91Ó°ÊÓ

The power rule for integration is a fundamental concept in calculus that allows us to find the antiderivative of functions that involve powers of variables. Specifically, the rule is applied to functions of the form \( x^n \), where \( n \) is any real number except \(-1\). The rule states that:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

For integration, \( C \) represents the constant of integration, but when evaluating definite integrals, this constant isn't necessary. In our integral \( \int 2x^{5} \, dx \), we use this rule to move from the expression with \( x^5 \) to a simpler form. By multiplying by the constant 2, we have:

• \( \int 2x^{5} \, dx = 2 \cdot \frac{x^{6}}{6} \)
• This simplifies to \( \frac{1}{3}x^{6} \)

This step is crucial as it provides the antiderivative needed to evaluate the definite integral later.

The integrand is a key part of any integral and is simply the function we are integrating. In the given problem, the integrand is \(2x^{5}\). Recognizing the integrand allows us to understand what transformation our variable of integration undergoes.

• The integrand here is the expression \( 2x^{5} \), which involves a power of \( x \).
• Determining the integrand is an essential first step before applying any integration techniques, such as the power rule.

Remember, the integrand is the content inside the integral symbol \( \int \), and understanding it helps set the stage for how the integration process will proceed.

In this problem, the integrand \( 2x^{5} \) is an odd function, creating symmetry about the origin. A symmetric function can lead to intriguing results when integrated over symmetric bounds, such as \(-2\) to \(2\).An odd function is one that satisfies \( f(-x) = -f(x) \). Let's see why this is important:

• The symmetry means that the function's positive area on one side of the y-axis is canceled by its equal negative area on the other side.
• Integration over symmetric intervals like \(-a\) to \(a\) for odd functions can lead directly to a zero result.

In our case, because \( 2x^{5} \) is odd and the interval is symmetric, the areas cancel out, resulting in a definite integral of zero.

Evaluating the bounds of a definite integral is the process of determining the value of the integration result over specific limits. In our example, the bounds were set at \(-2\) and \(2\).

• This involves substituting the upper and lower bounds into the antiderivative we've found: \( \frac{1}{3}x^{6} \).
• Substitute \(2\) and \(-2\): \( \left[ \frac{1}{3}x^{6} \right]_{-2}^{2} = \frac{1}{3}(2)^{6} - \frac{1}{3}(-2)^{6} \).
• Calculate each term: both \( (2)^6 \) and \( (-2)^6 \) equal 64, leading to: \( \frac{64}{3} - \frac{64}{3} = 0\).

This step reveals how integrating a symmetric function over symmetric bounds leads to the integral canceling itself out.