∫02∫01(sinx²+xy)dydx=∫02[ysinx2+2xy2]01dx
∫02∫01(sinx²+xy)dydx=∫02[sinx2+2x]dx
since this integral (Fresnel Sine Integral) does not have a closed form then:
FresneliSineiIntegral(S(u))=∫0usin(2πx2)dx
hence
∫sin(x2)dx=[22πS(π2x)+A
therefore
∫02[sinx2+2x]dx=[22πS(π2x)+4x2]02
∫02[sinx2+2x]dx=[22πS(π2(2))+4(2)2]−[22πS(π2(0))+4(0)2]
hence
∫02[sinx2+2x]dx=1.80477649
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