Answer
$\int_{0}^{1}(2 \sqrt{-x^{2}+1}+x) d x=\left[\frac{1+\pi}{2}\right]$
Work Step by Step
By using Theorem ( 5.5.4 )
\[
\int_{0}^{1}(x+2 \sqrt{1-x^{2}}) d x=\int_{0}^{1} 2+x d x \int_{0}^{1} \sqrt{1-x^{2}} d x
\]
By using the area of the triangle, we will find $\int_{0}^{1} x d x$
\[
\int_{0}^{1} x d x=A=1 \cdot 1 \cdot \frac{1}{2} =\frac{1}{2}
\]
We'll find $\int_{0}^{1} \sqrt{-x^{2}+1} d x$ by using the area of the quarter - circle
\[
\int_{0}^{1} \sqrt{-x^{2}+1} d x=\frac{1}{4} \pi \cdot 1=\frac{\pi}{4}
\]
$\int_{0}^{1}(2 \sqrt{-x^{2}+1}+x) d x=\int_{0}^{1} x d x+2 \int_{0}^{1} \sqrt{-x^{2}+1} d x$
$=\frac{1}{2}+2 \cdot \frac{\pi}{4}$
$=\left[\frac{1+\pi}{2}\right]$
