Question

вычислить двойной интеграл по области d (cos2x+siny)dxdy ограничены х=0 у=0 4x+4y-pi=0

Answer 1

$int intlimits^._D {(Cos2x+Siny)} , dx dy= intlimits^ frac{ pi}{4} _0 { } , dx intlimits^{frac{ pi}{4}-x}_0 {(Cos2x+Siny)} , dy= \= intlimits^ frac{ pi}{4} _0 { [ yCos2x-Cosy]_0^{frac{ pi}{4}-x} } , dx = \ intlimits^ frac{ pi}{4} _0 { [ ({frac{ pi}{4}-x} })Cos2x-Cos({frac{ pi}{4}-x} })- 0*Cos2x+Cos0], dx= \ intlimits^ frac{ pi}{4} _0 { [ frac{ pi}{4} Cos2x-xCos2x- Cos frac{ pi}{4}Cosx+Sinfrac{ pi}{4}Sinx+1 ], dx=$$\ intlimits^ frac{ pi}{4} _0 { [ frac{ pi}{4} Cos2x-xCos2x- frac{ sqrt{2} }{2} Cosx+ frac{ sqrt{2} }{2} Sinx+1 ], dx$$=[frac{ pi}{8} Sin2x- frac{ sqrt{2} }{2} Sinx-frac{ sqrt{2} }{2} Cosx+x]_0^{ frac{ pi}{4}}- intlimits^{ frac{ pi}{4}}_0 {xCos2x} , dx =$$[frac{ pi}{8} Sin frac{ pi}{2}- frac{ sqrt{2} }{2} Sin frac{ pi}{4}-frac{ sqrt{2} }{2} Cos frac{ pi}{4}+ frac{ pi}{4}]- \ - [frac{ pi}{8} Sin0- frac{ sqrt{2} }{2} Sin0-frac{ sqrt{2} }{2} Cos0+0]- frac{1}{2} intlimits^ frac{ pi }{4}_0 {x} , d(Sin2x) = \ = frac{3 pi}{8} -1+frac{ sqrt{2} }{2} - frac{x}{2}Sin2x|_0^{ frac{ pi }{4}}+ frac{1}{2}intlimits^{ frac{ pi }{4} }_0 {Sin2x} , dx =$
$= frac{3 pi}{8} -1+frac{ sqrt{2} }{2} - frac{ pi }{8}Sin frac{ pi }{2}+0Sin0- frac{1}{4} Cos2x|_0^{ frac{ pi }{4} }= \ frac{3 pi}{8} -1+frac{ sqrt{2} }{2} - frac{ pi }{8}- frac{1}{4} Cos frac{ pi}{2}+ frac{1}{4} Cos0= \ = frac{3 pi}{8} -1+frac{ sqrt{2} }{2} - frac{ pi }{8}+ frac{1}{4}= frac{ pi }{4}- frac{3}{4}+frac{ sqrt{2} }{2} = frac{ pi -3+2 sqrt{2} }{4}$