Problem 22 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int \frac{\ln t}{\sqrt{t}} d t\)

Step-by-Step Solution

Answer

The integral is \( 2\sqrt{t} \ln t + 4\sqrt{t} + C \).

Step 1: Choose Functions for Integration by Parts

Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). We need to select \( u \) and \( dv \) such that it's easier to integrate and differentiate. Here, let \( u = \ln t \) and \( dv = \frac{1}{\sqrt{t}} \, dt \).

Step 2: Differentiate and Integrate Parts

Differentiate \( u \) and integrate \( dv \) to find \( du \) and \( v \). So, \( du = \frac{1}{t} \, dt \) and \( v = \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t} \).

Step 3: Substitute into Integration by Parts Formula

Substitute the expressions for \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \( \int \frac{\ln t}{\sqrt{t}} \, dt = uv - \int v \, du = 2\sqrt{t} \ln t - \int 2\sqrt{t} \cdot \frac{1}{t} \, dt \).

Step 4: Simplify and Solve Remaining Integral

Simplify the integral \( \int 2\sqrt{t} \cdot \frac{1}{t} \, dt \) by writing it as \( \int \frac{2}{\sqrt{t}} \, dt \). Solving this gives \( -4\sqrt{t} \).

Step 5: Final Expression for the Integral

Put together the parts to get the final expression for the integral: \( \int \frac{\ln t}{\sqrt{t}} \, dt = 2\sqrt{t} \ln t + 4\sqrt{t} + C \), where \( C \) is the constant of integration.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the concept of integration, which is the process of finding the antiderivative or the "integral" of a function. It is performed to determine quantities like area, volume, and other quantities that accumulate continuously. There are two primary operations in calculus: differentiation and integration. Differentiation explores how functions change, while integration focuses on the accumulation of quantities.

In the context of calculus, integrals are classified into two main types:

• Indefinite Integral: Represents a family of functions, integrating without limits and including a constant of integration, denoted as \( C \). For example, the indefinite integral of \( f(x) = x^2 \) is \( F(x) = \frac{x^3}{3} + C \).
• Definite Integral: Calculates the net "area under a curve" between two bounds, providing a specific numerical result. It does not require a constant of integration because the bounds define the result uniquely. For example, the definite integral \( \int_0^1 x^2 \, dx \) results in a numerical area value, \( \frac{1}{3} \).

Integral calculus uses various methods for solving integrals, and one powerful method is known as "integration by parts," which is particularly useful for functions expressed as a product of two terms. It follows the formula \( \int u \, dv = uv - \int v \, du \), enabling the transformation of a complex integral into simpler forms.

Logarithmic Integration

Logarithmic integration is a technique that often involves the integration of functions containing logarithms. A common situation encountered is when an integral has a function like \( \ln x \) as part of the integrand. It can create complexity due to the natural logarithm, but there are strategies to manage this complexity.

One essential method is using integration by parts, which allows us to integrate the product of functions efficiently. Integration by parts is particularly handy for integrating terms with logarithmic functions because it uses the formula:

• Choose \( u \) and \( dv \) cleverly, typically setting \( u \) as the logarithmic term (like \( \ln t \) as seen in our exercise).
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to identify \( v \).
• Substitute these into the integration by parts equation and simplify accordingly.

For example, in the problem \( \int \frac{\ln t}{\sqrt{t}} dt \), \( u = \ln t \) and \( dv = \frac{1}{\sqrt{t}} dt \) are chosen. By differentiating \( u \) and integrating \( dv \), we can apply the integration by parts formula, simplifying the integral significantly and managing the complexity presented by the logarithmic term.

Definite and Indefinite Integrals

Definite and indefinite integrals are fundamental concepts in integral calculus that describe different aspects of integration. Understanding the distinction between these two types is crucial for solving a wide range of problems:

• Definite Integrals: These are expressed as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the lower and upper bounds of integration, respectively. Definite integrals provide the net area under the curve of a graph between these two points. They are particularly useful in physical applications, like calculating displacement or total distance traveled over a specific period.
• Indefinite Integrals: These are represented as \( \int f(x) \, dx \), where there are no bounds specified. The result is a general family of functions with a constant of integration, \( C \), such as \( F(x) = \int f(x) \, dx + C \). Indefinite integrals provide the antiderivative and are critical in solving problems involving accumulation, such as finding a function whose derivative matches a given rate of change.

In practice, finding the integral of a function may start with finding the indefinite integral (which includes the constant of integration) and then using boundary values to compute a definite integral, if required. The constant \( C \) in indefinite integrals represents an infinite number of possible vertical shifts of the function, addressing any initial value or starting point ambiguities.