To determine
To evaluate:
∫π/6πsinθ dθ
Answer
∫π/6πsinθ dθ=1+32
Explanation
1) Concept:
The Fundamental Theorem of Calculus(part 2) : Suppose f is continuous on [a, b], then
∫abfxdx=Fb-F(a), where F is antiderivative of f, that is F'=f.
2) Calculation:
By using the antiderivative of sinθ and fundamental theorem of calculus
∫π/6πsinθ dθ=-cosθππ/6
=-cosπ--cosπ6=-cosπ+cosπ6
After substituting values of cos(π) and cosπ6,
∫π/6πsinθ dθ=--1+32=1+32
Conclusion:
Therefore,
∫π/6πsinθ dθ=1+32