Problem 43 - Calculus

Question

Use symmetry to help you evaluate the given integral. $$ \int_{-1}^{1}\left(\left|x^{3}\right|+x^{3}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{1}{2}\).

Step 1: Identify the Symmetric Property

Examine the function inside the integral: \(|x^3| + x^3\). Notice how this can be divided into its components: \(|x^3|\) is always non-negative, and \(x^3\) changes sign based on \(x\). This hints at symmetry in the integral.

Step 2: Analyze Even and Odd Parts

Decompose \(|x^3| + x^3\). Here, \(|x^3|\) is an even function since \(|x^3| = (-x)^3| = x^3|\), and \(x^3\) is an odd function since \(-x^3 = -(x^3)\). Recognize that the integral of an odd function over a symmetric interval around zero is zero.

Step 3: Determine Contribution of Odd Function

Since \(x^3\) is an odd function, its integral over \([-1, 1]\) is zero. Mathematically, \(\int_{-1}^{1} x^3 \, dx = 0\). Therefore, only the even part of the function, \(|x^3|\), needs to be integrated.

Step 4: Focus on Even Function Integration

Now consider the integral of the even function \(|x^3|\) over \([-1, 1]\). Due to symmetry, this is twice the integral from 0 to 1. Hence, \(\int_{-1}^{1} |x^3| \, dx = 2 \int_{0}^{1} x^3 \, dx\).

Step 5: Compute the Integral

Calculate the integral of \(x^3\) from 0 to 1: \(\int_{0}^{1} x^3 \, dx = \left. \frac{x^4}{4} \right|_0^1 = \frac{1}{4} - 0 = \frac{1}{4}\).

Step 6: Conclude Total Integral Value

Using the results from above, the evaluated integral is \(2 \times \frac{1}{4} = \frac{1}{2}\). Therefore, \(\int_{-1}^{1} (|x^3| + x^3) \, dx = \frac{1}{2}\).

Key Concepts

Even and Odd Functions

In calculus, understanding the nature of functions as even or odd is crucial, especially when evaluating integrals over symmetric intervals. These properties can simplify complex integrals or even lead to simple solutions without performing intricate calculations.

**Even Functions:**
- An even function is symmetric around the y-axis. This means it satisfies the property: \( f(-x) = f(x) \).- A common example is \( x^2 \) or \( |x^3| \) as shown in the exercise. These functions look the same on both sides of the y-axis.- For integrals, this symmetry implies that the total area from \(-a\) to \(a\) is double the area from \(0\) to \(a\). - Mathematically: \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \).

**Odd Functions:**
- Odd functions have rotational symmetry around the origin and satisfy: \( f(-x) = -f(x) \).- Examples include \( x^3 \), which changes sign between positive and negative x-values.- For integrals, the symmetry results in the opposite areas canceling each other out.- Mathematically, the integral over a symmetric range centered at zero will always be zero: \( \int_{-a}^{a} f(x) \, dx = 0 \).

In the exercise, recognizing \( x^3 \) as an odd function allowed for the simplification because its integral from \([-1, 1]\) equals zero. It was then enough to focus only on the even part, \(|x^3|\).

Definite Integral

The concept of a definite integral represents the accumulation of quantities, which could mean area under a curve, total change, and more for calculus students. It is bound by two limits, providing a specific numerical value as opposed to an indefinite integral, which gives a general form of antiderivative.

Some key points worth noting:

• The definite integral of a function \( f(x) \) from \( a \) to \( b \), written as \( \int_{a}^{b} f(x) \, dx \), computes the net area under the curve, factoring in the direction of the area (above or below the x-axis).
• The result can be used to find total quantities such as distance, mass, and other cumulative metrics in applied contexts.
• The fundamental theorem of calculus relates the definite integral to the antiderivative, stating that if \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

This integral concept was critical in solving the main problem by providing a concrete value, \( \frac{1}{2} \), representing the specific accumulation as defined by the symmetry and properties applied. By integrating from one symmetric limit to another, we applied these abstract ideas to a real calculation.

Properties of Integrals

Some integral properties help in simplifying calculations, making analysis easier, especially when dealing with symmetric limits or known function behaviors. In the exercise, several properties were utilized:

- **Additivity Over Intervals:** * The property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \) allows breaking down integration into simpler parts. * Though not directly used in this exercise, knowing this helps when integrals span multiple regions or segments. This leads to understanding integrals over different intervals.
- **Integrals of Even and Odd Functions:** * As mentioned, since the integral of an odd function over a symmetric interval is zero, one can disregard the odd term \( x^3 \). * The integral of the even portions is simplified to double the positive side: \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \). This application simplified the task to just evaluating \(|x^3|\) from 0 to 1.
These properties underpin many integration techniques. Understanding and correctly applying them streamlines computations, avoiding unnecessary work, and revealing elegant solutions to seemingly complex integrals.