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»APLICATII ALE INTEGRALEI DEFINITE-liceu

Data publicarii: 04 Noiembrie, 2014

EXERCITIUL 8

Suport teoretic:

Arii,functiile arcsin,arccos,integrale definite.

Enunt:

Sa se afle aria A a suprafetei delimitata de curbele reprezentative ale functiilor

f,g:[-1,+1] - > R, unde

f(x) = arcsinx

g(x) = arccosx

si dreptele de ecuatii

$x=0,\;x=\frac{\sqrt{3}}{2}.$ x=0,\;x=\frac{\sqrt{3}}{2}.

Raspuns:

$A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.$ A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.

Rezolvare:

$A=\int_{0}^{\frac{\sqrt{3}}{2}}{|{arcsinx}-{arccosx}|}{dx}=\int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcsinx}|}{dx},$ A=\int_{0}^{\frac{\sqrt{3}}{2}}{|{arcsinx}-{arccosx}|}{dx}=\int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcsinx}|}{dx},

in baza identitatii cunoscute:

${arcsinx+arccosx}=\frac{\pi}{2},\;\forall{x}\in{[-1,1]}.$ {arcsinx+arccosx}=\frac{\pi}{2},\;\forall{x}\in{[-1,1]}.

Tinand cont de monotonia functiei arcsinx si de definitia modulului, avem:

$\int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcssinx}|}{dx}=$ \int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcssinx}|}{dx}= $\int_{0}^{\frac{\sqrt{2}}{2}}{(\frac{\pi}{2}-2{arcsinx})}{dx}+$ \int_{0}^{\frac{\sqrt{2}}{2}}{(\frac{\pi}{2}-2{arcsinx})}{dx}+ $\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{(2{arcsinx}-\frac{\pi}{2})}{dx}.$ \int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{(2{arcsinx}-\frac{\pi}{2})}{dx}.

Deci:

$A={\frac{\pi}{2}}\cdot{\int_{0}^{\frac{\sqrt{2}}{2}}{dx}}-2\int_{0}^{\frac{\sqrt{2}}{2}}{arcsinx}{dx}+2\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{arcsinx}{dx}-$ A={\frac{\pi}{2}}\cdot{\int_{0}^{\frac{\sqrt{2}}{2}}{dx}}-2\int_{0}^{\frac{\sqrt{2}}{2}}{arcsinx}{dx}+2\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{arcsinx}{dx}- ${\frac{\pi}{2}}\cdot{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{dx}}.$ {\frac{\pi}{2}}\cdot{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{dx}}.

Cum

$\int{arcsinx}{dx}=\int{x^{$ \int{arcsinx}{dx}=\int{x^{'}}{arcsinx}{dx}={xarcsinx}+\int{\frac{{(1-x^2)}^{'}}{2\sqrt{1-x^2}}}{dx}=

${xarcsinx}+\sqrt{1-x^2}+\mathcal{C},$ {xarcsinx}+\sqrt{1-x^2}+\mathcal{C},

rezulta:

$A={\frac{\pi}{2}}x|_{0}^{\frac{\sqrt{2}}{2}}-2({xarcsinx}+\sqrt{1-x^2})|_{0}^{\frac{\sqrt{2}}{2}}+2({xarcsinx}+\sqrt{1-x^2})|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}-\frac{\pi}{2}x|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}.$ A={\frac{\pi}{2}}x|_{0}^{\frac{\sqrt{2}}{2}}-2({xarcsinx}+\sqrt{1-x^2})|_{0}^{\frac{\sqrt{2}}{2}}+2({xarcsinx}+\sqrt{1-x^2})|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}-\frac{\pi}{2}x|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}.

In urma catorva calcule de rutina, se obtine in final:

$A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.$ A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.

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EXERCITIUL 8

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