Problem 31 - Calculus: Early Transcendentals

Question

Use a power series to approximate the definite integral to six decimal places. \( \int^{0.2}_0 x \ln (1 + x^2) dx \)

Step-by-Step Solution

Answer

The integral approximates to 0.000395.

Step 1: Understand the integral

We need to approximate the definite integral \( \int^{0.2}_0 x \ln(1 + x^2) \ dx \). We will use a power series expansion for the function \( \ln(1 + x^2) \).

Step 2: Expand \( \ln(1 + x^2) \) using a power series

Recall the power series for \( \ln(1 + u) \) which is \( \ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots \) valid for \(|u| < 1\). Substituting \( u = x^2 \), we get: \[\ln(1 + x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \cdots\]

Step 3: Substitute the power series into the integral

Replace \( \ln(1 + x^2) \) in the integral with the power series expansion:\[\int_0^{0.2} x(x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \frac{x^8}{4} + \cdots) dx\]Expand and simplify:\[\int_0^{0.2} (x^3 - \frac{x^5}{2} + \frac{x^7}{3} - \frac{x^9}{4} + \cdots) dx\]

Step 4: Integrate term by term

Integrate each term separately:\[\int x^3 \ dx = \frac{x^4}{4}, \quad \int x^5 \ dx = \frac{x^6}{6}, \quad \int x^7 \ dx = \frac{x^8}{8}, \quad \int x^9 \ dx = \frac{x^{10}}{10}\]Applying integration, we have:\[\int_0^{0.2} x^3 \ dx = \left[ \frac{x^4}{4} \right]_0^{0.2},\int_0^{0.2} \frac{x^5}{2} \ dx = \frac{1}{2} \left[ \frac{x^6}{6} \right]_0^{0.2},\]\[\int_0^{0.2} \frac{x^7}{3} \ dx = \frac{1}{3} \left[ \frac{x^8}{8} \right]_0^{0.2},\int_0^{0.2} \frac{x^9}{4} \ dx = \frac{1}{4} \left[ \frac{x^{10}}{10} \right]_0^{0.2}\]

Step 5: Evaluate each integral from 0 to 0.2

Now we evaluate each definite integral:\[\frac{(0.2)^4}{4} = \frac{0.0016}{4} = 0.0004,\frac{1}{2} \times \frac{(0.2)^6}{6} = \frac{1}{2} \times \frac{0.000064}{6} = \frac{0.000032}{6} \approx 0.00000533,\]\[\frac{1}{3} \times \frac{(0.2)^8}{8} = \frac{1}{3} \times \frac{0.00000256}{8} = \frac{0.000000853}{3} \approx 0.000000284,\frac{1}{4} \times \frac{(0.2)^{10}}{10} = \frac{1}{4} \times \frac{0.0000001024}{10} = \frac{0.0000000256}{4} = 0.0000000064\]

Step 6: Sum the evaluated integrals

Add the results from Step 5 to approximate the integral:\[0.0004 - 0.00000533 + 0.000000284 - 0.0000000064 \approx 0.00039495\]

Step 7: Conclusion

The 6-decimal place approximation of the integral \( \int^{0.2}_0 x \ln (1 + x^2) dx \) using the power series method is approximately \( 0.000395 \).

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that helps us find the accumulated quantity under a curve, between two specific points. This technique distinctly measures the total area under a curve from one point to another. In our problem, we evaluated the definite integral from 0 to 0.2, denoted as \( \int_{0}^{0.2} f(x) \ dx \), where \( f(x) = x \ln(1 + x^2) \).

• The integral effectively provides a numerical method to sum up infinitesimally small quantities, especially when an analytical solution is complex.
• In our exercise, we're approximating this integral using power series due to the complexity of directly integrating the given function.

The primary goal with such integrals, especially in the context provided, is to approximate the integral's value, which can be particularly useful in physics and engineering, where exact solutions might be difficult to derive.

Power Series Expansion

A power series expansion is a way to represent functions as an infinite sum of terms. Each term in the series is derived from the function's derivatives at a specific point. For the exercise, we specifically look at the power series for \( \ln(1 + x^2) \).

• This series is an example of how complex functions can often be broken down into simpler polynomial terms for approximation.
• The power series for \( \ln(1 + u) \) is \( u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots \).

In simpler terms, using a power series enables us to treat complex logarithmic expressions as easier to handle polynomials, which are much simpler to integrate.

Approximating Integrals

Approximating integrals involves estimating the value of an integral using various methods. One common method is to use a power series expansion, as seen in the exercise.

• Through power series, we rewrite complex integrals in terms of polynomials, which are easier to manage mathematically.
• This approach is particularly useful when the function is difficult or impossible to integrate analytically.

By integrating the series term by term within our integral bounds, we achieve a close numerical approximation. It's crucial to consider how many terms you include in the series, as more terms lend greater accuracy, but with diminishing returns.

Calculus

Calculus encompasses the mathematical study of continuous change and is divided into two main branches: differential and integral calculus. Our focus here is on using integral calculus to solve complex real-life problems through approximation techniques.

• Integral calculus helps calculate areas and volumes, and is essential in fields like physics and engineering.
• The exercise demonstrates how integration by parts and series expansions, both key calculus techniques, can solve what otherwise would be cumbersome integrals.

Understanding calculus and its concepts opens the door to solving a wide range of practical problems, from understanding acceleration in physics to calculating irregular structures’ volumes and much more.