`int_(pi/6)^pi(sin(theta))d theta` Evaluate the integral.

Textbook Question

Chapter 5, 5.3 - Problem 25 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the definite integral using the fundamental theorem of calculus such that `int_a^b f(x)dx = F(b) - F(a)`

`int_(pi/6)^pi sin theta d theta = -cos theta|_(pi/6)^pi`

`int_(pi/6)^pi sin theta d theta = -cos pi + cos (pi/6)`

`int_(pi/6)^pi sin theta d theta = 1 + sqrt3/2`

Hence, evaluating the definite integral yields

` int_(pi/6)^pi sin theta d theta = 1 + sqrt3/2.`




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