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. \(y=\sqrt{x}, y=x-4, x=0\)

Short answer

The total area of the bounded region is \(\frac{76}{9}\) square units.

Step-by-step solution

Understand the Graphs and Intersection Points

Plot the graphs of the equations given: \(y=\sqrt{x}\) and \(y=x-4\). The line \(x=0\) is the y-axis, which will serve as a boundary. First, find where these curves intersect by solving the equations \(\sqrt{x} = x-4\). Solving this will give the intersection points.

Solve for Intersection Points

To find the intersection points, set \(\sqrt{x} = x - 4\). Squaring both sides gives \(x = (x-4)^2\), expanding and rearranging gives \(x^2 - 9x + 16 = 0\). Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) will provide the x-values for intersection points: \(x = 1\) and \(x = 8\).

Identify the Bounded Region

The bounded region is between \(x = 0\) and \(x = 8\); from \(y = \sqrt{x}\) to \(y = x - 4\). We focus on the area between these curves from their intersection point at \(x = 1\) to \(x = 4\) for \(x-4 \leq 0\) and \(x=4\) to \(x = 8\) for \(x-4 > 0\)

Approximate and Set Up the Integral

For the interval \([0, 1]\), the area is between the curve \(y = \sqrt{x}\) and \(y = 0\). For \([1, 8]\), it switches to between \(y = x - 4\) and \(y = \sqrt{x}\). The areas can be represented by integrals:\[ A_1 = \int_0^1 \sqrt{x} \; dx \] \[ A_2 = \int_1^4 \sqrt{x} \; dx - \int_1^4 (x - 4) \; dx \]\[ A_3 = \int_4^8 (x - 4) - \sqrt{x} \; dx \]

Calculate the Area Using Integrals

Calculate the areas:1. \[ A_1 = \int_0^1 \sqrt{x} \; dx = \left[ \frac{2}{3}x^{3/2} \right]_0^1 = \frac{2}{3} \]2. \[ A_2 = \int_1^4 \sqrt{x} \; dx - \int_1^4 (x-4) \; dx \] - \(\int_1^4 \sqrt{x} \; dx = \left[ \frac{2}{3}x^{3/2} \right]_1^4 = \frac{14}{3} \) - \(\int_1^4 (x-4) \; dx = \left[ \frac{x^2}{2} - 4x \right]_1^4 = -\frac{9}{2} \) - \(A_2 = \frac{14}{3} + \frac{9}{2} = \frac{61}{6} \)3. \[ A_3 = \int_4^8 (x-4) - \sqrt{x} \; dx \] - \(= \left[ \frac{x^2}{2} - 4x \right]_4^8 - \left[ \frac{2}{3}x^{3/2} \right]_4^8 = \frac{36}{2} - \frac{32}{3} = \frac{10}{3} \) Sum these areas: \( A_1 + A_2 + A_3 = \frac{2}{3} + \frac{61}{6} + \frac{10}{3} = \frac{76}{9}\)

Confirm the Approximation

Estimate the bounded area using geometric intuition, such as considering trapezoids formed by the straight line and the curved line. This verification ensures that your integration covered the accurate areas. Ensure the results are congruent with estimates such as checking using graphing tools to visually compare the regions.

Key concepts

Definite Integrals

Definite integrals are fundamental in calculus for finding the total accumulation of quantities, such as area. When computing the area between curves, we use definite integrals to consolidate our findings into a single value within a specified interval. This process involves:

• Identifying the limits of integration, which are determined by the x-values of the points where the curves intersect.
• Calculating the integral of the top function minus the bottom function, given that the area between two curves is essentially the difference in their y-values across the interval.

For example, when finding the area between two curves such as the given functions, it's crucial to determine the upper and lower curves in the interval for setting up your integral correctly.

Area Between Curves

Calculating the area between two curves involves analyzing the graphs of the equations and determining the section where they overlap. First, sketch the curves, identifying intersection points, which serve as key boundaries. The process generally follows these steps:

• Determine the upper and lower curves between the intersection points.
• Set up the integral by subtracting the lower curve function from the upper curve function.
• Evaluate this integral between the intersection limits.

For curves like \(y = \sqrt{x} \) and \(y = x - 4\), begin by identifying intersection points; here, these points are \(x = 1\) and \(x = 8\). Then, divide the problem into segments where different curves act as upper or lower bounds. This ensures accurate integration across varied sections.

Quadratic Equations

Quadratic equations are crucial in solving intersections of curves. They appear when equations like \(\sqrt{x} = x - 4\) are rearranged and manipulated. Squaring both sides transforms these into quadratic equations like \(x^2 - 9x + 16 = 0\), which are solvable with the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides roots of the quadratic, giving us the x-coordinates where the curves intersect. In the given exercise, solving the equation yields x-values of \(x = 1\) and \(x = 8\), essential for defining the integration limits and partitioning the area calculation.

Graphing Functions

Graphing functions assists in visualizing the arrangement of curves spatially. By plotting the equations, you can determine which function is on top, guiding your integration setup. