Problem 20 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int_{1}^{e} \ln x^{2} d x $$

Step-by-Step Solution

Answer

The integral is 2.

Step 1: Identify parts for integration by parts

Recall that integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int \ln x^2 \ dx \), let \( u = \ln x^2 \) and \( dv = dx \). Then we need to find \( du \) and \( v \).

Step 2: Differentiate u and integrate dv

Differentiate \( u = \ln x^2 \) to find \( du \). Use the chain rule: \( \ln x^2 = 2\ln x \), so \( du = \frac{2}{x} \, dx \). Integrate \( dv = dx \) to find \( v \): \( v = x \).

Step 3: Apply the integration by parts formula

Substitute \( u = \ln x^2 \), \( du = \frac{2}{x} \, dx \), \( v = x \), and \( dv = dx \) into the integration by parts formula: \[ \int \ln x^2 \, dx = x \ln x^2 - \int x \cdot \frac{2}{x} \, dx. \] Simplify the expression: \[ x \ln x^2 - 2 \int dx = x \ln x^2 - 2x + C. \]

Step 4: Evaluate the definite integral from 1 to e

Now, evaluate \( x \ln x^2 - 2x \) from \( x = 1 \) to \( x = e \). Substitute and calculate: \[ \left[ x \ln x^2 - 2x \right]_1^e = \left[ e \ln e^2 - 2e \right] - \left[ 1 \ln 1^2 - 2 \cdot 1 \right]. \]

Step 5: Simplify the definite integral results

Calculate each part of the expression: For \( x = e \), \( e \ln e^2 - 2e = 2e - 2e = 0 \). For \( x = 1 \), \( 1 \ln 1^2 - 2 = -2 \). Therefore, \( 0 - (-2) = 2 \).

Key Concepts

Definite Integral

The definite integral of a function provides the net area under the curve between two specified bounds. In the case of evaluating \( \int_{1}^{e} \ln x^{2} \, dx \), you're tasked with finding this net area from \(x = 1\) to \(x = e\).
Definite integrals take into account two things:

• The antiderivative of the function.
• The values of this antiderivative at the upper and lower limits of integration.

In the completed example, once the antiderivative was found using integration by parts, the definite integral was evaluated by applying:\[ \left[ F(x) \right]_a^b = F(b) - F(a) \]where \( F(x) \) is the antiderivative.
Calculating definite integrals is essential in many applications, such as finding areas and solving problems related to physics and engineering.

Integration Techniques

Integration techniques are strategies to help solve complex integrals. One fundamental technique used in the exercise is **Integration by Parts**. This method is derived from the product rule of differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du \]When you have an integral that involves a product of functions, particularly logarithms or inverse trigonometric functions paired with polynomials, integration by parts is often effective.
In the exercise, the application involved:

• Identifying \( u = \ln x^2 \) and \( dv = dx \).
• Computing \( du = \frac{2}{x} \, dx \) by using the chain rule for differentiation.
• Finding \( v = x \) by integrating \( dv \) with respect to \( x \).

This technique leverages differentiation and integration characters to transform and simplify the integral into a solvable form. As a result, one can solve integrals that initially seem intractable.

Calculus Integration

Calculus integration involves various methods to find the integral of functions, which is the reverse process of differentiation. In our case, we're using integration by parts, but many techniques exist:

• **Substitution:** Useful when an integral contains a function and its derivative.
• **Partial Fractions:** Suited for rational functions where the degree of the numerator is less than the degree of the denominator.
• **Trigonometric Integrals:** Applied to integrals involving trigonometric functions.
• **Numerical Integration:** Approximates integrals when an analytical solution is challenging to derive.

Each integration technique exploits different characteristics of functions to simplify integrals into a form that we can easily evaluate. Mastery of these techniques empowers solving a wide range of problems in calculus. Furthermore, being proficient in integration forms the foundation of much of calculus, vital in scientific disciplines and engineering.