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. \(y=(x-4)(x+2), y=0\), between \(x=0\) and \(x=3\)

Short answer

The area of the region is 24 square units.

Step-by-step solution

Sketch the Graphs

First, plot the equations to visualize the region bounded by them. The equation \(y = (x-4)(x+2)\) is a quadratic function opening upwards, intersecting the x-axis at \(x=4\) and \(x=-2\). The line \(y=0\) represents the x-axis. We consider the region between \(x=0\) and \(x=3\), so sketch portions of the graph that lie within these x-values. This will give the parabola that intersects the x-axis in this interval.

Typical Slice

Draw a vertical slice (representing a typical strip for integration) within the interval from \(x=0\) to \(x=3\). The top of the strip will be bounded by the function \(y=(x-4)(x+2)\), and it will touch the x-axis forming the base of the slice. The height of the slice at any point \(x\) is given by \((x-4)(x+2)\) - 0, the function value minus zero.

Approximate the Area

Use a series of vertical slices to approximate the area. The approximate total area can be calculated by summing up the areas of these slices (i.e., rectangles) over the interval from \(x=0\) to \(x=3\). The area of each slice is roughly \((x-4)(x+2) \, \Delta x\), where \(\Delta x\) is a small change in x.

Set Up the Integral

The integral to find the area under the curve from \(x=0\) to \(x=3\) replaces the approximate sum for the exact calculation. Set up the integral as: \[ A = \int_{0}^{3} (x-4)(x+2) \, dx \]This integral will give the exact area between the curve and the x-axis over the specified interval.

Calculate the Integral

Expand the product in the integrand: \[ (x-4)(x+2) = x^2 + 2x - 4x -8 = x^2 - 2x - 8 \].Substitute back into the integral to get: \[ A = \int_{0}^{3} (x^2 - 2x - 8) \, dx \].Evaluate the integral: \[ A = \left[ \frac{x^3}{3} - x^2 - 8x \right]_{0}^{3} \].Compute the definite integral boundaries by substituting \(x=3\) and \(x=0\):\[ A = \left( \frac{3^3}{3} - 3^2 - 8 \times 3 \right) - \left( \frac{0^3}{3} - 0^2 - 8 \times 0 \right) \].Calculate this to get:\[ A = (9 - 9 - 24) - (0) = -24 \].Since we are calculating area, take the absolute value: \[ A = 24 \].

Confirm Estimate

Checking the work by estimating visually or calculating with a simpler approach suggests that the approximate area is indeed correct when we consider the behavior of the parabolic curve in this interval. Thus, the integral calculation correctly approximates the confined area under the parabola and above the x-axis from \(x=0\) to \(x=3\) by confirming against an estimate or intuitive reasoning based on visual inspection.

Key concepts

Area Under a Curve

In calculus, finding the area under a curve is a fundamental use of integration. This concept involves calculating the total space that lies between a curve and a given axis, often over a specific interval of the x-axis. Imagine shading the space under a bridge's arc; this shaded area is what we're aiming to determine when integrating. The area provides valuable information like physical displacement and total revenue in real-world contexts. Essentially, when we set up an integral, we're using it to accumulate the infinitely thin slices (rectangles) that make up the entire region under a curve. In this case, for the equation \(y = (x-4)(x+2)\), and over the interval \(x = 0\) to \(x = 3\), the integral allows us to precisely calculate the area under the curve, which lies above the x-axis within this range.

Definite Integral

A definite integral is a type of integral with set boundary values on an interval. It gives us the exact accumulated value, like area under a curve, for functions between two points. Unlike an indefinite integral, which represents a family of functions, a definite integral provides a specific numerical value. The expression for a definite integral is written with limits of integration; for example, \(\int_{a}^{b} f(x) \, dx\). These limits \(a\) and \(b\) represent the interval on the x-axis over which we perform the integration. In the provided solution, the integral \(\int_{0}^{3} (x^2 - 2x - 8) \, dx\) is calculated. After integrating and evaluating this from \(x=0\) to \(x=3\), we derive the exact area beneath the curve, gaining insight into the behavior of the graph within those bounds. Such applications are pivotal in fields like engineering and finance, where precise calculations are vital.

Parabolic Regions

The term 'parabolic regions' refers to areas bound by curves of a parabola, which is shaped like an arch and described by quadratic equations. Our focus is on the curve formed by \(y = (x-4)(x+2)\), a parabola opening upwards. This equation is a product of two linear factors, and its graph acts as a bowl or cup in which the region we're interested in lies. Understanding parabolic regions is essential, as they appear frequently in physics and economics, modeling objects in motion under gravity or utility maximization scenarios. The curve intersects the x-axis at roots \(x = 4\) and \(x = -2\), creating sections above or below the x-axis. By focusing on \(x = 0\) to \(x = 3\), we isolate the specific sub-area of the parabola, aiding precise calculations of confined space using integration methods.

Graphing Techniques

Graphing techniques are invaluable for visualizing functions and understanding their underlying behavior. They assist in identifying key features like roots, intersections, and regions of interest such as maximum or minimum points. When graphing the function \(y = (x-4)(x+2)\), we recognize it as a parabola intersecting the x-axis at its roots. By sketching this with a clear view of its span over the x-axis from 0 to 3, we pinpoint the specific portion where integration applies. This area, above the x-axis under the parabola within our specified interval, becomes crucial for predicting the function's behavior and real-world applications. Graphs simplify complex algebraic expressions by presenting them visually, enabling more intuitive and quicker insights. Whether for academic exercises or practical problems, graphing techniques act as a bridge, connecting abstract mathematical concepts to tangible understanding.