Answer
$\int_{0}^{1} (x+\sqrt{1-x^2})~dx = \frac{\pi+2}{4}$
Work Step by Step
$\int_{0}^{1} (x+\sqrt{1-x^2})~dx = \int_{0}^{1} x~dx+\int_{0}^{1} \sqrt{1-x^2}~dx$
On the interval $0 \leq x \leq 1$, the graph of $f(x) = x$ forms a triangle above the x-axis. We can find the area of this triangle:
$A = \frac{1}{2}(1)(1) = \frac{1}{2}$
On the interval $0 \leq x \leq 1$, the graph of $\sqrt{1-x^2}$ forms a quarter of a circle above the x-axis. We can find the area:
$A = \frac{\pi~r^2}{4} = \frac{\pi(1)^2}{4} = \frac{\pi}{4} $
Therefore:
$\int_{0}^{1} (x+\sqrt{1-x^2})~dx = \frac{1}{2}+ \frac{\pi}{4} = \frac{\pi+2}{4}$