Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 327: 25

$1 +\frac{\sqrt 3}{2}$

Evaluate the integral: $\int^{\pi}_{\pi/6}sin(θ)dθ$ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(x)$: $F(x)=\int sin(θ)dθ$ $F(x)=-cos(θ)$ Now evaluate $F(b)−F(a)$: $F(\pi)−F(\frac{\pi}{6})$ $-cos(\pi) - (-cos(\frac{\pi}{6}))$ $-(-1) - - \frac{\sqrt 3}{2}$ $1 +\frac{\sqrt 3}{2}$