(x\right)$ has, on the interval $[-2\pi ;2\pi ]$, the zeros $-2\pi$, $-\pi$, $0$, $\pi$, and $2\pi$. Since the graph of $f$ is centrally symmetric with respect to the origin, for the area we obtain:


Of course, the integral can also be calculated without noting that the graph of $f$ is centrally symmetric with respect to the origin.

To calculate the area $I_A$ of the region between the graphs of the functions $f$ and $g$, the difference $f-g$ with $f(x)-g(x)=3-(x-2{)}^2-2·(x-2{)}^4$ on the interval $[1;3]$ is considered.
From the drawing of the graphs of the functions we see that the difference $f(x)-g(x)$ is greater or equal to zero for $x\in [1;3]$. This can also be seen by calculation: According to the assumption, we have $1\le x\le 3$ and thus $-1\le x-2\le 1$. Hence, $(x-2{)}^2\le 1$ and thus $-(x-2{)}^2\ge -1$ such that

${\displaystyle f(x)-g(x)\ge 3-1-2·1=0}$
for all $1\le x\le 3$.
Thus, for the calculation of the area, we have to evaluate the integral ${\displaystyle {\int }_1^3(f(x)-g(x))dx}$. For this purpose, the terms can be multiplied out and integrated according to the sum rule. Another way is to consider the terms of the functions in more detail: in this situation we have two terms, namely $(x-2{)}^2$ and $(x-2{)}^4$, that result from shifting the known terms $z^2$ and $z^4$ according to $z=x-2$. An antiderivative of $h$ with $h(z)=z^2$ is $H$ with $H(z)=\frac{1}{3}·z^3$. If we now consider $F$ with $F(x)=3x-\frac{1}{3}·(x-2{)}^3$ accordingly, then we have (from the chain rule) $F\text{'}(x)=3-\frac{1}{3}·3·(x-2{)}^2·1=f(x)$. Here, the last factor results from the derivative of the inner function $u$ with $u(x)=x-2$. Therefore, $F$ is an antiderivative of $f$. Likewise, it can be checked that $G$ with $G(x)=\frac{2}{5}·(x-2{)}^5$ is a antiderivative of $g$.
Thus, for the area $I_A$ of the region between the graphs of the functions, we have

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{I_A}& =\hfill & \left|{\int }_1^3(3-(x-2{)}^2-2·(x-2{)}^4)dx\right|\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left|{[3x-\frac{1}{3}(x-2{)}^3-\frac{2}{5}(x-2{)}^5]}_1^3\right|\hfill \\ \multicolumn{1}{c}{}& =\hfill & |27-\frac{1}{3}-\frac{2}{5}-(3+\frac{1}{3}+\frac{2}{5})|\hfill \\ \multicolumn{1}{c}{}& =\hfill & 22+\frac{8}{15}\hspace{0.5em}.\hfill \end{array}}$

The force $F_s$ that acts along the path on the small body $k$ with mass $m$ to lift it from the surface of the body $K$ is directed against the gravitational force $F$. Thus, we have $F_s=-F$.
Hence, the work $W$ done by the force in lifting the small body $k$ from $r_1=1$ to $r_2=4$ is

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{W={\int }_1^4{F}_s(r)\hspace{0.17em}dr=-{\int }_1^{4}F(r)\hspace{0.17em}dr}& =\hfill & -{\int }_1^4-\gamma ·\frac{2·m}{r^2}\hspace{0.17em}dr\hfill \\ \multicolumn{1}{c}{}& =\hfill & {\int }_1^4\gamma ·\frac{2·m}{r^2}\hspace{0.17em}dr\hfill \\ \multicolumn{1}{c}{}& =\hfill & {[-\gamma ·\frac{2·m}{r}]}_1^4\hfill \\ \multicolumn{1}{c}{}& =\hfill & -\gamma ·2·m(\frac{1}{4}-\frac{1}{1})\hfill \\ \multicolumn{1}{c}{}& =\hfill & \gamma ·\frac{3m}{2}\hspace{0.5em}.\hfill \end{array}}$