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Data publicarii: 06 August, 2016

EXERCITIUL 11

Suport teoretic:

Integrale definite,schimbari de variabile,functii trigonometrice,identitati 

trigonometrice.

Enunt:

Sa se calculeze:

I=\int_0^1{x^2\sqrt{4-x^2}dx}\;.I=\int_0^1{x^2\sqrt{4-x^2}dx}\;.

Raspuns:

I=\frac{4\pi-3\sqrt{3}}{12}\;.I=\frac{4\pi-3\sqrt{3}}{12}\;.

Rezolvare:

Efectuam schimbarea de variabila prin substitutia x = 2sint = > dx = 2(cost)dt .

Evident, xЄ[0;1] = > tЄ[0;π/6] . Rezulta, succesiv:

I=\int_0^{\frac{\pi}{6}}{4(sin^2t){\sqrt{4-4sin^2t}}(2cost)dt}=\cdots=4\cdot{\int_0^{\frac{\pi}{6}}{(sin^22t)}dt}=I=\int_0^{\frac{\pi}{6}}{4(sin^2t){\sqrt{4-4sin^2t}}(2cost)dt}=\cdots=4\cdot{\int_0^{\frac{\pi}{6}}{(sin^22t)}dt}=

=4\cdot{\int_0^{\frac{\pi}{6}}{(\frac{1-cos4t}{2})}dt}=4(\frac{t}{2}-\frac{sin4t}{8})|_0^{\frac{\pi}{6}}=\cdots=\frac{4\pi-3\sqrt{3}}{12}\;.=4\cdot{\int_0^{\frac{\pi}{6}}{(\frac{1-cos4t}{2})}dt}=4(\frac{t}{2}-\frac{sin4t}{8})|_0^{\frac{\pi}{6}}=\cdots=\frac{4\pi-3\sqrt{3}}{12}\;.


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Sonny

38MCAZx6Qx9l, 15.08.2016 20:51

Full of salient points. Don't stop benleviig or writing!

 

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