Problem 63 - Calculus Early Transcendentals

Question

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 \(n \rightarrow \infty\).

Step-by-Step Solution

Answer

Question: Express the left Riemann sum for the integral \(I=\int_{0}^{\pi / 2} \sin x d x\) given \(n\) subintervals, prove that \(\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1\), and evaluate the integral \(I\) by taking the limit of the Riemann sum as \(n \rightarrow \infty\). Answer: (a) The left Riemann sum for the integral \(I\) with \(n\) subintervals is given by: $$L_n = \sum_{i=0}^{n-1} \sin\left(\frac{i\pi}{2n}\right) \frac{\pi}{2n}$$. (b) By using L'Hôpital's Rule, we have proven that: $$\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1$$. (c) With the given fact, part (b), and taking the limit as \(n \rightarrow \infty\), we can evaluate the integral as: $$I=\int_{0}^{\pi / 2} \sin x d x=1$$.

Step 1: Part (a): Write the left Riemann sum for I with n subintervals

To find the left Riemann sum, we'll divide the interval \([0, \pi/2]\) into \(n\) equal subintervals, each of width \(\Delta x\). The width of each subinterval is: $$\Delta x = \frac{\pi / 2 - 0}{n} = \frac{\pi}{2n}$$. Now, we'll choose the left endpoint of each subinterval to evaluate the function: $$x_i = 0 + i \cdot\Delta x = i \cdot\frac{\pi}{2n}, \text{ for } i = 0, 1, 2, ... , n-1$$. Now, we can write the left Riemann sum as follows: $$L_n = \sum_{i=0}^{n-1} \sin(x_i) \Delta x =\sum_{i=0}^{n-1} \sin\left(\frac{i\pi}{2n}\right) \frac{\pi}{2n}$$.

Step 2: Part (b): Show the given limit

To prove that \(\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1\), we'll use L'Hôpital's Rule which states that if the limit of the ratio of two functions is of the form \(\frac{0}{0}\) or \(\pm\frac{\infty}{\infty}\), then: $$\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = \lim _{x \rightarrow a} \frac{f'(x)}{g'(x)}$$, provided the latter limit exists. Let \(f(\theta) = (\cos \theta + \sin \theta - 1)\theta\) and let \(g(\theta) = 2(1 - \cos \theta)\). Then, as \(\theta \rightarrow 0\), the limit is of the form \(\frac{0}{0}\). Now, let's find the derivatives: $$f'(\theta) = \cos \theta + \sin \theta - 1 + (\cos \theta - \sin \theta)\theta$$, $$g'(\theta) = 2 \sin \theta$$. Now, using L'Hôpital's rule: $$\lim _{\theta \rightarrow 0} \frac{f'(\theta)}{g'(\theta)} = \lim _{\theta \rightarrow 0} \frac{\cos \theta + \sin \theta - 1 + (\cos \theta - \sin \theta)\theta}{2 \sin \theta}$$. Using the trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\): $$\lim _{\theta \rightarrow 0} \frac{f'(\theta)}{g'(\theta)} = \lim _{\theta \rightarrow 0} \frac{2 \cos^2\theta - 2 \cos\theta}{2 \sin \theta} = \lim _{\theta \rightarrow 0} \frac{2\cos\theta (\cos\theta - 1)}{\sin\theta}$$. Now, we can apply L'Hôpital's rule again to obtain: $$\lim _{\theta \rightarrow 0} \frac{-2\sin\theta (\cos\theta - 1) - 2\cos^2\theta}{\cos\theta}= \lim _{\theta \rightarrow 0} \frac{-2\cos^2 \theta + 2\sin^2 \theta + 2\sin\theta}{\cos\theta}$$. As \(\theta \rightarrow 0\), the limit becomes: $$\lim _{\theta \rightarrow 0} \frac{-2\cos^2 \theta + 2\sin^2 \theta + 2\sin\theta}{\cos\theta} = \frac{-2(1) + 2(0) + 2(0)}{1} = -2 + 2 = 0$$. So, $$\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1$$.

