Given the flat curve and the polar graph of the polar equation $r=5 \cos (2 \theta)-2.5$. Let $\int_{a}^{b} \frac{1}{2}(5 \cos (2 \theta)-2.5)^{2} d \theta$ indicates the area of the shaded region. 1) $a=$ I (enter a fraction involving pi) $b=$ (enter a fraction involving pi)
Solution
Step 1 ：Given the polar equation $r=5 \cos (2 \theta)-2.5$, we need to find the area of the shaded region which is given by the integral $\int_{a}^{b} \frac{1}{2}(5 \cos (2 \theta)-2.5)^{2} d \theta$. To find the limits of integration $a$ and $b$, we need to find the values of $\theta$ for which $r=0$.
Step 2 ：Setting $r=0$ in the polar equation, we get $5 \cos (2 \theta)-2.5=0$. Solving this equation for $\theta$ will give us the limits of integration.
Step 3 ：The solutions to the equation $5 \cos (2 \theta)-2.5=0$ are $\theta = \frac{0.523598775598299}{\pi}$ and $\theta = \frac{2.61799387799149}{\pi}$. These are the points where the polar curve intersects the origin, and hence they are the limits of integration for the area of the shaded region.
Step 4 ：Final Answer: $a=\frac{0.523598775598299}{\pi}$ and $b=\frac{2.61799387799149}{\pi}$, which can be simplified to $a=\frac{1}{6}$ and $b=\frac{5}{6}$ in terms of $\pi$. So, $a=\boxed{\frac{1}{6}\pi}$ and $b=\boxed{\frac{5}{6}\pi}$.
