8.2.5 Exercises

The area is subdivided by vertical lines at $x_0=-3$, $x_1=-1$, $x_2=0$, $x_3=1$, $x_4=\frac{3}{2}$, $x_5=2$, $x_6=3$, and $x_7=4$ into regions that are bounded each by the graph of $f$, the $x$-axis and the lines at $x_{k-1}$ and $x_k$ for $1\le k\le 7$.
Then the corresponding signed areas are summed up resulting in the value of the integral:

${\displaystyle {\int }_{-3}^{4}f(x)\hspace{0.17em}dx=-\frac{2·2}{2}+\frac{1·1}{2}+1·1+\frac{\frac{1}{2}·1}{2}-\frac{\frac{1}{2}·1}{2}-1·1-\frac{1·1}{2}=-2\hspace{0.5em}.}$
Of course, the area can also be subdivided into other regions. If, for example, the area is subdivided by vertical lines at $z_0=-3$, $z_1=-1$, $z_2=\frac{3}{2}$, and $z_3=4$ into three regions, the area of the region $B_2$ between the lines at $z_1$ and $z_2$ equals the area of the region $B_3$ between the lines at $z_2$ and $z_3$. However, the signs of the areas of $B_2$ and $B_3$ are opposite since $B_2$ lies above the $x$-axis and $B_3$ below. Thus, the value of the integral equals the negative area of the region $B_1$ between the lines $z_0$ and $z_1$, which lies below the $x$-axis.

From the fact that the integrand is odd and the interval of integration is symmetric with respect to the origin $(\mathrm{0,0})$, we can deduce that the value of the integral is $0$. Alternatively, we can calculate the integral using the fundamental theorem of calculus:

${\displaystyle {\int }_{-\pi }^{\pi }(5x^3-4\sin (x))dx={[\frac{5}{4}·x^4+4\cos (x)]}_{-\pi }^{\pi }={[\frac{5}{4}·x^4+4\cos (x)]}_{-\pi }^{\pi }=0\hspace{0.5em}.}$

If we consider $z$ as an unknown constant, we have

${\displaystyle {\int }_0^2(x^2+z·x+1)dx\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}{[\frac{1}{3}{x}^3+\frac{1}{2}z{x}^2+x]}_0^2\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}\frac{8}{3}+2z+2\hspace{0.5em}.}$

Hence, $z=-\frac{14}{6}=-\frac{7}{3}$ is the required value.

The integrand $f$ with $f(x)=\frac{1}{2x}=\frac{1}{2}·\frac{1}{x}$ for $x<0$ has an antiderivative $F$ with $F(x)=\frac{1}{2}\ln |x|$. From the fundamental theorem, we have

${\displaystyle {\int }_{-24}^{-6}\frac{1}{2x}\hspace{0.17em}dx={[\frac{1}{2}\ln |x|]}_{-24}^{-6}=\frac{1}{2}(\ln |-6|-\ln |-24|)=\frac{1}{2}\ln \left(\frac{6}{24}\right)=\frac{1}{2}\ln \left(2^{-2}\right)=-\ln (2)\hspace{0.5em}.}$

The integrand $f$ with $f(x)=\frac{3\pi }{x^2}-4\sin (x)$ has an antiderivative $F$ with $F(x)=-\frac{3\pi }{x}+4\cos (x)$. From the fundamental theorem, we have

${\displaystyle {\int }_{\pi }^{3\pi }(\frac{3\pi }{x^2}-4\sin (x))dx={[-\frac{3\pi }{x}+4\cos (x)]}_{\pi }^{3\pi }=(-1-4)-(-3-4)=2\hspace{0.5em}.}$
Remark: The integral over a period $2\pi$ of the two periodic functions $\sin$ and $\cos$ equals zero. However, for other periodic functions, as for example $f(x)=\sin (x)+\frac{1}{2}$, the value of the integral over a period can differ from zero.