Problem 75 - Calculus Early Transcendentals: Pearson New International Edition

Question

first recognize the given limit as a definite integral and then evaluate that integral by the Second Fundamental Theorem of Calculus. $$ \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[\sin \left(\frac{\pi i}{n}\right)\right] \frac{\pi}{n} $$

Step-by-Step Solution

Answer

The limit of the Riemann sum is 2.

Step 1: Recognize the expression as a Riemann sum

The given limit expression \( \lim _{n \to \infty} \sum_{i=1}^{n}\left[ \sin \left(\frac{\pi i}{n}\right)\right] \frac{\pi}{n} \) resembles a Riemann sum of a function over an interval. Here, the function is \( f(x) = \sin(x) \) evaluated at points \( x_i = \frac{\pi i}{n} \), and \( \Delta x = \frac{\pi}{n} \) is the width of each subinterval.

Step 2: Write the Riemann sum as a definite integral

The Riemann sum \( \sum_{i=1}^{n} f\left(\frac{\pi i}{n}\right) \frac{\pi}{n} \) approximates the definite integral \( \int_{0}^{\pi} \sin(x) \, dx \) as \( n \to \infty \). Therefore, the limit expression is equivalent to this integral.

Step 3: Evaluate the definite integral

Use the Second Fundamental Theorem of Calculus, which states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( f(x) = \sin(x) \) and an antiderivative is \( F(x) = -\cos(x) \). So the integral \( \int_{0}^{\pi} \sin(x) \, dx \) becomes:\[-\cos(x) \bigg|_0^\pi = [-\cos(\pi) - (-\cos(0))] = [-(-1) - (-1)] = 2 \]

Step 4: Conclude the solution

The definite integral \( \int_{0}^{\pi} \sin(x) \, dx = 2 \) represents the limit of the Riemann sum as \( n \to \infty \). Therefore, the value of the original limit expression is \( 2 \).

Key Concepts

Riemann Sum

Understanding the concept of a Riemann Sum is key to grasping how definite integrals work. Imagine slicing a curve into many rectangular pieces. The Riemann Sum essentially approximates the area under the curve by adding up the areas of these rectangles. It takes the form:

• Function value at points: Here we use function values such as \( f(x_i) = \sin\left(\frac{\pi i}{n}\right) \).
• Width of rectangles: Given by \( \Delta x = \frac{\pi}{n} \).

As \( n \) (the number of rectangles) becomes larger, these rectangles become thinner and cover the area under the curve more accurately. Hence, the Riemann Sum turns into a \limiting process, forming a definite integral as \( n \to \infty \). In our problem, this process changes \the Riemann Sum into the integral \( \int_{0}^{\pi} \sin(x) \, dx \). By understanding this, \we see how small approximations add up to give an almost precise area.

Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus is a powerful tool in evaluating definite integrals. It states that if you have a continuous function \( f \), and you find its antiderivative (let's call it \( F \)) \, the integral of \( f \) from \( a \) to \( b \) is given by \( F(b) - F(a) \). \This theorem connects the concept of antiderivatives directly to the computation of areas:

• It allows us to calculate definite integrals quite easily by finding antiderivatives.
• To find \( \int_{0}^{\pi} \sin(x) \, dx \), we first determine an antiderivative of \( \sin(x) \), which is \( -\cos(x) \).

The theorem gives you a quick method to compute the exact area under a curve (integral), transforming calculus \from purely theoretical to practically computational. This method relies on neatly linked relations between derivatives and integrals. \Applying it to the integral \( \int_{0}^{\pi} \sin(x) \, dx \) yields the value \( 2 \), representing the limit expression's actual value.

Antiderivative

Antiderivatives or indefinite integrals are essentially the reverse of derivatives. Finding an antiderivative means \identifying a function whose derivative is the given function. \To compute a definite integral, we often need to identify the antiderivative of the integrand:

• For the function \( f(x) = \sin(x) \), an antiderivative is \( F(x) = -\cos(x) \).
• To solve the integral \( \int_{0}^{\pi} \sin(x) \, dx \), using this antiderivative simplifies our \computation massively.

Antiderivatives link closely with the Second Fundamental Theorem of Calculus, as they offer a way to "build back" \a function from its rate of change. This linkage turns integral computation into a straightforward subtraction problem, \where one calculates the difference in function values at the boundary points of integration. \Thus, finding an antiderivative for \( \sin(x) \) allows us efficiently to solve the exercise through fundamental calculus principles.