[inlmath]2.[/inlmath]
[dispmath]\begin{array}{l}
y=\ln x\\
y=\frac{1}{5}x^2-\frac{6}{5}x+1\\
y=1\\
T\left(3,0\right)
\end{array}[/dispmath]
Presek [inlmath]y=\frac{1}{5}x^2-\frac{6}{5}x+1[/inlmath] s [inlmath]x[/inlmath]-osom:
[dispmath]\frac{1}{5}x^2-\frac{6}{5}x+1=0[/dispmath][dispmath]x_{1,2}=\frac{\frac{6}{5}\pm\sqrt{\frac{36}{25}-\frac{4}{5}}}{\frac{2}{5}}\quad /\cdot\frac{5}{5}[/dispmath][dispmath]x_{1,2}=\frac{6\pm\sqrt{36-20}}{2}[/dispmath][dispmath]x_{1,2}=3\pm 2[/dispmath][dispmath]x_1=1,\quad x_2=5[/dispmath]
Pošto i [inlmath]\ln x[/inlmath] seče [inlmath]x[/inlmath]-osu u tački [inlmath]x=1[/inlmath], zaključujemo da će se [inlmath]y=\ln x[/inlmath] i [inlmath]y=\frac{1}{5}x^2-\frac{6}{5}x+1[/inlmath] seći u tački koja je na [inlmath]x[/inlmath]-osi, u tački [inlmath]\left(1,0\right)[/inlmath].
Presek [inlmath]y=\ln x[/inlmath] i [inlmath]y=1[/inlmath]:
[dispmath]\ln x=1\quad\Rightarrow\quad x=e[/dispmath]
Presek [inlmath]y=\frac{1}{5}x^2-\frac{6}{5}x+1[/inlmath] i [inlmath]y=1[/inlmath]:
[dispmath]\frac{1}{5}x^2-\frac{6}{5}x+\cancel 1=\cancel 1\quad /\cdot 5[/dispmath][dispmath]x^2=6x[/dispmath][dispmath]x=0\quad\lor\quad x=6[/dispmath]
- povrsina.png (3.56 KiB) Pogledano 302 puta
[dispmath]P=\int\limits_1^e\left[\ln x-\left(\frac{1}{5}x^2-\frac{6}{5}x+1\right)\right]\mathrm dx+\int\limits_e^6\left[1-\left(\frac{1}{5}x^2-\frac{6}{5}x+1\right)\right]\mathrm dx[/dispmath][dispmath]P=\int\limits_1^e\left(\ln x-\frac{1}{5}x^2+\frac{6}{5}x-1\right)\mathrm dx+\int\limits_e^6\left(\cancel 1-\frac{1}{5}x^2+\frac{6}{5}x-\cancel 1\right)\mathrm dx[/dispmath][dispmath]P=\int\limits_1^e\ln x\mathrm dx-\frac{1}{5}\int\limits_1^e x^2\mathrm dx+\frac{6}{5}\int\limits_1^e x\mathrm dx-\int\limits_1^e\mathrm dx-\frac{1}{5}\int\limits_e^6 x^2\mathrm dx+\frac{6}{5}\int\limits_e^6 x\mathrm dx[/dispmath]
[inlmath]\int\ln x\mathrm dx=[/inlmath]
[inlmath]\begin{array}{l}
u=\ln x\\
\mathrm du=\frac{\mathrm dx}{x}\\
\mathrm dv=\mathrm dx\\
v=x
\end{array}[/inlmath]
[inlmath]x\ln x-\int\cancel x\frac{\mathrm dx}{\cancel x}=x\ln x-x[/inlmath]
[dispmath]P=\left.\left(x\ln x-x\right)\right|_1^e-\frac{1}{5}\cdot\left.\frac{x^3}{3}\right|_1^e+\frac{6}{5}\cdot\left.\frac{x^2}{2}\right|_1^e-\left.x\right|_1^e-\frac{1}{5}\cdot\left.\frac{x^3}{3}\right|_e^6+\frac{6}{5}\cdot\left.\frac{x^2}{2}\right|_e^6[/dispmath][dispmath]P=e\cancelto{1}{\ln e}-e-1\cdot\cancelto{0}{\ln 1}+1-\cancel{\frac{1}{5}\cdot\frac{e^3}{3}}+\frac{1}{5}\cdot\frac{1}{3}+\cancel{\frac{6}{5}\cdot\frac{e^2}{2}}-\frac{6}{5}\cdot\frac{1}{2}-\\
-e+1-\frac{1}{5}\cdot\frac{6^3}{3}+\cancel{\frac{1}{5}\cdot\frac{e^3}{3}}+\frac{6}{5}\cdot\frac{6^2}{2}-\cancel{\frac{6}{5}\cdot\frac{e^2}{2}}[/dispmath][dispmath]P=\cancel e-\cancel e+1+\frac{1}{15}-\frac{3}{5}-e+1-\frac{72}{5}+\frac{108}{5}[/dispmath][dispmath]P=2-e+\frac{1}{15}+\frac{33}{5}[/dispmath][dispmath]\enclose{box}{P=\frac{26}{3}-e}[/dispmath]