Problem 14 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x^{2} \ln x^{2} d x $$

Step-by-Step Solution

Answer

\( \frac{x^3}{3} \ln x^2 - \frac{2x^3}{9} + C \).

Step 1: Identify Parts for Integration by Parts

To apply integration by parts, we need to identify the parts of the integral according to the formula \( \int u \, dv = uv - \int v \, du \). Choose \( u = \ln x^2 \) because the derivative of a logarithm simplifies, and \( dv = x^2 \, dx \) because we can easily integrate a power of \( x \).

Step 2: Differentiate and Integrate Individual Parts

Differentiate \( u = \ln x^2 \) with respect to \( x \), which yields \( du = \frac{2}{x} \, dx \). Integrate \( dv = x^2 \, dx \) to get \( v = \frac{x^3}{3} \).

Step 3: Substitute into Integration by Parts Formula

Substituting \( u \), \( du \), \( v \) into the integration by parts formula gives: \[ \int x^2 \ln x^2 \, dx = \frac{x^3}{3} \ln x^2 - \int \left( \frac{x^3}{3} \right) \left( \frac{2}{x} \right) dx \].

Step 4: Simplify and Solve the Remaining Integral

Simplify the remaining integral: \( \int \frac{2x^2}{3} \, dx \). Integrate to get \( \frac{2}{3} \cdot \frac{x^3}{3} = \frac{2x^3}{9} \).

Step 5: Write Final Expression

Substitute and simplify: \[ \frac{x^3}{3} \ln x^2 - \frac{2x^3}{9} + C \]. Combine like terms if possible. No further combination is possible, so the final answer is \( \frac{x^3}{3} \ln x^2 - \frac{2x^3}{9} + C \).

Key Concepts

Definite and Indefinite Integrals

In calculus, integrals are used to find various quantities, such as areas, volumes, and central points. There are two primary types of integrals: definite and indefinite. Indefinite integrals, often referred to as antiderivatives, do not have specified upper and lower limits. They include a constant of integration, often represented by "C," which accounts for the fact that there are infinitely many antiderivatives differing only by a constant. For example, \[\int x^2 \, dx = \frac{x^3}{3} + C\].

• This expression represents a family of functions, each shifted vertically by different amounts.
• The constant "C" is crucial in expressing the most general form of an antiderivative.

Definite integrals, on the other hand, are evaluated over a specific interval [a, b], and they yield a numerical value representing the area under the curve of a function between the limits. They do not include a constant of integration. An example is \[\int_{1}^{3} x^2 \, dx = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}\].

• The notation \[\int_{a}^{b} f(x) \, dx\] inherently includes the evaluation limits.
• This value represents a definite quantity, such as area, without any arbitrary constants.

In problems where integration by parts is needed, understanding these concepts is crucial to correctly applying the method and deriving accurate results.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are products or quotients of other functions, especially when they involve logarithms. It's particularly useful when dealing with functions in the form \(y = u(x)^{v(x)}\), where both exponents involve variables. Instead of directly differentiating, we take the natural logarithm of both sides to simplify the differentiation process.
For example, given a function \(y = x^x\), we can apply it as follows:

• Take the natural log of both sides: \(\ln y = \ln(x^x) = x \ln x\).
• Differentiate implicitly with respect to \(x\): \(\frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x)\).
• Apply the derivative: \(\frac{1}{y} \cdot \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} ) = \ln x + 1\).
• Solve for \(dy/dx\): \(dy/dx = y(\ln x + 1)\).
• Substitute back \(y = x^x\): \(dy/dx = x^x (\ln x + 1)\).

This technique was used partially in the solution process to manage the \(\ln x^2\) term when identifying the parts for integration by parts, as it simplifies the differentiation significantly by avoiding cumbersome product or quotient rules.
Understanding logarithmic differentiation helps in handling complex differentiations efficiently.

Polynomial Functions

Polynomial functions are expressions involving variables raised to whole number exponents and coefficients. They have extensive applications in calculus, as they are continuous and differentiable over real numbers. A typical polynomial looks like \(f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\), where \(a_n, a_{n-1}, ...\) are coefficients.

• Polynomials are fundamental in calculus because they are simple to integrate and differentiate.
• The powers of \(x\) decrease by one each time you differentiate, allowing repeated application of simple power rule integration or differentiation.

In integration by parts, as used in our exercise, polynomials play a pivotal role because:

• They are easy to integrate or differentiate, making them handy workhorses in integration techniques.
• The integral of a monomial \(x^n\) yields: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), straightforward compared to more complex functions.
• In our exercise, selecting \(dv = x^2 \, dx\) was strategic because it's a straightforward power of \(x\), simplifying our integration process.

Thus, understanding and working with polynomial functions are essential skills in calculus; they simplify many processes and allow us to handle more complex expressions, like logarithms and trigonometric functions.