91Ó°ÊÓ

The definite integral is a fundamental concept in calculus. It helps us find the area under a curve over a specific interval.
For the curve given by the equation \(y = 4 - x^2\) over the interval \([-2, 2]\), we want to calculate the area beneath this curve.
This can be done using the definite integral. The notation for a definite integral is: \[ \int_{a}^{b} f(x) \, dx \] Here, `\(a\)` and `\(b\)` are the bounds of the interval, and `\(f(x)\)` is the function we are integrating.
Plugging in the given values, we get: \[ \text{Area} = \int_{-2}^{2} (4 - x^2) \, dx \]

The graph of the function \(y = 4 - x^2\) is a parabola. A parabola is a symmetrical curve, and in this case, it opens downward.
The general form of a parabola is \(y = ax^2 + bx + c\). Here, we have: \[ y = -x^2 + 4 \] This means:

• The coefficient of \(x^2\) is negative, so the parabola opens downward.
• The vertex of the parabola is at the maximum point.
• The function describes a symmetrical curve about the y-axis.

To better understand the area under the parabola, sketch the graph. The interval \([-2, 2]\) covers the entire base of the inverted parabola.

To evaluate the integral, we use basic integration techniques.
First, we express the integral as the sum of simpler integrals: \[ \int_{-2}^{2} (4 - x^2) \, dx = \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} x^2 \, dx \] These two integrals can be calculated separately:
1. For the constant term \(4\): \[ \int_{-2}^{2} 4 \, dx = 4 \left[ x \right]_{-2}^{2} = 4(2 - (-2)) = 4 \cdot 4 = 16 \] 2. For the term \(x^2\): \[ \int_{-2}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{2^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \left( \frac{-8}{3} \right) = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \] Now, subtract these results to find the area: \[ \text{Area} = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \] This gives the area under the curve as \( \frac{32}{3} \) square units.