Problem 76 - 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. $$ \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[1+\frac{2 i}{n}+\left(\frac{2 i}{n}\right)^{2}\right] \frac{2}{n} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{20}{3} \).

Step 1: Recognize the Limit as a Riemann Sum

The given limit is of the form \( \lim_{n \to \infty} \sum_{i=1}^{n} f\left(x_i^*\right) \Delta x \). In this case, \( \Delta x = \frac{2}{n} \) and \( x_i^* = \frac{2i}{n} \). The expression inside the sum is \( 1 + x_i^* + (x_i^*)^2 \). Therefore, the function is \( f(x) = 1 + x + x^2 \). The sum is a Riemann sum for the integral of \( f(x) \) over the interval \([0, 2]\).

Step 2: Write the Definite Integral

From the Riemann sum, we identify the definite integral as \( \int_{0}^{2} (1 + x + x^2)\, dx \). This represents the area under the curve \( y = 1 + x + x^2 \) between \( x = 0 \) and \( x = 2 \).

Step 3: Apply the Second Fundamental Theorem of Calculus

To evaluate the integral \( \int_{0}^{2} (1 + x + x^2)\, dx \), we first find an antiderivative of \( f(x) = 1 + x + x^2 \). The antiderivative is \( F(x) = x + \frac{x^2}{2} + \frac{x^3}{3} \).

Step 4: Evaluate the Antiderivative at the Bounds

Using the Second Fundamental Theorem of Calculus, we evaluate the antiderivative at the upper and lower bounds: \[ F(2) - F(0) = \left( 2 + \frac{2^2}{2} + \frac{2^3}{3} \right) - \left( 0 + \frac{0^2}{2} + \frac{0^3}{3} \right) = 2 + 2 + \frac{8}{3}. \]

Step 5: Simplify the Expression

Simplify the expression \( 2 + 2 + \frac{8}{3} \) to get \( 4 + \frac{8}{3} = \frac{12}{3} + \frac{8}{3} = \frac{20}{3} \). Thus, the value of the integral is \( \frac{20}{3} \).

Key Concepts

Riemann Sum

A Riemann Sum is a method for estimating the total area underneath a curve on a graph, otherwise known as the integral of a function over an interval. In our problem, we are given a limit form that is a Riemann sum. It is a sum of the form \( \lim_{n \to \infty} \sum_{i=1}^{n} f\left(x_i^*\right) \Delta x \). Here, \( \Delta x = \frac{2}{n} \) represents the width of each sub-interval, and \( x_i^* = \frac{2i}{n} \) is a point within each sub-interval used to evaluate the function.

• In the exercise, the expression inside the sum is \( 1 + x_i^* + (x_i^*)^2 \).
• This denotes the function \( f(x) = 1 + x + x^2 \) over the interval \([0, 2]\).

By recognizing this form, we understand the sum as an approximation of the integral of the given function over the specified range. As \( n \) approaches infinity, the approximation becomes exact, converting the sum into a definite integral.

Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus is a cornerstone in the world of integrals. It bridges the concept of differentiation with that of integration. It tells us that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then:\[\int_{a}^{b} f(x)\,dx = F(b) - F(a)\]

• This theorem allows us to compute definite integrals, transforming the problem of finding areas under curves into a simple subtraction of anti-derivates at boundaries.
• In our exercise, we use it to evaluate the integral \( \int_{0}^{2} (1 + x + x^2) \, dx \).

Once we find \( F(x) = x + \frac{x^2}{2} + \frac{x^3}{3} \), the second fundamental theorem directs us to evaluate \( F(2) - F(0) \) to obtain the value of the definite integral.

Antiderivative

Understanding antiderivatives is essential to solving integrals. An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that the derivative of \( F \) is \( f \). In simpler terms, it is the opposite of differentiation.

• In the solution, we find the antiderivative of \( f(x) = 1 + x + x^2 \), which is \( F(x) = x + \frac{x^2}{2} + \frac{x^3}{3} \).
• Each term is integrated separately: \(1\), \(x\), and \(x^2\).

The key step here involves increasing the exponent of each term by one and dividing by the new exponent. Mastering this process is crucial for carrying out definite integrals, as it allows us to apply the Second Fundamental Theorem of Calculus effectively.

Limit Evaluation

Limit evaluation is a pivotal concept in calculus, particularly in the transition from a Riemann sum to a definite integral. Limits allow us to find values that functions approach as the input gets closer to a particular point or infinity.

• In the case of the given problem, as \( n \to \infty \), the number of sub-intervals increases, reducing \( \Delta x \) to an infinitesimally small size, thus forming a continuous integration.
• This move from the discrete sum (Riemann sum) to the continuous integral requires evaluating the limits precisely.

This final step ensures that the approximation of areas becomes exact, resulting in an efficient and reliable means of calculating the exact area under a curve without ever needing to estimate sums directly.