Problem 11 - Calculus: Early Transcendental Functions

Question

Write the given (total) area as an integral or sum of integrals. The area above the \(x\) -axis and below \(y=4-x^{2}\).

Step-by-Step Solution

Answer

The total area as an integral is \(A=\int_{-2}^{2}(4-x^{2}) dx\) or numerically \(A=\frac{16}{3}\)

Step 1: Find the x-intercepts

Set \(y=0\) in the function \(y=4-x^{2}\) and solve for \(x\) to find the x-intercepts. That gives \(0=4-x^{2}\) which can be solved by adding \(x^{2}\) to both sides and then taking square root: \(x^{2}=4\), so \(x=-2,2\). Those are the limits of the integral.

Step 2: Write the integral

The area is given by the definite integral of the function from -2 to 2. So the area \(A\) can be written as: \(A=\int_{-2}^{2}(4-x^{2}) dx\).

Step 3: If necessary, compute the integral

In cases where it's necessary to find the numerical value of the area, this integral can be computed. First, find an antiderivative of \(4-x^{2}\), which is \(4x-\frac{1}{3}x^{3}\), then evaluate it at 2 and -2 and subtract. This yields \(\frac{16}{3}\).

Key Concepts

Calculating X-Intercepts

When dealing with functions and graphs, finding the x-intercepts is often our starting point for understanding the behavior of the function. The x-intercepts are the points where the curve crosses the x-axis, meaning these are the values of x for which the function equals zero. To calculate these intercepts, simply set the function equal to zero and solve for x.

In the context of the given exercise, the function is a quadratic function represented by the equation y=4-x^2. By setting y to zero, you obtain the equation 0=4-x^2. Solving for x then entails factoring or applying algebraic techniques to find the x-values where the curve intersects the x-axis. In this case, by adding x^2 on both sides, you would then take the square root, which yields the solutions x=-2 and x=2. These intercepts are crucial as they define the limits of the integral when calculating the area beneath the curve.

Setting Up a Definite Integral

A definite integral is at the core of calculating areas under curves in calculus. To set up a definite integral, we need to identify the function that we are integrating, as well as the limits of integration, which are often determined by the x-intercepts found in the earlier step.

In the given exercise, the function y=4-x^2 defines a parabola, and you’ve already determined that the curve crosses the x-axis at x=-2 and x=2. This means your limits of integration are -2 and 2. The area under the curve and above the x-axis is then represented by the integral \( A = \int_{-2}^{2}(4-x^2) dx \). Here, the expression (4-x^2) is the integrand – the function that describes the shape above the x-axis – and dx indicates that you are integrating with respect to x, moving from left to right between the two x-intercepts.

Antiderivative Evaluation

Evaluating antiderivatives is a fundamental part of solving definite integrals. This involves finding the function's antiderivative, also known as the indefinite integral, which reverses differentiation. Once the antiderivative is known, it can be evaluated at the upper and lower limits of integration to solve the definite integral. This process is encapsulated in the Fundamental Theorem of Calculus.

For the parabola defined by 4-x^2, an antiderivative would be the function obtained by integrating the term 4-x^2 with respect to x, which is 4x - \frac{1}{3}x^3. To find the area using the definite integral, evaluate the antiderivative at the upper limit of 2 and the lower limit of -2, and subtract the latter from the former. Performing the evaluation gives us \( \left[4(2) - \frac{1}{3}(2)^3\right] - \left[4(-2) - \frac{1}{3}(-2)^3\right] \), resulting in the numeric value of the area, \( \frac{16}{3} \). It’s critical to pay close attention to the signs and the subtraction during this final evaluation step to ensure the accuracy of the area calculation.