Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(y=\frac{1}{4}\left(x^{2}-7\right), y=0\), between \(x=0\) and \(x=2\)

Short answer

The area of the region is \( \frac{17}{6} \).

Step-by-step solution

Identify the Bounded Region

We are given the functions \( y = \frac{1}{4}(x^2 - 7) \) and \( y = 0 \), and we need to find the region between these curves from \( x = 0 \) to \( x = 2 \). The line \( y=0 \) is the x-axis, and the parabolic curve opens upward. We need to sketch these graphs to identify the area of interest.

Sketch the Graphs

Plot the graph of the parabola \( y = \frac{1}{4}(x^2 - 7) \), which intersects the x-axis at points where \( x^2 = 7 \). Thus, the intercepts are \( x = \pm \sqrt{7} \). For our region of interest, we only consider the interval from \( x = 0 \) to \( x = 2 \). The line \( y = 0 \) is the x-axis. Sketch these curves to visualize the enclosed region.

Choose a Typical Slice

A typical slice perpendicular to the x-axis is a vertical strip extending from \( y=0 \) to \( y = \frac{1}{4}(x^2 - 7) \). This slice's height is given by \( \frac{1}{4}(x^2 - 7) \), and its width is \( \Delta x \), an infinitesimally small change in x.

Approximate the Area

Approximate the area using a Riemann sum. The area of one slice is the height \( \frac{1}{4}(x^2 - 7) \) times the width \( \Delta x \). Sum these areas from \( x=0 \) to \( x=2 \) to approximate the entire region: \( \sum \frac{1}{4}(x^2 - 7) \Delta x \).

Set Up the Integral

To find the exact area, set up an integral with limits from \( x = 0 \) to \( x = 2 \). The integral becomes: \[ \int_{0}^{2} \left( \frac{1}{4}(x^2 - 7) \right) \, dx \]. Simplify and evaluate this integral to find the exact area.

Evaluate the Integral

Evaluate the integral \( \int_{0}^{2} \frac{1}{4}(x^2 - 7) \, dx \). This simplifies to \( \frac{1}{4} \left( \int_{0}^{2} x^2 \, dx - \int_{0}^{2} 7 \, dx \right) \). Calculate each part: \[ \frac{1}{4} \left( \left[ \frac{x^3}{3} \right]_{0}^{2} - 7[x]_{0}^{2} \right) \]. This simplifies to \[ \frac{1}{4} \left( \frac{8}{3} - 14 \right) = \frac{1}{4} \left( -\frac{34}{3} \right) = -\frac{34}{12} = -\frac{17}{6} \]. Hence, since the region is below \( y = 0 \), use the absolute value of the integral for area: \( \frac{17}{6} \).

Estimate the Area

Confirm the answer by roughly estimating. Divide the x-interval from 0 to 2 into smaller intervals, approximate heights at these points using the parabola equation, average these heights, and multiply by the base length. Check this estimate against your calculated integral to ensure consistency.

Key concepts

Riemann Sum

The concept of a Riemann Sum is an essential tool for approximating the area under a curve, especially when dealing with irregular shapes. In our exercise, we use a Riemann Sum to estimate the area between the curve of the parabola and the x-axis from \( x = 0 \) to \( x = 2 \). To do this, we divide the area into several thin vertical strips, each having width \( \Delta x \) and multiplying by the function's height at that point.

This method involves summing up the products between the strip's height, \( \frac{1}{4}(x^2 - 7) \), and the width. Mathematically, we express this as:

• The Riemann Sum: \( \sum \frac{1}{4}(x^2 - 7) \Delta x \)
• As \( \Delta x \rightarrow 0 \), our approximation approaches the exact area.

By using the Riemann Sum, we effectively prepare for the transition towards calculating the exact area using the Definite Integral.

Definite Integral

Once we've approximated an area using a Riemann Sum, the next step is to calculate the exact area with a Definite Integral. In the context of our problem, we set up the integral to precisely measure the area under the curve \( y = \frac{1}{4}(x^2 - 7) \) from \( x = 0 \) to \( x = 2 \).

The Definite Integral provides a complete calculation by integrating the function over the specified interval:

• Integral Setup: \[ \int_{0}^{2} \left( \frac{1}{4}(x^2 - 7) \right) \, dx \]
• This represents the sum of all infinitesimal strips from \( x = 0 \) to \( x = 2 \).
• The solution \(-\frac{34}{12}\) becomes positive upon recognizing the area below the x-axis, resulting in \( \frac{17}{6} \).

With this integral, we confirm that the exact bounded area is \( \frac{17}{6} \) units squared.

Parabola

A parabola is a symmetrical curve that plays a crucial role in our exercise. Our specific equation, \( y = \frac{1}{4}(x^2 - 7) \), defines the shape and orientation of the parabola within the x-y coordinate plane.

Key features of a parabola include:

• Vertex: the point of minimum or maximum curvature, here above \( x = 0 \).
• Orientation: opens upwards, as indicated by the positive fraction \( \frac{1}{4} \).
• Intercepts: intersects the x-axis at \( x = \pm \sqrt{7} \), though we only consider up to \( x = 2 \) here.

Understanding these features aids in sketching and helps visualize the region whose area we're calculating.

Sketching Graphs

Sketching graphs is an invaluable step when solving calculus problems, especially to capture the region of interest accurately. For our problem, two graphs are involved: the parabola \( y = \frac{1}{4}(x^2 - 7) \) and the line \( y = 0 \), which is the x-axis.

Here's how to effectively sketch these graphs:

• Define the vertex and orientation of the parabola first.
• Mark intercepts clearly, knowing the parabola crosses the x-axis at \( x = 0 \) to \( x = 2 \).
• Plot both curves on the same graph, identifying the region bound above by the parabola and below by the x-axis.

This visual step provides clarity, ensures accuracy in your calculations, and confirms the area being evaluated within set bounds.