Evaluating a definite integral of sin(x) + cos(x)

An antiderivative of $\sin(x) + \cos(x)$ is $F(x) = -\cos(x) + \sin(x)$, since differentiating $-\cos(x)$ gives $\sin(x)$ and differentiating $\sin(x)$ gives $\cos(x)$. By FTC Part 2, $\int_0^{\pi/2} (\sin(x)+\cos(x))\,dx = F(\pi/2) - F(0)$.

Evaluating, $F(\pi/2) = -\cos(\pi/2) + \sin(\pi/2) = -0 + 1 = 1$, and $F(0) = -\cos(0) + \sin(0) = -1 + 0 = -1$. Using the bar notation, $\big[-\cos(x)+\sin(x)\big]_0^{\pi/2} = 1 - (-1) = 2$.