Problem 5 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=8-2 x^{2}, y=0$$

Step-by-Step Solution

Answer

The area bounded by the curves is \( \frac{64}{3} \) square units.

Step 1: Identify the curves

We are given the two equations: \( y = 8 - 2x^2 \) and \( y = 0 \). The equation \( y = 8 - 2x^2 \) represents a downward-opening parabola, while \( y = 0 \) is the x-axis.

Step 2: Find the intersection points

To find the points of intersection, set the equations equal to each other: \[ 8 - 2x^2 = 0 \] Solve for \( x \).\[ 8 = 2x^2 \] Divide by 2:\[ 4 = x^2 \]Take the square root:\[ x = \pm 2 \]Thus, the points of intersection are \( x = -2 \) and \( x = 2 \).

Step 3: Set up the integral

The area between the curves from \( x = -2 \) to \( x = 2 \) is found using the definite integral:\[ \int_{-2}^{2} (8 - 2x^2) \, dx \]This integral represents the area from the parabola down to the x-axis.

Step 4: Calculate the integral

Integrate the function:\[ \int (8 - 2x^2) \, dx = 8x - \frac{2}{3}x^3 \]Evaluate from \( x = -2 \) to \( x = 2 \).

Step 5: Evaluate the antiderivative

Substitute \( x = 2 \) and \( x = -2 \) into the antiderivative, then find the difference:\[ \left[ 8(2) - \frac{2}{3}(2)^3 \right] - \left[ 8(-2) - \frac{2}{3}(-2)^3 \right] \]Simplify each part:\[ \left( 16 - \frac{16}{3} \right) - \left( -16 + \frac{16}{3} \right) \]

Step 6: Calculate the simplified areas

Simplfy further:\[ \left( 16 - \frac{16}{3} \right) + 16 - \frac{16}{3} \]Convert to a common denominator:\[ \frac{48}{3} - \frac{16}{3} + \frac{48}{3} - \frac{16}{3} = \frac{64}{3} \]

Step 7: Final evaluation

The total area bounded by the curves is: \[ \frac{64}{3} \] square units.

Key Concepts

Finding Area

In calculus, finding the area between curves is a widely used application of the definite integral. To do this, we calculate the integral of the top function minus the bottom function in the given interval. Here, the task involves the parabola represented by the function \( y = 8 - 2x^2 \) and the x-axis, represented by \( y = 0 \).To find the area, we use the expression for the definite integral: \[\int_{a}^{b} f(x) \, dx\]where \(f(x)\) is the function defining the curve and \(a\) and \(b\) are the intersection points of the curves, or in simple terms, the limits of integration.

• Set up the proper integral by determining the limits, which in this case are the intersection points found to be \(x = -2\) and \(x = 2\).
• Integrate the function, substituting the limits of integration to find the exact area.

The area represents the space trapped between the curve of the parabola and the x-axis, providing a view of how the function behaves within the interval.

Intersection Points

Determining intersection points is crucial when calculating the area between curves, as they define the boundaries of integration.For the equations \(y = 8 - 2x^2\) and \(y = 0\), we find the points of intersection by setting them equal to each other:\[8 - 2x^2 = 0\]Solving this equation involves a few simple algebraic manipulations:

• First, move constants to one side to set the equation to zero.
• Divide by the coefficient of \(x^2\) to simplify.
• Take the square root of both sides to find the values of \(x\).

In this case, solving gives us \(x = -2\) and \(x = 2\), which are the intersection points. They represent the x-values where the parabola touches the x-axis, helping to define the limits for the integral that finds the area.

Parabola

Parabolas are one of the simplest types of curves, having a characteristic U-shape. The given equation \( y = 8 - 2x^2 \) specifically describes a downward-opening parabola.The key features to note about a parabola include:

• Vertex: This is the highest point when the parabola opens downward, located at \(y = 8\) when \(x = 0\).
• Axis of Symmetry: For this parabola, it is the vertical line \(x = 0\).
• Intercepts: The point \(x = 0\) provides a y-intercept, and solving for the intersection with \(y = 0\) gives the x-intercepts at \(x = -2\) and \(x = 2\).

Understanding these properties can make graphing and dealing with parabolas much more straightforward, aiding in visualizing the shape and behavior of the curve. Such understanding is particularly useful when applying these concepts to integrals, as it provides a clear picture of the regions whose area is being calculated.