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

"\\int^{2}_{0}\\int^{1}_{0}(sinx\u00b2 + xy)dydx=\\int^{2}_{0} [ysinx^{2}+\\frac{xy^{2}}{2}]^{1}_{0}dx"


"\\int^{2}_{0}\\int^{1}_{0}(sinx\u00b2 + xy)dydx=\\int^{2}_{0} [sinx^{2}+\\frac{x}{2}]dx"


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


"Fresnel \\phantom{i}Sine\\phantom{i} Integral(S(u))=\\int_{0}^{u}sin(\\frac{\\pi x^{2}}{2})dx"

hence

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

therefore

"\\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}"


"\\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

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


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