Problem 31 - Essential Calculus Early Transcendentals

Question

Use a power series to approximate the definite integral to six decimal places. $$ \int_{0}^{0.1} x \arctan (3 x) d x $$

Step-by-Step Solution

Answer

Approximately, the integral is 0.000982 to six decimal places.

Step 1: Understand the Function to Approximate

Given the function to integrate is \( x \arctan(3x) \), the first step is to note that this is a product of \( x \) and \( \arctan(3x) \). We want to express \( \arctan(3x) \) as a power series expansion so we can then multiply by \( x \).

Step 2: Expand \( \arctan(3x) \) as a Power Series

The Taylor series for \( \arctan(u) \) about 0 is \( u - \frac{u^3}{3} + \frac{u^5}{5} - \cdots \). So for \( \arctan(3x) \), substitute \( u = 3x \):\[ \arctan(3x) = 3x - \frac{(3x)^3}{3} + \frac{(3x)^5}{5} - \cdots = 3x - \frac{27x^3}{3} + \frac{243x^5}{5} - \cdots \]This series becomes:\[ 3x - 9x^3 + \frac{243x^5}{5} - \cdots \]

Step 3: Multiply the Series by \( x \)

To obtain the integrand, multiply each term in the series by \( x \):\[ x \arctan(3x) = x(3x) - x(9x^3) + x\left(\frac{243x^5}{5}\right) - \cdots \]This simplifies to:\[ 3x^2 - 9x^4 + \frac{243x^6}{5} - \cdots \]

Step 4: Integrate Term by Term

Integrate the series term by term within the limits 0 to 0.1:- \( \int 3x^2 \, dx = x^3 = \left[ x^3 \right]_0^{0.1} = (0.1)^3 = 0.001 \)- \( \int 9x^4 \, dx = \frac{9x^5}{5} = \left[ \frac{9x^5}{5} \right]_0^{0.1} = \frac{9 \times (0.1)^5}{5} = 0.000018 \)- \( \int \frac{243x^6}{5} \, dx = \frac{243x^7}{35} = \left[ \frac{243x^7}{35} \right]_0^{0.1} = \frac{243 \times (0.1)^7}{35} = 0.000000243 \)

Step 5: Sum the Integrals to Approximate

Sum the results obtained from the integrals of each term:\[ 0.001 - 0.000018 + 0.000000243 \approx 0.000982243 \]This is the approximate value of the integral using the first few terms of the power series.

Key Concepts

Power Series Approximation

A power series is a way to express functions as an infinite sum of terms based on powers of a variable. It's a crucial technique in calculus to approximate functions that are otherwise difficult to compute directly.
To approximate the definite integral of a function involving trigonometric functions, powers, or other complex forms, you'll often convert the function into a power series.

• The power series allows for integration term by term, making calculations simpler and more manageable.
• This method works especially well for functions that are continuous and differentiable.
• Though infinite, we can truncate the series to a few leading terms for a decent approximation.

Each term of the power series contributes to the overall function approximation, helping us understand how each power affects the shape of the function. For example, in this exercise, we expressed \( \arctan(3x) \) using a power series: \( 3x - 9x^3 + \frac{243x^5}{5} - \cdots \). This series lets us evaluate the integral via a polynomial of several terms rather than one complex expression.

Taylor Series in Calculus

The Taylor series is a specific type of power series that approximates a function around a specific point, usually 0. It's represented mathematically as \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \] This series can represent various functions, including \( \arctan(3x) \), which we expressed around 0.

• The choice of point \( a \) and the number of terms included significantly impact accuracy.
• For small \( x \), fewer terms are generally required for an accurate approximation.
• Taylor series expand any function provided it is smooth (differentiable) enough.

The series breaks down complex behaviors into an approachable form by using the function's derivatives at a single point. This makes essential features of the function, like curvature or oscillation, clear and mathematically tractable. In terms of \( \arctan(3x) \), the Taylor series expansion becomes \( 3x - \frac{(3x)^3}{3} + \frac{(3x)^5}{5} \), making it easier to handle within calculus problems like integration.

Definite Integral Approximation

Definite integrals calculate the area under a curve represented by a function over a closed interval. When the function is complex, we can use power series to make the integration more straightforward.
Using a power series, we replace the complex function with a simpler polynomial approximation, integrate each term separately, and sum up the results to get an overall approximate value.

• Each term \( a_nx^n \) in the series is integrated individually.
• The integration limits guide us in substituting the values post-integration.
• Accuracy depends on the number of terms used in the series expansion.

For the integral \( \int_{0}^{0.1} x \arctan(3x) \, dx \), we used the power series to transform and integrate term by term:

• \( \int 3x^2 \, dx = 0.001 \)
• \( \int 9x^4 \, dx = 0.000018 \)
• \( \int \frac{243x^6}{5} \, dx = 0.000000243 \)

Summing these results gives us the approximate definite integral value \( \approx 0.000982243 \), reflecting our approach's blend of approximation and exact calculation along small intervals.