Question

Area Find (i) the net area and (ii) the area of the following regions. Graph the function and indicate the region in question. The region above the \(x\) -axis bounded by \(y=4-x^{2}\)

Short answer

Answer: The net area of the region above the x-axis bounded by the curve \(y=4-x^2\) is \(\frac{32}{3}\) square units. To depict this region in a graph, sketch a downward-opening parabola with a vertex at \((0,4)\), x-intercepts at \(x=-2\) and \(x=2\), and shade the area enclosed by the curve and the x-axis, bounded by the vertical lines at \(x=-2\) and \(x=2\).

Step-by-step solution

Determine intersection points with x-axis

To find the points where the curve intersects the x-axis, set y equal to 0 and solve for x in the equation \(y=4-x^{2}\): $$ 0 = 4-x^{2} $$ Add \(x^2\) to both sides and factor: $$ x^{2} = 4 \\ x = \pm \sqrt{4} \\ x = \pm 2 $$ The curve intersects the x-axis at \(x=-2\) and \(x=2\).

Set up the integral

To find the area of the region above the x-axis, integrate the function \(y=4-x^{2}\) from \(x=-2\) to \(x=2\). This will give us the area under the curve and above the x-axis: $$ \text{Area} = \int_{-2}^{2} (4-x^{2}) dx $$

Evaluate the integral

Now, evaluate the integral by calculating the antiderivative of the integrand and applying the Fundamental Theorem of Calculus: $$ \int (4-x^{2}) dx = 4x - \frac{x^{3}}{3} \\ \text{Area} = \left[4x - \frac{x^{3}}{3}\right]_{-2}^{2} $$ Evaluate the definite integral by subtracting the function value at the lower limit from the function value at the upper limit: $$ \text{Area} = \left(4(2) - \frac{2^3}{3}\right) - \left(4(-2) - \frac{(-2)^3}{3}\right) \\ \text{Area} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) \\ \text{Area} = 16 - \frac{16}{3} $$

Simplify for the net area

Finally, simplify the expression for the net area: $$ \text{Net Area} = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} $$ The net area of the region above the x-axis bounded by \(y=4-x^{2}\) is \(\frac{32}{3}\) square units.

Graph the function and indicate the region

To graph the function \(y=4-x^{2}\), sketch a downward-opening parabola with a vertex at \((0,4)\), and x-intercepts at \(x=-2\) and \(x=2\). The region of interest is the area enclosed by the curve and the x-axis. To indicate the region, shade the area bounded by the curve \(y=4-x^{2}\), the x-axis, and the vertical lines at \(x=-2\) and \(x=2\). So, the net area of the region above the x-axis bounded by the curve \(y=4-x^{2}\) is (i) \(\frac{32}{3}\) square units and (ii) the region shown in the graph that meets given conditions.

Key concepts

Definite Integral

In Integral Calculus, the concept of the definite integral is fundamental. Imagine you want to calculate the area under a curve on a graph between two points. The definite integral makes this possible. A definite integral takes a function and finds the total area from one point to another on a graph.

For example, consider the function given by the equation \( y = 4 - x^2 \). To find the area under this curve from \( x = -2 \) to \( x = 2 \), we make use of a definite integral. This means calculating:

• \( \int_{-2}^{2} (4-x^{2}) \, dx \)

• The numbers \(-2\) and \(2\) are called the limits of integration. They define the interval over which you find the area under the curve.
• The result is known as the net area, which considers both the above and below-axis sections by providing the signed area.

Antiderivative

An antiderivative is a function whose derivative gives back the original function. When doing a definite integral, calculating an antiderivative is essential. The antiderivative is used to evaluate the area under the curve.

For the function \(y = 4 - x^2\), the antiderivative is found to be:

• \( 4x - \frac{x^{3}}{3} \)

This means if you differentiate \( 4x - \frac{x^{3}}{3} \), you should return to the original function \( 4 - x^2 \).
Moreover, applying the Fundamental Theorem of Calculus helps in finding the exact area. You evaluate the antiderivative at the upper limit and subtract the antiderivative evaluated at the lower limit:

• \( \left[4x - \frac{x^{3}}{3}\right]_{-2}^{2} \)

Area under a Curve

Finding the area under a curve on a graph is a central concept in calculus. It helps in determining actual physical values like distance, mass, or total expenditure, depending on what the function represents.

In the exercise given, understanding the area under the curve translates to finding where the curve \(y = 4 - x^2\) lies above the x-axis. This step is crucial for visual learners as it helps them see the problem in a more tangible way.

• In this case, since it's symmetric around the y-axis, both sides from \(-2\) to \(2\) reflect the exact area but in positive measure.
• By integrating from \(-2\) to \(2\), the area under the curve comes out to be \( \frac{32}{3} \) square units.

Such calculations simplify complex real-world problems into manageable mathematical solutions.