Answer to Question #110059 in Calculus for Nimra

Question

∫¹ ∫² (sinx² + xy)dx

Answer 1

This integral (Fresnel Sine Integral) does not have a closed form

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

hence:

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

therefore:

$\int^{2}_{1}(sin(x^2)+xy)dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}x}{\sqrt{\pi}})+\frac{x^{2}y}{2}]^{2}_{1}\\$

$\int^{2}_{1}(sin(x^2)+xy)dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}(2)}{\sqrt{\pi}})+\frac{(2)^{2}y}{2}]-[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}(1)}{\sqrt{\pi}})+\frac{(1)^{2}y}{2}]\\$

$\int^{2}_{1}(sin(x^2)+xy)dx=0.4945081876+1.5y$