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To solve integrals and find the area under a curve, we use the concept of an antiderivative. An antiderivative is a function whose derivative gives us the original function we want to integrate. It's essentially the reverse of taking a derivative.
For example, if our function is given by the expression \( y = 4 - x^2 \), we can determine its antiderivative. The process involves looking for functions whose derivatives will yield each term in our original function.

• The antiderivative of a constant like \( 4 \) is \( 4x \), since the derivative of \( 4x \) is \( 4 \).
• The term \( -x^2 \) becomes \( -\frac{x^3}{3} \) because taking the derivative of \( -\frac{x^3}{3} \) gives us back \( -x^2 \).
  Thus, combining these, the antiderivative of \( 4 - x^2 \) is \( 4x - \frac{x^3}{3} \). This formula represents a family of functions that, when differentiated, return the integrand \( 4 - x^2 \). Understanding how to find antiderivatives is crucial for solving integrals.
  Whenever you have a polynomial function like this, you can often find its antiderivative by reversing basic derivative rules.

The Fundamental Theorem of Calculus links the concept of a definite integral with antiderivatives, thereby simplifying the process of finding the area under a curve. This theorem consists of two main parts, but here, we are primarily concerned with the evaluation part.
The essence of this theorem is that to find the definite integral of a function over a specific interval, you can use its antiderivative. Once you have determined an antiderivative for the function, such as \( F(x) = 4x - \frac{x^3}{3} \) for \( y = 4 - x^2 \), you can then evaluate this antiderivative at the bounds of the integral.

• Calculate the antiderivative at the upper limit of the integral, \( x = 2 \).
• Calculate the antiderivative at the lower limit, \( x = -2 \).
• Subtract the value of the antiderivative at the lower limit from the value at the upper limit.

This can be represented mathematically as: \[\int_{-2}^{2} (4 - x^2) \, dx = \left[ F(x) \right]_{-2}^{2} = F(2) - F(-2)\] By this calculation, you efficiently determine the area under the curve between two points, thanks to the Fundamental Theorem of Calculus.

Finding the area under a curve is a common application of integrals in calculus. It allows us to quantify the space between the curve and the x-axis, within a specified interval.
In this exercise, we identify the area under the curve \( y = 4 - x^2 \) from \( x = -2 \) to \( x = 2 \). The curve is a downward-opening parabola, and we seek the total region it encompasses with the x-axis over this interval.
To achieve this, we set up the definite integral \( \int_{-2}^{2} (4 - x^2) \, dx \). Solving this integral using the previously determined antiderivative (\( 4x - \frac{x^3}{3} \)) and applying the Fundamental Theorem of Calculus, we determine that the area is computed to be 16 square units.

• This method ensures a precise calculation of area, accounting for areas above and below the x-axis being positive.
• It emphasizes how calculus provides tools to calculate areas that are not straightforward to measure geometrically.

Finding the area under a curve with integrals is not just limited to simple shapes like rectangles or triangles but extends to any continuous function over a given interval.