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A sine integral by Riemann sums Consider the integral $I=\int_{0}^{\pi / 2} \sin x d x$a. Write the left Riemann sum for $I$ with $n$ subintervals.b. Show that $\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1$c. It is a fact that $\sum_{k=0}^{n-1} \sin \left(\frac{\pi k}{2 n}\right)=\frac{\cos \left(\frac{\pi}{2 n}\right)+\sin \left(\frac{\pi}{2 n}\right)-1}{2\left[1-\cos \left(\frac{\pi}{2 n}\right)\right]}$.Use this fact and part (b) to evaluate $I$ by taking the limit of the Riemann sum as $n \rightarrow \infty$.

a. $\frac{\pi}{2 n} \sum_{k=0}^{n-1} \sin \frac{k \pi}{2 n}$b. 1c. $I=\int_{0}^{\pi / 2} \sin x d x=1$

Calculus 1 / AB

Chapter 5

Integration

Section 4

Working with Integrals

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Harvey Mudd College

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