Problem 2 - Advanced Engineering Mathematics

Question

$$\int_{-1}^{1} x^{3}\left(x^{2}+1\right) d x=\left.\frac{1}{6} x^{6}\right|_{-1} ^{1}+\left.\frac{1}{4} x^{4}\right|_{-1} ^{1}=0$$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understand the Exercise

We are given a definite integral to solve: \( \int_{-1}^{1} x^{3}(x^{2}+1) dx \). The goal is to determine if the evaluated expression is 0.

Step 2: Write the Integral in a Solvable Form

Rewrite the integrand by expanding the expression: \( x^3(x^2 + 1) = x^5 + x^3 \). So the integral simplifies to: \( \int_{-1}^{1} (x^5 + x^3) \, dx \).

Step 3: Integrate Term by Term

Integrate each term separately: \[ \int (x^5 + x^3) \, dx = \int x^5 \, dx + \int x^3 \, dx. \] Integrating each term, we get: \( \frac{x^6}{6} + \frac{x^4}{4} \).

Step 4: Evaluate the Indefinite Integral at Bounds

Substitute the bounds \([-1, 1]\) into \( \left. \frac{x^6}{6} + \frac{x^4}{4} \right|_{-1}^{1} \). Evaluate this expression: \( \left( \frac{1}{6} + \frac{1}{4} \right) - \left( \frac{1}{6} + \frac{1}{4} \right) \).

Step 5: Simplify the Evaluated Expression

Both terms from the previous step result in the same quantity for \(x = 1\) and \(x = -1\), so the expression evaluated at the upper and lower bounds cancel each other out.

Key Concepts

Indefinite Integrals

Indefinite integrals, sometimes referred to as antiderivatives, are essential in calculus, representing a family of functions whose derivative is the given function. These integrals do not have specific upper or lower limits, which distinguishes them from definite integrals.
Here's how they work:

• If you have a function \( f(x) \), its indefinite integral is written as \( \int f(x) \, dx \), and it represents all the antiderivatives of \( f(x) \).


• The result of an indefinite integral often includes a constant of integration \( C \), because when differentiating, any constant goes away (since the derivative of a constant is zero). Thus, the family of functions is infinite.

The indefinite integral is a foundation for solving definite integrals. Knowing how to find an indefinite integral (by antiderivation) allows you to accumulate small quantities to find total quantities over an interval, precisely how we handled the function in our original problem before applying bounds to obtain a definite integral result.

Integration Techniques

Integration techniques are methods used to solve integrals, and they come in different forms depending on the kind of function you're dealing with. In our exercise, a polynomial function was expanded and then integrated term by term.
Some commonly used techniques include:

• Substitution: Often used when an integral contains a composite function. We substitute part of the integral with a new variable to simplify the process.


• Integration by Parts: This technique is useful for integrating products of functions. It's derived from the product rule for differentiation.


• Partial Fractions: Used when dealing with rational functions, which can be broken down into simpler fractions before integrating.


• Polynomial Integration: For polynomial functions, each term can be integrated separately as we did above, using the formula \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).

Understanding and mastering different integration techniques enable you to tackle a wide variety of problems effectively, turning seemingly complex equations into solvable pieces.

Polynomial Functions

Polynomial functions are expressions involving variables raised to whole number powers, and they may involve one or more terms. An integral part of calculus involves working with such functions because of their simple structure.
In the exercise, the polynomial was originally in the form \( x^3(x^2 + 1) \). By expanding this expression, we converted it to \( x^5 + x^3 \), which simplified the integration process.
Important properties of polynomial functions include:

• Degrees: This is the highest power of the variable in the polynomial. In \( x^5 + x^3 \), the degree is 5.


• Term-wise integration: Because the integral of a sum is the sum of the integrals, each term within a polynomial can be individually integrated and then summed.


• Simplicity in Calculation: Polynomials' structural simplicity makes them easy to work with using basic rules of integration and arithmetic operations.

By understanding how to manipulate and integrate polynomial functions, you can solve a broader range of problems, turning comprehensive exercises into manageable steps.