Answer to Question #110057 in Calculus for Nimra

Question #110057

∫²∫1 (sinx² + xy)dx


1
Expert's answer
2020-04-30T19:46:45-0400

0201(sinx²+xy)dydx=02[ysinx2+xy22]01dx\int^{2}_{0}\int^{1}_{0}(sinx² + xy)dydx=\int^{2}_{0} [ysinx^{2}+\frac{xy^{2}}{2}]^{1}_{0}dx


0201(sinx²+xy)dydx=02[sinx2+x2]dx\int^{2}_{0}\int^{1}_{0}(sinx² + xy)dydx=\int^{2}_{0} [sinx^{2}+\frac{x}{2}]dx


since this integral (Fresnel Sine Integral) does not have a closed form then:


FresneliSineiIntegral(S(u))=0usin(πx22)dxFresnel \phantom{i}Sine\phantom{i} Integral(S(u))=\int_{0}^{u}sin(\frac{\pi x^{2}}{2})dx

hence

sin(x2)dx=[2π2S(2xπ)+A\int sin(x^2)dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}x}{\sqrt{\pi}})+A

therefore

02[sinx2+x2]dx=[2π2S(2xπ)+x24]02\int^{2}_{0} [sinx^{2}+\frac{x}{2}]dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}x}{\sqrt{\pi}})+\frac{x^{2}}{4}]^{2}_{0}


02[sinx2+x2]dx=[2π2S(2(2)π)+(2)24][2π2S(2(0)π)+(0)24]\int^{2}_{0} [sinx^{2}+\frac{x}{2}]dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}(2)}{\sqrt{\pi}})+\frac{(2)^{2}}{4}]-[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}(0)}{\sqrt{\pi}})+\frac{(0)^{2}}{4}]


hence

02[sinx2+x2]dx=1.80477649\int^{2}_{0} [sinx^{2}+\frac{x}{2}]dx=1.80477649


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