Problem 62 - Calculus and its applications

Question

Occasionally, integration by parts yields an integral of the form \(\int u d v\) that is identical to the original integral. In some cases, we can then solve for \(\int u d v\) algebraically. For example, to find \(\int 2^{x} e^{x} d x,\) we let \(u=2^{x}\) and \(d v=e^{x},\) so \(d u=(\ln 2) 2^{x} d x\) and \(v=e^{x} .\) Using integration by parts, we have $$ \int 2^{x} e^{x} d x=2^{x} e^{x}-\ln 2 \int 2^{x} e^{x} d x $$ Note that \(\int 2^{x} e^{x} d x\) appears twice. Adding \(\ln 2 \int 2^{x} e^{x} d x\) to $$ \begin{aligned} \int 2^{x} e^{x} d x+\ln 2 \int 2^{x} e^{x} d x &=2^{x} e^{x} \\ (1+\ln 2) \int 2^{x} e^{x} d x &=2^{x} e^{x} \\ \int 2^{x} e^{x} d x &=\frac{2^{x} e^{x}}{1+\ln 2}+C \end{aligned} $$ Use this method to evaluate the integrals in Exercises \(59-62\) $$ \begin{aligned} &\int x \ln x d x\\\ &\text { and } d v=d x \text { . Assume }\\\ &x>0 .) \end{aligned} $$

Step-by-Step Solution

Answer

The integral \( \int x \ln x \, dx \) evaluates to \( \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C \).

Step 1: Choose Functions for Integration by Parts

We are given the integral \( \int x \ln x \, dx \). For integration by parts, choose \( u = \ln x \) and \( dv = x \, dx \). This gives: \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).

Step 2: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute the chosen functions: \( \int x \ln x \, dx = \ln x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \).

Step 3: Simplify the Resulting Expression

Simplify the expression by distributing and simplifying terms: \( \int x \ln x \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx \).

Step 4: Integrate Simplified Expression

Now, integrate the simplified expression: \( \int \frac{x}{2} \, dx = \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \).

Step 5: Solve for the Original Integral

Combine the results: \( \int x \ln x \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C \), where \( C \) is the constant of integration.

Key Concepts

Definite and Indefinite Integrals

In calculus, integrals are a fundamental concept essential for solving problems related to areas, volumes, and many physical properties. But what's the difference between definite and indefinite integrals?
Indefinite integrals are expressed with an integration symbol followed by the function and DX, for example, \( \int f(x) \, dx \). It represents a family of functions and includes a constant \( C \). This is because the derivative of a constant is zero, and when doing the reverse operation (integration), we might "lose" the constant.
A definite integral, on the other hand, is used to calculate the area under a curve from one point to another and is expressed with limits, such as \( \int_{a}^{b} f(x) \, dx \). The definite integral results in a number, not a function. It represents the accumulation of quantities, often visualized as the area under the curve.
Both types of integrals share rules and properties, but they are used in different contexts to solve real-world problems, making them pivotal in calculus.

Algebraic Manipulation

Algebraic manipulation is a technique used to rearrange and simplify mathematical expressions. It is essential in calculus to make complex integrals solvable.
In the context of integration by parts, algebraic manipulation can involve combining like terms, factor integers, or strategically rearranging elements to make integrals easier to calculate. For example, if the solution to an integral results in the same expression appearing on both sides of the equation, we can use algebra to isolate the integral, as seen in the example \( (1+\ln 2) \int 2^{x} e^{x} \, dx = 2^{x} e^{x} \). By dividing through by \( (1+\ln 2) \), the integral can be isolated and resolved.
Mastery of algebraic manipulation enables students to tackle more complex calculus exercises by effectively simplifying expressions and solving equations.

Calculus Exercises

Calculus exercises are designed to solidify the understanding of concepts such as limits, derivatives, and integrals. They often involve applying formulas and theorems, like the integration by parts formula used here \( \int u \, dv = uv - \int v \, du \).
Being systematic with each step in such exercises is crucial. Let's take the integral \( \int x \ln x \, dx \) as an example. The exercise involves:

• Choosing the correct functions: Assign \( u = \ln x \) and \( dv = x \, dx \) to set the integration by parts process in motion.
• Applying the integration by parts formula to break down the integral into simpler components.
• Simplifying the resulting expression and solving it step-by-step: For instance, integrating \( \int \frac{x}{2} \, dx \) and combining this with previous results to obtain the integral for \( \int x \ln x \, dx \).

Calculus exercises are not just about getting the right answer but are essential for building a deep understanding of concepts and methods that are applicable in various fields, including physics and engineering.