#______ $f(x)=2x\\cos(x^{2}),f(\\sqrt{\\frac{7\\pi}{6}})=2$ to advance in the circuit, locate…

Explanation:

Step1: Integrate (f^{\prime}(x))

We know that if (f^{\prime}(x)=2x\cos(x^{2})), then by the substitution (u = x^{2}), (du=2xdx). (\int f^{\prime}(x)dx=\int 2x\cos(x^{2})dx=\int\cos(u)du=\sin(u)+C=\sin(x^{2})+C) So (f(x)=\sin(x^{2})+C)

Step2: Find the value of (C)

Since (f\left(\sqrt{\frac{7\pi}{6}}\right) = 2), we substitute (x = \sqrt{\frac{7\pi}{6}}) into (f(x)=\sin(x^{2})+C) (x^{2}=\frac{7\pi}{6}), then (f\left(\sqrt{\frac{7\pi}{6}}\right)=\sin\left(\frac{7\pi}{6}\right)+C) We know that (\sin\left(\frac{7\pi}{6}\right)=-\frac{1}{2}) So (2=-\frac{1}{2}+C), then (C = 2+\frac{1}{2}=\frac{5}{2})

Step3: Calculate (f\left(\frac{\sqrt{\pi}}{2}\right))

Substitute (x=\frac{\sqrt{\pi}}{2}) into (f(x)=\sin(x^{2})+\frac{5}{2}) (x^{2}=\frac{\pi}{4}), then (f\left(\frac{\sqrt{\pi}}{2}\right)=\sin\left(\frac{\pi}{4}\right)+\frac{5}{2}) Since (\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}) (f\left(\frac{\sqrt{\pi}}{2}\right)=\frac{\sqrt{2}}{2}+\frac{5}{2}=\frac{5 + \sqrt{2}}{2})

Answer:

(\frac{5+\sqrt{2}}{2})