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Index stranica MATEMATIČKA ANALIZA INTEGRALI

Integrali – pismeni ispit iz matematike, Osijek 2013.

[inlmath]\int xe^x\mathrm dx[/inlmath]

Integrali – pismeni ispit iz matematike, Osijek 2013.

Postod slavonija035 » Petak, 28. Jun 2013, 19:04

Slika

ako može pomoć sa prva dva zadatka, ili sa svime ovim što je napisano iako je to za jednu drugu temu :)
 
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Re: Integrali – pismeni ispit iz matematike, Osijek 2013.

Postod Daniel » Petak, 28. Jun 2013, 21:15

[inlmath]1.[/inlmath]
[dispmath]\int\frac{\sin 2x}{\sqrt{\sin x+2}}\mathrm dx=2\int\frac{\sin x\cos x}{\sqrt{\sin x+2}}\mathrm dx=[/dispmath]
Smena:
[inlmath]\sin x=t\\
\cos x\mathrm dx=\mathrm dt[/inlmath]
[dispmath]=2\int\frac{t}{\sqrt{t+2}}\mathrm dt=2\int\frac{t+2-2}{\sqrt{t+2}}\mathrm dt=2\int\frac{t+2}{\sqrt{t+2}}\mathrm dt-2\int\frac{2}{\sqrt{t+2}}\mathrm dt=[/dispmath][dispmath]=2\int\sqrt{t+2}\:\mathrm dt-4\int\frac{\mathrm dt}{\sqrt{t+2}}=2\cdot\frac{2}{3}\left(t+2\right)^\frac{3}{2}-4\cdot 2\sqrt{t+2}+c=[/dispmath][dispmath]=\frac{4}{3}\left(t+2\right)^\frac{3}{2}-8\sqrt{t+2}+c=\frac{4}{3}\left(\sin x+2\right)^\frac{3}{2}-8\sqrt{\sin x+2}+c[/dispmath][dispmath]=\frac{4}{3}\sqrt{\sin x+2}\:\left[\left(\sin x+2\right)-6\right]+c=\frac{4}{3}\sqrt{\sin x+2}\:\left(\sin x-4\right)+c[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain
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Re: Integrali – pismeni ispit iz matematike, Osijek 2013.

Postod Daniel » Subota, 29. Jun 2013, 20:24

[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
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]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain
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