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Data publicarii: 06 August, 2016

EXERCITIUL 12

Suport teoretic:

Integrale definite,schimbari de variabila,functiile sh,ch.

Enunt:

Sa se calculeze:

I=\int_0^1{\frac{x^2}{\sqrt{x^2+1}}dx}\;.I=\int_0^1{\frac{x^2}{\sqrt{x^2+1}}dx}\;.

Raspuns:

I=\sqrt{2}-(\frac{1}{8})\cdot[e^{2(1+\sqrt{2})}+4(1+\sqrt{2})-e^{-2(1+\sqrt{2})}]\;.I=\sqrt{2}-(\frac{1}{8})\cdot[e^{2(1+\sqrt{2})}+4(1+\sqrt{2})-e^{-2(1+\sqrt{2})}]\;.  

Rezolvare:

I=\int_0^1{x\cdot{\frac{2x}{2\sqrt{x^2+1}}}dx}=\int_0^1{x\cdot{(\sqrt{x^2+1})^{I=\int_0^1{x\cdot{\frac{2x}{2\sqrt{x^2+1}}}dx}=\int_0^1{x\cdot{(\sqrt{x^2+1})^{'}}dx}=

=x\cdot{\sqrt{x^2+1}}|_0^1-\int_0^1{(\sqrt{x^2+1})}dx==x\cdot{\sqrt{x^2+1}}|_0^1-\int_0^1{(\sqrt{x^2+1})}dx=  

=\sqrt{2}-\int_0^1{(\sqrt{x^2+1})}dx=\sqrt{2}-J\;.(1)=\sqrt{2}-\int_0^1{(\sqrt{x^2+1})}dx=\sqrt{2}-J\;.(1)

Pentru calculul integralei J, folosim substitutia x = sht = > dx = chtdt, unde

sht=\frac{e^t-e^{-t}}{2}\;si\;cht=\frac{e^t+e^{-t}}{2}\;.sht=\frac{e^t-e^{-t}}{2}\;si\;cht=\frac{e^t+e^{-t}}{2}\;. (vezi aici)

Calcule simple arata ca xЄ[0;1] = > t Є[0;1+√2], deci:

J=\int_0^{1+\sqrt{2}}{(\sqrt{sh^2t+1})(cht)dt}=\int_0^{1+\sqrt{2}}{\sqrt{(\frac{e^{2t}-2+e^{-2t}}{4}+1})}\cdot{(\frac{e^t+e^{-t}}{2})}dt=J=\int_0^{1+\sqrt{2}}{(\sqrt{sh^2t+1})(cht)dt}=\int_0^{1+\sqrt{2}}{\sqrt{(\frac{e^{2t}-2+e^{-2t}}{4}+1})}\cdot{(\frac{e^t+e^{-t}}{2})}dt=

=\cdots=\int_0^{1+\sqrt{2}}{(\frac{e^t+e^{-t}}{2})^2}dt=\frac{1}{4}(\frac{e^{2t}}{2}+2t-\frac{e^{-2t}}{2})|_0^{1+\sqrt{2}}==\cdots=\int_0^{1+\sqrt{2}}{(\frac{e^t+e^{-t}}{2})^2}dt=\frac{1}{4}(\frac{e^{2t}}{2}+2t-\frac{e^{-2t}}{2})|_0^{1+\sqrt{2}}=

=(\frac{1}{8})\cdot[e^{2(1+\sqrt{2})}+4(1+\sqrt{2})-e^{-2(1+\sqrt{2})}]\;.(2)=(\frac{1}{8})\cdot[e^{2(1+\sqrt{2})}+4(1+\sqrt{2})-e^{-2(1+\sqrt{2})}]\;.(2)

Din (1) si (2) rezulta imediat: 

I=\sqrt{2}-(\frac{1}{8})\cdot[e^{2(1+\sqrt{2})}+4(1+\sqrt{2})-e^{-2(1+\sqrt{2})}]\;.I=\sqrt{2}-(\frac{1}{8})\cdot[e^{2(1+\sqrt{2})}+4(1+\sqrt{2})-e^{-2(1+\sqrt{2})}]\;.

Postat în: SCHIMBARI DE VARIABILA-liceu

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