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

Step-by-Step Solution

Answer

The value of the limit and definite integral is \( \frac{26}{3} \).

Step 1: Recognize the Limit as a Riemann Sum

The given limit \( \lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[1+\frac{2i}{n}+\left(\frac{2i}{n}\right)^{2}\right] \frac{2}{n} \) resembles the Riemann sum for a definite integral. Here, \( \Delta x = \frac{2}{n} \) and the function \( f(x) = 1 + 2x + x^2 \). The interval of integration is from \( a = 0 \) to \( b = 2 \).

Step 2: Set Up the Definite Integral

Convert the Riemann sum to the definite integral: the integral from 0 to 2 of \( f(x) = 1 + 2x + x^2 \). Thus the integral is \( \int_{0}^{2}(1 + 2x + x^2) \; dx \).

Step 3: Apply the Second Fundamental Theorem of Calculus

To find the integral, apply the Second Fundamental Theorem of Calculus. First, find the antiderivative \( F(x) \) of \( f(x) = 1 + 2x + x^2 \). The antiderivative is \( F(x) = x + x^2 + \frac{x^3}{3} \).

Step 4: Evaluate the Antiderivative at the Limits

Evaluate \( F(x) \) at the upper limit, \( x = 2 \), and then at the lower limit, \( x = 0 \). Compute \( F(2) = 2 + 2^2 + \frac{2^3}{3} = 2 + 4 + \frac{8}{3} = 6 + \frac{8}{3} = \frac{18}{3} + \frac{8}{3} = \frac{26}{3} \). Next, compute \( F(0) = 0 + 0^2 + \frac{0^3}{3} = 0 \).

Step 5: Apply the Fundamental Theorem

Use the Fundamental Theorem of Calculus to find the definite integral: \( F(b) - F(a) = F(2) - F(0) = \frac{26}{3} - 0 = \frac{26}{3} \).

Key Concepts

Riemann Sum

The concept of a Riemann Sum is foundational in understanding definite integrals. In calculus, a Riemann Sum approximates the area under a curve by dividing the area into thin rectangles and summing their areas. This process transforms a known curve into a sum of discrete values that can be evaluated as a limit.

Here's how it works:

• Consider a function, say \( f(x) \), defined on an interval \([a, b]\).
• To form a Riemann Sum, divide the interval into \( n \) subintervals of equal width \( \Delta x \).
• At each subinterval \( [x_i, x_{i+1}] \), pick a sample point, \( x_i^* \), and calculate \( f(x_i^*) \Delta x \).
• Sum all these to form the Riemann Sum: \( \sum_{i=1}^{n} f(x_i^*) \Delta x \).
• The limit of this sum as \( n \rightarrow \infty \) gives the definite integral \( \int_{a}^{b} f(x) \, dx \).

In the original exercise, the Riemann Sum represents the function \( 1 + 2x + x^2 \) over the interval \( [0, 2] \), showing how mathematicians can turn a continuous function into manageable parts.

Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus is a key element in evaluating integrals efficiently. It states that if \( F(x) \) is an antiderivative of \( f(x) \) over an interval, then the definite integral of \( f(x) \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).

Here’s why this is crucial:

• The theorem provides a direct way to compute definite integrals without the need for calculating limits like in Riemann Sums.
• Once the antiderivative \( F(x) \) of a function \( f(x) \) is found, determining the integral over an interval simply involves subtraction of boundary values.
• It bridges the gap between the derivative and the integral, connecting two major calculus concepts.

In the given problem, this theorem allowed the rapid evaluation of the integral by using the antiderivative \( F(x) = x + x^2 + \frac{x^3}{3} \) and calculating \( F(2) - F(0) \). The antiderivative evaluation at the bounds simplifies finding the exact area under the function from \( 0 \) to \( 2 \).

Antiderivative

An antiderivative is essentially the reverse of taking a derivative. If you have a function \( f(x) \), an antiderivative \( F(x) \) is a function whose derivative is \( f(x) \). In other words, \( \frac{d}{dx}F(x) = f(x) \).

Understanding antiderivatives is crucial because:

• They allow us to find the integral of a function over an interval.
• Antiderivatives represent the accumulation of the quantities that the original function describes.
• This concept helps determine areas under curves, solve differential equations, and much more.

In the context of the exercise, the function \( f(x) = 1 + 2x + x^2 \) had an antiderivative \( F(x) = x + x^2 + \frac{x^3}{3} \). Finding this was pivotal in using the Second Fundamental Theorem of Calculus to evaluate the integral from \( 0 \) to \( 2 \). It implies you can move from the rate of change back to the total amount.