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Data publicarii: 14 Iulie, 2011

EXERCITIUL 5

Suport teoretic:

Integrale definite,functii irationale,schimbari variabila,identitati trigonometrice.

Enunt:

Sa se calculeze aria domeniului marginit de graficul functiei

f:[-3;3]\rightarrow{R},\;f(x)={(x+1)}\cdot{\sqrt{9-x^2}},f:[-3;3]\rightarrow{R},\;f(x)={(x+1)}\cdot{\sqrt{9-x^2}},

axa Ox si dreptele de ecuatii

x=0\;si\;x=\sqrt{2}.x=0\;si\;x=\sqrt{2}.

Raspuns:

A=\frac{10-4\sqrt{2}+\pi}{2}.A=\frac{10-4\sqrt{2}+\pi}{2}.

Rezolvare:

Se constata usor ca functia f, restrictionata la intervalul

[0;\sqrt{2}][0;\sqrt{2}]

este pozitiva, deci aria domeniului respectiv este data de formula:

A=\int_{0}^{\sqrt{2}}{(x+1)}\cdot{\sqrt{4-x^2}}{dx}.A=\int_{0}^{\sqrt{2}}{(x+1)}\cdot{\sqrt{4-x^2}}{dx}.

Folosind substitutia x = 2sint, obtinem dx = 2cost·dt si 

x\in{[0;\sqrt{2}]}\Rightarrow{t}\in{[0;\frac{\pi}{4}]}.x\in{[0;\sqrt{2}]}\Rightarrow{t}\in{[0;\frac{\pi}{4}]}.

Rezulta:

A=\int_{0}^{\frac{\pi}{4}}{{(2sint+1)}\cdot{\sqrt{4-4sin^2t}}\cdot{2cost}{dt}}=\cdots=A=\int_{0}^{\frac{\pi}{4}}{{(2sint+1)}\cdot{\sqrt{4-4sin^2t}}\cdot{2cost}{dt}}=\cdots= {4}\cdot{\int_{0}^{\frac{\pi}{4}}{{(2sint+1)}\cdot{cos^2t}}{dt}}={4}\cdot{\int_{0}^{\frac{\pi}{4}}{{(2sint+1)}\cdot{cos^2t}}{dt}}=

={2}\cdot{\int_{0}^{\frac{\pi}{4}}{(2sint+1)(1+cos2t)}{dt}}=\cdots=={2}\cdot{\int_{0}^{\frac{\pi}{4}}{(2sint+1)(1+cos2t)}{dt}}=\cdots=

={4}\cdot{\int_{0}^{\frac{\pi}{4}}{sintdt}}+{2}\cdot{\int_{0}^{\frac{\pi}{4}}{cos2tdt}}+{2}\cdot{\int_{0}^{\frac{\pi}{4}}{2sintcos2tdt}}+{2}\cdot{\int_{0}^{\frac{\pi}{4}}{1}{dt}}=={4}\cdot{\int_{0}^{\frac{\pi}{4}}{sintdt}}+{2}\cdot{\int_{0}^{\frac{\pi}{4}}{cos2tdt}}+{2}\cdot{\int_{0}^{\frac{\pi}{4}}{2sintcos2tdt}}+{2}\cdot{\int_{0}^{\frac{\pi}{4}}{1}{dt}}=

={-4}\cdot{cost}|_0^{\frac{\pi}{4}}+{sin2t}|_0^{\frac{\pi}{4}}+{2}\cdot{\int_0^{\frac{\pi}{4}}{(sin3t-sint){dt}}}+2t|_0^{\frac{\pi}{4}}=\cdots=\frac{10-4\sqrt{2}+\pi}{2}.={-4}\cdot{cost}|_0^{\frac{\pi}{4}}+{sin2t}|_0^{\frac{\pi}{4}}+{2}\cdot{\int_0^{\frac{\pi}{4}}{(sin3t-sint){dt}}}+2t|_0^{\frac{\pi}{4}}=\cdots=\frac{10-4\sqrt{2}+\pi}{2}.


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