Problem 73 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 73-76, first recognize the given limit as a definite integral and then evaluate that integral by the Second Fundamental Theorem of Calculus \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{3 i}{n}\right)^{2} \frac{3}{n}\)

Step-by-Step Solution

Answer

The limit evaluates to 9.

Step 1: Identify as a Riemann Sum

The given limit \( \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{3 i}{n}\right)^{2}\frac{3}{n} \) resembles the form of a Riemann sum, which approximates a definite integral. Notice that \( \frac{3}{n} \) acts as \( \Delta x \) and \( \left(\frac{3i}{n}\right)^{2} \) represents the function \( f(x) \) evaluated at \( x_i = \frac{3i}{n} \).

Step 2: Define the Integral Limits

Recognize that the sum \( \sum_{i=1}^{n}\frac{3}{n}\left(\frac{3i}{n}\right)^{2} \) integrates the function \( f(x) = x^2 \). The values of \( x_i \) and \Delta x \ give us the interval \([a, b]\) from \(0\) to \(3\).

Step 3: Set Up the Definite Integral

By the definition of the Riemann sum and recognizing it as a definite integral, the limit can be written as the integral: \[ \int_{0}^{3} x^2 \, dx \].

Step 4: Apply the Second Fundamental Theorem of Calculus

Use the Second Fundamental Theorem of Calculus, which states that if a function \(f\) is continuous on an interval \([a, b]\), then the integral \( \int_{a}^{b} f(x)\, dx \) can be evaluated by finding an antiderivative \(F(x)\) of \(f(x)\), and computing \( F(b) - F(a) \).

Step 5: Calculate the Antiderivative

The antiderivative of \( f(x) = x^2 \) is \( F(x) = \frac{x^3}{3} \).

Step 6: Evaluate the Integral

Substitute the limits into the antiderivative: \[ \left. \frac{x^3}{3} \right|_0^3 = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} = 9. \]

Step 7: Present Conclusion

After evaluating the integral, the value is \(9\). Hence, the original limit evaluates to 9.

Key Concepts

Riemann Sum

A Riemann Sum is a way to approximate the area under a curve by dividing it into small rectangles and summing their areas. Each rectangle's height is determined by the value of the function at a specific point within the interval, and its width is a small segment of the x-axis, denoted as \( \Delta x \). As these segments get smaller, and more rectangles are used, the Riemann Sum becomes a better approximation.
In the context of the given problem, the expression \( \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(\frac{3 i}{n}\right)^{2}\frac{3}{n} \) represents a Riemann Sum. Here, \( \left(\frac{3i}{n}\right)^{2} \) is the height of each rectangle (i.e., \( f(x) \)), and \( \frac{3}{n} \) is the width (i.e., \( \Delta x \)). As \( n \) approaches infinity, these rectangles approximate the curve between the points \( x=0 \) and \( x=3 \). This concept lays the groundwork for understanding how definite integrals are formed.

Definite Integral

The Definite Integral provides the exact area under a curve between two points, \( a \) and \( b \), on the x-axis. It is often represented as \( \int_{a}^{b} f(x) \, dx \) and can be thought of as the limit of a Riemann Sum as the number of rectangles approaches infinity.
In this exercise, identifying the limit as a Riemann Sum allows us to recognize it as a definite integral. After determining the interval of integration from \( x=0 \) to \( x=3 \) and the function \( f(x) = x^2 \), the sum is transformed into the integral \( \int_{0}^{3} x^2 \, dx \). This integral is then evaluated by using the Second Fundamental Theorem of Calculus, which connects the concept of integration with antiderivatives.

Antiderivative

An Antiderivative is a function \( F(x) \) whose derivative is the original function \( f(x) \). Finding an antiderivative is essential for determining the value of a definite integral.
In our example, we need to find an antiderivative of \( f(x) = x^2 \). The antiderivative \( F(x) \) can be found by reversing the process of differentiation, resulting in \( F(x) = \frac{x^3}{3} \). Once the antiderivative is known, the definite integral can be computed by substituting the upper and lower limits of the interval into \( F(x) \), and subtracting: \( F(3) - F(0) = \frac{27}{3} - 0 = 9 \). This shows how the second part of the Fundamental Theorem of Calculus is applied to find the exact area under the curve.

Continuity

Continuity of a function is crucial when applying the Fundamental Theorem of Calculus. A function is continuous on an interval if it is smooth and unbroken over that interval. This characteristic ensures there are no gaps or jumps in the graph of a function.
For the Fundamental Theorem of Calculus to be applicable, the function \( f(x) \) must be continuous over the interval \([a, b]\). In our example, the function \( f(x) = x^2 \) is a polynomial. Polynomials are continuous everywhere, which means \( f(x) \) is continuous over \([0, 3]\).
This continuity allows us to confidently apply the theorem, ensuring that the definite integral calculation provides an accurate and reliable result. Understanding continuity helps in identifying when and how integration techniques can be used effectively, especially in the presence of theorems like the Fundamental Theorem of Calculus.