Problem 64 - Calculus Single Variable

Question

Integrate by parts. This will result in an integrand of the form \(P(x) / Q(x)\) where \(P(x)\) and \(Q(x)\) are polynomials with the degree of \(P(x)\) greater than or equal to the degree of \(Q(x)\). Such an integrand is handled by performing polynomial division to put \(P(x) / Q(x)\) into the form \(r(x)+s(x) / Q(x)\) where \(r(x)\) and \(s(x)\) are polynomials with the degree of \(s(x)\) less than the degree of \(Q(x)\) $$ \int_{0}^{1} x \arctan (x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately 0.285.

Step 1: Choose Functions for Integration by Parts

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int x \arctan(x) \, dx \), let's choose \( u = \arctan(x) \) and \( dv = x \, dx \). Then, calculate \( du = \frac{1}{1+x^2} \, dx \) and \( v = \frac{x^2}{2} \).

Step 2: Apply Integration by Parts Formula

Substitute the values into the integration by parts formula: \[\int x \arctan(x) \, dx = \arctan(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx\]. The first term simplifies to \(\frac{x^2}{2} \arctan(x)\).

Step 3: Simplify Second Integral

Simplify the second integral \( \int \frac{x^2}{2(1+x^2)} \, dx \). This can be rewritten as \( \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) \, dx \).

Step 4: Integrate and Simplify Further

The integral now separates into \( \frac{1}{2} \left( \int 1 \, dx - \int \frac{1}{1+x^2} \, dx \right) \). These integrals are straightforward: \( \int 1 \, dx = x \) and \( \int \frac{1}{1+x^2} \, dx = \arctan(x) \). Thus, \( \int \frac{x^2}{2(1+x^2)} \, dx = \frac{x}{2} - \frac{\arctan(x)}{2} \).

Step 5: Combine Results and Evaluate Definite Integral

Reassemble the integral expression using parts and evaluate the definite integral from 0 to 1. The integral result inside the integral becomes: \[\frac{x^2}{2} \arctan(x) - \left( \frac{x}{2} - \frac{\arctan(x)}{2} \right)\].Simplify to: \[\frac{x^2}{2} \arctan(x) + \frac{\arctan(x)}{2} - \frac{x}{2}\].Now evaluate from 0 to 1: Substituting 1: \(\left(\frac{1}{2}\right)\cdot\frac{\pi}{4} + \frac{\pi}{8} - \frac{1}{2} \approx 0.285\).Substituting 0: yields 0.Thus, the definite integral evaluates to \( 0.285 \).

Key Concepts

Polynomial Division

Polynomial division is a helpful technique when you're dealing with the integration of rational functions where the numerator's degree is equal to or greater than the denominator's. In these cases, polynomial division helps simplify the fraction into a more manageable form. The process is somewhat similar to long division in arithmetic.

Here's what you generally do:

• Divide the highest degree term of the numerator by the highest degree term of the denominator.
• Multiply the entire denominator by this term and subtract the result from the original polynomial.
• Repeat these steps with the resulting polynomial until the degree of the remainder is less than the degree of the denominator.

The outcome will be an expression called "improper fraction," typically in the form of some polynomial plus a proper fraction. This makes further integration steps, whether using partial fractions or other techniques, much easier to manage.

Definite Integrals

In calculus, a definite integral represents the area under a curve between two specified points, usually on the x-axis. The result is a specific numerical value, unlike the indefinite integral, which includes an arbitrary constant of integration.

For the integral \[ \int_{0}^{1} x \arctan(x) \, dx \],we are looking at the area from x = 0 to x = 1 under the curve defined by the function. The definite integral can be evaluated using several methods like substitution, integration by parts, and sometimes numerical approximation techniques, depending on the complexity of the function.
In our specific problem, using integration by parts was advantageous. After obtaining the indefinite integral, we substitute the upper and lower limits to find the numerical value. This provides us with insight into the total "accumulated change" represented by the integral over that specific interval.

Trigonometric Functions

Trigonometric functions like sine, cosine, and arctan often appear in integration problems, mainly because they frequently describe phenomena in physics, engineering, and other fields. The arcsine, arccosine, and arctangent functions are the inverse operations of the basic trigonometric functions.

The function \( \arctan(x) \) is crucial in our integral. Its derivative is \( \frac{1}{1+x^2} \), a fact that is incredibly useful when applying integration by parts. The arctan function maps a real number domain into an interval of the angle, providing a bridge between algebraic expressions and geometrical interpretations.
Understanding how to manipulate these functions helps not just in solving integrals but also in understanding the underlying geometry and cycles, especially in periodic phenomena.

Polynomial Functions

Polynomial functions are expressions that consist of variables raised to powers, combined using addition, subtraction, and multiplication. These functions come in various degrees, depending on the highest power present in the expression.

In the context of calculus, the integration of polynomial functions is generally straightforward. Each term is integrated separately, raising the power by one and dividing by the new power. This rule is simple yet powerful: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \ n eq -1 \].

The integral we solved includes dealing with a polynomial function, which was integrated after simplifying using trigonometric identities and substitution. Understanding the behavior of these polynomial components in a function can greatly simplify the integration process, especially when dealing with complex expressions composed of polyterms.