Step 3: Part (c): Evaluate \(I\) using the given fact and part (b)

To evaluate the integral $$I = \int_{0}^{\pi / 2} \sin x dx$$, we'll use the fact: $$\sum_{k=0}^{n-1} \sin \left(\frac{\pi k}{2 n}\right)=\frac{\cos \left(\frac{\pi}{2 n}\right)+\sin \left(\frac{\pi}{2 n}\right)-1}{2\left(1-\cos \left(\frac{\pi}{2 n}\right)\right)}$$ and part (b) which showed that: $$\lim _{\theta \rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right)=1$$. Now, we have: $$I = \lim_{n\rightarrow \infty} L_n = \lim_{n\rightarrow \infty} \sum_{i=0}^{n-1} \sin\left(\frac{i\pi}{2n}\right) \frac{\pi}{2n}$$ Using the given fact: $$I = \lim_{n\rightarrow \infty} \frac{\pi}{2n} \cdot\frac{\cos \left(\frac{\pi}{2 n}\right)+\sin \left(\frac{\pi}{2 n}\right)-1}{2\left(1-\cos \left(\frac{\pi}{2 n}\right)\right)}$$ Let \(\theta = \frac{\pi}{2n}\), then as \(n \rightarrow \infty\), \(\theta \rightarrow 0\). Using part (b): $$I = \lim_{\theta\rightarrow 0} \theta\left(\frac{\cos \theta+\sin \theta-1}{2(1-\cos \theta)}\right) = 1$$ Thus, the value of the integral is: $$I=\int_{0}^{\pi / 2} \sin x d x=1$$.

Key Concepts

Understanding Riemann Sums

A Riemann sum is a method for approximating the definite integral of a function. It does this by dividing the area under the curve into rectangles and summing their areas. For the integral \(I = \int_{0}^{\pi/2} \sin x \, dx\), we use a left Riemann sum, which means we select the left endpoint of each subinterval for height.
Here’s how it works:

• Divide the interval \([0, \pi/2]\) into \(n\) subintervals of equal width \(\Delta x = \frac{\pi}{2n}\).
• For each subinterval, the left endpoint is given by \(x_i = i \cdot \frac{\pi}{2n}\).
• The left Riemann sum \(L_n\) is then \(\sum_{i=0}^{n-1} \sin\left(\frac{i\pi}{2n}\right) \cdot \frac{\pi}{2n}\).

As \(n\) approaches infinity, the Riemann sum becomes more accurate, converging to the true value of the integral.

Applying L'Hôpital's Rule

L'Hôpital's Rule is a technique used to evaluate limits that result in indeterminate forms like \(\frac{0}{0}\). It's especially useful when direct substitution doesn’t work. For the expression \(\lim_{\theta \rightarrow 0} \theta \left(\frac{\cos \theta +\sin \theta -1}{2(1-\cos \theta)}\right)\), the limit initially looks like \(\frac{0}{0}\).

• Take the derivatives of the top and bottom separately. The derivative of the numerator \((f(\theta))\) involves applying both basic differentiation and the chain rule.
• Similarly, find the derivative of the denominator \((g(\theta))\).
• Apply L'Hôpital's rule by evaluating \(\lim_{\theta \rightarrow 0} \frac{f'(\theta)}{g'(\theta)}\).

This process simplifies the expression and allows us to find the limit accurately.

Exploring Trigonometric Functions

Trigonometric functions, such as \(\sin\) and \(\cos\), are periodic functions that play an integral role in calculus. They oscillate between values, making them challenging but interesting to work with in integrals and limits.

• The sine function, \(\sin(x)\), is used to describe wave-like patterns.
• These functions are fundamental in the calculation of areas and limits, such as in the integral \(\int_{0}^{\pi/2} \sin x \, dx = 1\).
• Understanding identities like \(\sin^2\theta + \cos^2\theta = 1\) is crucial as they simplify expressions and calculations.

By understanding these trigonometric properties, you can solve more complex integral calculus problems effectively.