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 \( 0.385 \) approximates the integral \( 0.333... \), with accuracy improving as \( n \to \infty \).

Step 1: Understanding the Integral

The integral \( \int_{0}^{1} x^{2} \, dx \) represents the area under the curve \( y = x^2 \) from \( x = 0 \) to \( x = 1 \). This integral finds the net area, which can be evaluated using the Second Fundamental Theorem of Calculus.

Step 2: Evaluating the Integral

To evaluate \( \int_{0}^{1} x^{2} \, dx \), find the antiderivative of \( x^2 \), which is \( \frac{x^3}{3} \). Then, apply the limits of integration:\[\int_{0}^{1} x^{2} \, dx = \left. \frac{x^3}{3} \right|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}.\]

Step 3: Understanding the Summation

The summation \( \frac{1}{n^3} \sum_{i=1}^{n} i^{2} \) is an approximation to the integral by using the method of Riemann sums. As \( n \to \infty \), the width \( \frac{1}{n} \) of each rectangle becomes smaller, more closely approximating the integral.

Step 4: Calculating the Summation for n = 10

Calculate \( \frac{1}{10^3} \sum_{i=1}^{10} i^{2} \). The formula for the sum of squares is \( \sum_{i=1}^{n} i^{2} = \frac{n(n + 1)(2n + 1)}{6} \). For \( n = 10 \):\[\sum_{i=1}^{10} i^{2} = \frac{10 \times 11 \times 21}{6} = 385.\]Then,\[\frac{1}{10^3} \cdot 385 = \frac{385}{1000} = 0.385.\]

Step 5: Comparing Values

The calculated integral is \( \frac{1}{3} \approx 0.333... \) and the summation for \( n = 10 \) is \( 0.385 \). The approximation becomes more accurate as \( n \) increases, because the rectangles used in the Riemann sum become narrower, better approximating the curve \( y = x^2 \).

Key Concepts

Riemann Sums

Riemann sums are a foundational concept in calculus used to approximate the area under a curve. Imagine you want to find the area under the curve of a function, such as \( y = x^2 \) from \( x = 0 \) to \( x = 1 \). We do this by breaking the area into rectangles, calculating their area, and summing them up.
By dividing the interval \( [0, 1] \) into \( n \) equal parts, we construct rectangles whose heights are determined by the function values. For large \( n \), these rectangles become thinner and fill in more of the area under the curve.

• The width of each rectangle is \( \frac{1}{n} \).
• The height is given by the function value at a particular point in each subinterval.

In the given expression \( \frac{1}{n^3} \sum_{i=1}^{n} i^2 \), the summation approximates the integral. As \( n \to \infty \), the approximation's accuracy improves, giving insights into how Riemann sums work as a method for integration.

Definite Integrals

Definite integrals calculate the exact area under a curve between two specified limits. For example, \( \int_{0}^{1} x^2 \, dx \) computes the area under the curve \( y = x^2 \) from \( x = 0 \) to \( x = 1 \). This process is more precise than Riemann sums because integration considers infinitely small slices between the limits.

• The integral's value is \( \frac{1}{3} \), as shown in the exercise, providing an exact area.
• The limits of integration in this example are 0 and 1, defining the region under consideration.

Thus, definite integrals turn the approximation process into a precise method of calculation. They are vital in getting exact functional areas, helpful in various application fields like physics and engineering.

Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus is a powerful tool in finding definite integrals. It states that if you have a continuous function, the integral of that function over an interval can be found using its antiderivative.
For the function \( x^2 \), the antiderivative is \( \frac{x^3}{3} \). Using the theorem, we can evaluate \( \int_{0}^{1} x^2 \, dx \) by:

• Finding the antiderivative \( \frac{x^3}{3} \).
• Evaluating it at the upper limit \( x = 1 \) and the lower limit \( x = 0 \).
• Subtracting the lower limit result from the upper limit result.

The calculated area using this theorem is \( \frac{1}{3} \), confirming the precision of definite integrals. This theorem connects differentiation and integration beautifully and simplifies the process of finding areas and solving complex calculus problems.