f(x)=\\frac{72}{\\pi^{2}}x + \\cos x, f(\\frac{\\pi}{2}) = 10\nto advance in the circuit, locate…

Explanation:

Step1: Integrate (f'(x))

We know that if (f'(x)=\frac{72}{\pi^{2}}x+\cos x), then by the rules of integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)) and (\int\cos xdx=\sin x + C). [ \begin{align*} f(x)&=\int\left(\frac{72}{\pi^{2}}x+\cos x\right)dx\ &=\frac{72}{\pi^{2}}\int xdx+\int\cos xdx\ &=\frac{72}{\pi^{2}}\cdot\frac{x^{2}}{2}+\sin x + C\ &=\frac{36}{\pi^{2}}x^{2}+\sin x + C \end{align*} ]

Step2: Use the initial condition (f(\frac{\pi}{2}) = 10)

Substitute (x = \frac{\pi}{2}) into (f(x)=\frac{36}{\pi^{2}}x^{2}+\sin x + C). [ \begin{align*} f\left(\frac{\pi}{2}\right)&=\frac{36}{\pi^{2}}\cdot\left(\frac{\pi}{2}\right)^{2}+\sin\left(\frac{\pi}{2}\right)+C\ 10&=\frac{36}{\pi^{2}}\cdot\frac{\pi^{2}}{4}+1 + C\ 10&=9 + 1+C\ C&=0 \end{align*} ] So (f(x)=\frac{36}{\pi^{2}}x^{2}+\sin x)

Step3: Find (f(\frac{\pi}{6}))

Substitute (x=\frac{\pi}{6}) into (f(x)=\frac{36}{\pi^{2}}x^{2}+\sin x) [ \begin{align*} f\left(\frac{\pi}{6}\right)&=\frac{36}{\pi^{2}}\cdot\left(\frac{\pi}{6}\right)^{2}+\sin\left(\frac{\pi}{6}\right)\ &=\frac{36}{\pi^{2}}\cdot\frac{\pi^{2}}{36}+\frac{1}{2}\ &=1+\frac{1}{2} \end{align*} ]

Answer:

(\frac{3}{2})