Problem 77 - Calculus Early Transcendentals: Pearson New International Edition

Question

Explain why \(\left(1 / n^{3}\right) \sum_{i=1}^{n} i^{2}\) should be a good approximation to \(\int_{0}^{1} x^{2} d x\) for large \(n\). Now calculate the summation expression for \(n=10\), and evaluate the integral by the Second Fundamental Theorem of Calculus. Compare their values.

Step-by-Step Solution

Answer

For large \( n \), the summation approximates \( \frac{1}{3} \), consistent with the integral \( \int_{0}^{1} x^{2} dx = \frac{1}{3} \). For \( n=10 \), the summation value is 0.385.

Step 1: Analyzing the Summation Expression

The summation expression \( \left(1 / n^{3}\right) \sum_{i=1}^{n} i^{2} \) is equivalent to \( \frac{1}{n^3} \) multiplied by the sum of the squares of the first \( n \) natural numbers. The formula for this sum is \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \). This means the expression becomes \( \left(1 / n^{3}\right) \times \frac{n(n+1)(2n+1)}{6} \).

Step 2: Simplifying the Expression for Large n

For large \( n \), the expression \( \frac{n(n+1)(2n+1)}{6n^3} \) can be simplified. Using the properties of polynomials, this simplifies to approximately \( \frac{2}{6} = \frac{1}{3} \), as \( n \to \infty \), higher order terms dominate in the expression.

Step 3: Evaluating the Integral \( \int_{0}^{1} x^{2} dx \)

To evaluate the integral \( \int_{0}^{1} x^{2} dx \), we apply the Second Fundamental Theorem of Calculus. The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). Evaluating from 0 to 1 gives \( \left. \frac{x^3}{3} \right|_0^1 = \frac{1}{3} - 0 = \frac{1}{3} \).

Step 4: Calculating the Sum Expression for n=10

Substituting \( n = 10 \) into the sum expression, we get \( \frac{1}{10^3} \times \frac{10 \times 11 \times 21}{6} = \frac{1}{1000} \times 385 = 0.385 \).

Step 5: Comparing the Sum and Integral Values

For \( n = 10 \), the sum expression equals 0.385 while the actual integral is \( \frac{1}{3} \approx 0.333 \). For large \( n \), the sum converges to the value of the integral.

Key Concepts

Riemann Sum

In calculus, the Riemann Sum provides an approximate calculation of the area under a curve. It is used to estimate the value of a definite integral.
Meet the Riemann Sum, a tool that breaks down the area under a curve into small rectangles. These rectangles are then summed up to get an approximation of the actual area.
Here's how it typically works:

• The interval of integration is divided into smaller sub-intervals of equal width.
• For each sub-interval, a rectangle is drawn.
• The height of each rectangle is determined by the function's value at a specified point in the sub-interval.

The basic formula for a Riemann Sum can be written as:

• \[\sum_{i=1}^{n} f(x_i) \Delta x\]

Where \( f(x_i) \) is the function value at the chosen point and \( \Delta x \) represents the width of the sub-intervals. As the number of sub-intervals \( n \) becomes very large, the width \( \Delta x \) tends to zero. The sum then approaches the exact value of the definite integral. This demonstrates how a Riemann Sum is a stepping stone to the precise calculation of integrals.

Definite Integral

A definite integral calculates the actual area under a curve between two bounds. The concept is critical in understanding continuous functions. To find a definite integral, you usually need to determine the antiderivative of the function in question.
The calculation of a definite integral from point \( a \) to point \( b \) is expressed as:

• \[\int_{a}^{b} f(x) \, dx\]

The Second Fundamental Theorem of Calculus makes this relatively straightforward. This theorem states that if \( F \) is the antiderivative of a function \( f \), then the integral from \( a \) to \( b \) is given by:

• \[F(b) - F(a)\]

This calculation not only provides the exact area under the curve from \( a \) to \( b \) but also crucial numerical values for mathematical models and applications. In our exercise, the definite integral \( \int_{0}^{1} x^2 \, dx \) equals \( \frac{1}{3} \), which shows the precise area under the curve \( x^2 \) from 0 to 1.

Polynomial Approximation

Polynomials often serve as simple approximations for complex functions, making calculations manageable. In the context of Riemann Sums, polynomial approximations help predict definite integrals when dealing with large sums.
The value calculation exercise involving \( \left(1 / n^{3}\right) \sum_{i=1}^{n} i^{2} \) serves as such an approximation. Here, the approximation arises because the sum of a series of polynomial terms simplifies into a recognizable, integrable form:
The formula \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \) approximates the area under the curve \(x^2\).
We simplify further to approximate the actual integral's value more easily as \( n \) becomes large. This method of simplification is particularly useful where a closed-form exact integration is challenging or impossible. Thus, polynomial approximation bridges the gap between numerical simulation and exact integration.
By comparing polynomial approximations with definite integrals, you can understand the method's power and precision, observing how sums converge to integrals as \( n \) tends towards infinity.