91Ó°ÊÓ

When we talk about finding the area between curves, we're looking at the region that is "sandwiched" between two graphs. In the exercise provided, the curves are the line \( y = 4 \) and the parabola \( y = x^2 \). These two curves cross each other at points where they have the same \( y \)-values, specifically when \( x = -2 \) and \( x = 2 \). This is why they determine the boundaries for our integral.

To find the area, imagine drawing vertical lines between these two curves at each point of \( x \). The length of each line is the distance or difference in \( y \)-values between the two curves at that particular \( x \).

In mathematical terms, for any point between \( x = -2 \) and \( x = 2 \), the length of this line is \( (4 - x^2) \). By integrating these lengths over the entire range of \( x \), we add up all these tiny lines to get the total area between the curves.

Integration is the process of finding the total accumulation of a quantity. When applied to areas between curves, integration helps us total the lengths of those tiny vertical lines we discussed.

In the context of our problem, to compute the area between the curves, we set up this integral:
\[ \int_{-2}^{2} (4 - x^2) \, dx \]
This notation might look overwhelming at first, but here's what it does:

• The function \( 4 - x^2 \) inside the integral represents the length of the vertical lines between the two curves at each point \( x \).
• \( dx \) signifies a tiny change in \( x \), essentially choosing a narrow slice of the area.
• The numbers \(-2\) and \(2\) at the bottom and top of the integral sign indicate that \( x \) runs from \(-2\) to \(2\).

Through the magic of calculus, integration sums up those slices to find the total area.

In calculus, finding the antiderivative is the reverse process of taking a derivative. It's all about finding the original function that led to a given derivative.

For our exercise, to calculate the total area, we must first compute the antiderivatives of the terms in \((4 - x^2)\). The antiderivative of \(4\) is \(4x\), and the antiderivative of \(-x^2\) is \(\frac{-x^3}{3}\). Therefore:

• Antiderivative of \(4\) is \(4x\)
• Antiderivative of \(-x^2\) is \(-\frac{x^3}{3}\)

This allows us to write the integrated form as:
\[ 4x - \frac{x^3}{3} + C \]
C is a constant of integration, but it cancels out when evaluating a definite integral, so we often omit it in these calculations.

A definite integral calculates the net accumulation of a function over a specific interval. By using the antiderivative we just calculated, we find the definite integral to compute the total area between our curves.

In our problem, we evaluate this antiderivative from \( x = -2 \) to \( x = 2 \). This involves substituting these values into the antiderivative formula and subtracting:

• First, substitute \( x = 2 \) to get: \(4(2) - \frac{(2)^3}{3} = \frac{16}{3}\).
• Next, substitute \( x = -2 \) to get: \(4(-2) - \frac{(-2)^3}{3} = -\frac{16}{3}\).

Finally, we subtract the second result from the first:
\[ \frac{16}{3} - (-\frac{16}{3}) = \frac{32}{3} \]
This final number represents the total area between the two curves from \( x = -2 \) to \( x = 2 \). Understanding definite integrals allows you to solve many real-world problems involving total accumulations, like areas under curves or volumes of solids.