Answer to Question #110056 in Calculus for Nimra

Question

∫²∫¹ (sinx² + xy) dx

Answer 1

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

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

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

then:

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

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:

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