Answer
$A=\frac{\pi}{2}$
Work Step by Step
Setting up the integral using the formula for the area of a region in polar coordinates, we get:
$$\frac{1}{2}\int_{0}^{\pi} (sin\,x+cos\,x)^2 dx$$
Expanding the integrand, we can rewrite the expression as:
$$\frac{1}{2}\int_0^\pi(sin^2x+2sinx\,cosx+cos^2x)dx$$
We can rearrange the terms slightly and get:
$$\frac{1}{2}\int_0^\pi(sin^2x+cos^2x+2sinx\,cosx)dx$$
Using the Pythagorean trignometric identity and the double-angle identity for sine, we can rewrite the expression as:
$$\frac{1}{2}\int_0^\pi(1+sin\,2x)dx$$.
Evaluating the integral, we get:
$$\frac{1}{2}(x-\frac{1}{2}cos\,2x)\bigg\rvert_{0}^{\pi}=\frac{\pi}{2}$$
