Neposredno integriranje u dvostrukom integralu

Promijenite redoslijed integriranja i izračunajte vrijednost integrala

$\displaystyle \int\limits_2^4\int\limits_x^{2x}\frac{y}{x} dx dy$ .

Rješenje.

Područje integracije grafički je prikazano na slici 4.2.

**Slika:** Područje integracije $D=\{(x,y)\in\mathbb{R}^2:2\le x\le 4,\quad x\le y\le 2x\}$ .
$\begin{figure} \begin{center} \epsfig{file=sl16_dvostruki_integral.eps, width=8cm} \end{center} \end{figure}$

Promijenimo redoslijed integriranja. U tom slučaju integriramo po uniji dvaju područja (slika 4.3):

$\displaystyle D_1=\{(x,y)\in\mathbb{R}^2:2\le y\le 4,\quad 2\le x\le y\}$ ,

$\displaystyle D_2=\{(x,y)\in\mathbb{R}^2:4\le y\le 8,\quad \frac{y}{2}\le x\le 4\}$ .

**Slika:** Područje integracije $D=D_1\cup D_2=\{(x,y)\in\mathbb{R}^2:2\le y\le 4,\quad 2\le x\le y\} \cup \{(x,y)\in\mathbb{R}^2:4\le y\le 8,\quad \frac{y}{2}\le x\le 4\}$ .
$\begin{figure} \begin{center} \epsfig{file=sl17_dvostruki_integral.eps, width=8cm} \end{center} \end{figure}$

Izračunajmo vrijednost integrala uzastopnim računanjem dvaju jednostrukih integrala uz pomoć Newton-Leibnitzove formule ( [M2, poglavlje 2.2]):

$\displaystyle I=\int\limits_2^4\int\limits_x^{2x}\frac{y}{x} dx dy$	$\displaystyle = \int\limits_2^4 y dy\int\limits_2^{y}\frac{dx}{x}+\int\limits_4^8 y dy\int\limits_{\frac{y}{2}}^{4}\frac{dx}{x}$
	$\displaystyle =\int\limits_2^4 y dy\ln \vert x\vert\bigg\vert_2^y+\int\limits_4^8 y dy\ln \vert x\vert\bigg\vert_{\frac{y}{2}}^{4}$
	$\displaystyle =\int\limits_2^4 y\left(\ln \vert y\vert-\ln 2\right) dy+\int\limits_4^8 y\left(\ln 4-\ln \frac{\vert y\vert}{2}\right) dy$
	$\displaystyle \overset{y>0}{=}\int\limits_2^4 y\ln \frac{y}{2} dy+\ln 4\int\limits_4^8 y dy-\int\limits_4^8 y\ln \frac{y}{2} dy.$

Budući je

$\displaystyle \int y\ln \frac{y}{2} dy$	$\displaystyle =\left\{\begin{array}{cc} u=\ln\frac{y}{2} & du=\frac{2}{y}\frac{1}{2} dy=\frac{dy}{y} dv=y dy \quad & v=\frac{y^2}{2}\end{array}\right\}$
	$\displaystyle =\frac{y^2}{2}\ln\frac{y}{2}-\int \frac{y^2}{2}\frac{1}{y} dy=\frac{y^2}{2}\ln\frac{y}{2}-\frac{y^2}{4}+C$ ,

imamo

$\displaystyle I$	$\displaystyle =\left(\frac{y^2}{2}\ln\frac{y}{2}-\frac{y^2}{4}\right)\bigg\vert... ...g\vert_4^8-\left(\frac{y^2}{2}\ln\frac{y}{2}-\frac{y^2}{4}\right)\bigg\vert_4^8$
	$\displaystyle =8\ln 2-4-\left(2\ln 1-1\right)+32\ln 4-8\ln 4-\left(32\ln 4-16\right)+\left(8\ln 2-4\right)$
	$\displaystyle =8\ln 2-4+1+32\ln 4-16\ln 2-32\ln 4+16+8\ln 2-4=9.$