Problem 60 - Essential Calculus

Question

Use integration by parts to verify the formula. (For Exercises \(57-60\), assume that \(n\) is a positive integer.) $$ \int x^{n} e^{a x} d x=\frac{x^{n} e^{a x}}{a}-\frac{n}{a} \int x^{n-1} e^{a x} d x $$

Step-by-Step Solution

Answer

The given integral formula is verified using integration by parts. The integral \(\int x^{n} e^{a x} dx\) is indeed equal to \(\frac{x^{n} e^{a x}}{a}-\frac{n}{a} \int x^{n-1} e^{a x} dx\).

Step 1: Understand Integration by Parts

Integration by parts is a method for solving integrals, and is based on the rule of differentiation for the product of two functions. It is given by: \[\int u dv = uv - \int v du\].

Step 2: Assign functions to \(u\) and \(dv\)

We must assign the functions to \(u\) and \(dv\) to proceed with the integration by parts. Here, assign \(u = x^{n}\) and \(dv = e^{ax} dx\)

Step 3: Find \(du\) and \(v\)

After assigning \(u\) and \(dv\), calculate \(du\) and \(v\). The derivative of \(u\) is \(n x^{n - 1} dx\), so \(du = n x^{n - 1} dx\). Integration of \(dv = e^{ax} dx\) gives \(v = \frac{1}{a}e^{ax}\).

Step 4: Apply the formula of Integration by Parts

Substitute \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula \[\int u dv = uv - \int v du\]\n to get \[\int x^{n} e^{ax} dx = x^{n} \cdot \frac{1}{a} e^{ax} - \int \frac{1}{a} e^{ax} n x^{n - 1} dx\]. This simplifies to \[\frac{x^{n} e^{ax}}{a} - \frac{n}{a} \int x^{n-1} e^{ax} dx\], which verifies the given formula.

Key Concepts

Integration Techniques

Integration by parts is a handy tool within the realm of calculus, especially when direct integration isn't feasible. This technique is derived from the product rule of differentiation and allows us to break down a complicated integral by expressing it as a product of two functions. The core idea behind integration by parts is:

• Identify two parts of the integrand as the functions \( u \) and \( dv \).
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to obtain \( v \).
• Substitute into the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).

This process transforms the original integral into forms that are easier to evaluate. It's essential to wisely choose \( u \) and \( dv \), as a poor choice can complicate the solution. A helpful mnemonic to guide this choice is LIATE, standing for logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential functions, in this order to prioritize the selection of \( u \). When dealing with polynomials multiplied by exponentials or trigonometric functions, integration by parts often turns out to be particularly effective.

Differential Calculus

In differential calculus, differentiation plays a key role in finding how functions change. It concerns the rate at which quantities change and provides the foundation for integral calculus through the Fundamental Theorem of Calculus.The derivative tells us the slope of the tangent line to the curve of a function at any point and indicates the rate of change of the function's value. When performing integration by parts, differentiation is a pivotal step. In our problem, we assigned \( u = x^n \), which upon differentiation gives \( du = n x^{n-1} dx \).Differential rules, like the power rule, guide us in differentiating functions such as \( x^n \). This rule states that \( \frac{d}{dx}(x^n) = n x^{n-1} \). Using these rules, we prepare the components necessary for successful integration by parts.Having a robust understanding of differential calculus allows for strategic manipulation of functions to make the integration process more straightforward and manageable.

Mathematical Proofs

Mathematical proofs are essential for verifying the correctness and validity of mathematical statements. They involve a sequence of logical steps that demonstrate why a particular statement is true. In the given exercise, using integration by parts, we prove that:\[ \int x^{n} e^{a x} d x = \frac{x^{n} e^{a x}}{a} - \frac{n}{a} \int x^{n-1} e^{a x} d x \]To arrive at this proof, we:

• Break down the task into manageable steps by identifying appropriate functions \( u \) and \( dv \).
• Apply the integration by parts formula correctly.
• Simplify the expression systematically to reveal the structure of the original equation.

This exercise underscores the methodical nature of mathematical proofs – each step is rigorously derived from previous ones, ensuring the conclusion logically follows from the premises. Mastery of proofs not only confirms the results but deepens understanding, building a strong foundation in mathematical reasoning. Through such practice, mathematicians develop a confidence in their solutions, assured by the truth and logic of their proofs.