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The definite integral is a fundamental concept in calculus. It allows us to calculate the accumulated total of a function's values over a specific interval. This is particularly helpful for finding areas under curves on a graph. The definite integral is structured as \[ \int_{a}^{b} f(x) \, dx \], where \( a \) and \( b \) are the limits of the interval, and \( f(x) \) is the function being integrated.
For our given function, \( y = 1 - \frac{1}{x^2} \) over the interval \( \left[ \frac{1}{2}, 2 \right] \), the definite integral becomes \[ \int_{1/2}^{2} \left( 1 - \frac{1}{x^2} \right) \, dx \]. By evaluating this integral, we can determine the area enclosed between the curve and the x-axis within these bounds.
Here are the quick steps:

• First, integrate \( 1 \, dx \) which evaluates to \( [x]_{1/2}^{2} \), resulting in \( 3/2 \).
• Next, integrate \( -\frac{1}{x^2} \, dx \) which evaluates to \( [-\frac{1}{x}]_{1/2}^{2} \), resulting in \( 5/2 \).
• Finally, we add these results to find the total area, which is \( 4 \).

Understanding how to compute the definite integral is crucial in many practical applications of calculus.

Curve sketching is a method used to draw an approximation of a graph based on the behavior of a function. It entails understanding the general shape, orientation, and key features of the curve. For the function \( y = 1 - \frac{1}{x^2} \), simplifying to \( y = 1 - x^{-2} \), reveals essential information.

• As \( x \) becomes very large, \( y \) tends towards 1. This indicates a horizontal asymptote at \( y = 1 \).
• As \( x \) approaches zero from the positive side, \( y \) tends towards negative infinity. This suggests a vertical asymptote around \( x = 0 \).
• At the endpoints of our interval, namely \( x = \frac{1}{2} \) and \( x = 2 \), the values of \( y \) are \(-3\) and \(\frac{3}{4}\), respectively.

Visualizing these behaviors aids in constructing a representative sketch of the function over the interval. Observing how the curve dips steeply and ascends within the interval provides a more intuitive understanding of the function's performance.

Critical points are specific values of \( x \) where a function changes direction, potentially indicating a peak, trough, or inflection. To find these points, we compute the derivative of the function and determine where it equals zero or is undefined.
For our function \( y = 1 - \frac{1}{x^2} \), the derivative is \( y' = \frac{2}{x^3} \). Setting this derivative to zero, \( y' = 0 \), reveals no critical points where the derivative is zero because \( \frac{2}{x^3} \) is never zero. However, the function is undefined at \( x = 0 \).
In this exercise, critical points at the interval's endpoints are more relevant:

• Examine \( x = \frac{1}{2} \).
• Examine \( x = 2 \).

These analyses guide us in emphasizing the function's behavior over the selected interval and accounting for notable changes in the graph's direction.

Calculating the area under a curve involves using definite integration to sum up the infinite number of tiny areas beneath the curve, between given limits along the x-axis. This area represents the net sum of the function's values over the interval. Specifically, for our function, we are looking at the area between \( y = 1 - \frac{1}{x^2} \) and the x-axis from \( x = \frac{1}{2} \) to \( x = 2 \).

• The first step is setting up the integral: \[ \int_{1/2}^{2} \left( 1 - \frac{1}{x^2} \right) \, dx \].
• The second is calculating this integral, decomposed into two parts: \(3/2\) from the simple integral of \(1\), and \(5/2\) from the integral of \(-\frac{1}{x^2}\).
• Add these two values to determine the total area, yielding \(4\).

Calculating the area under the curve provides valuable insights into the function's accumulative behavior across the interval. It also connects integral calculus concepts to geometrical interpretations, aiding in better understanding mathematical properties and their real-world applications.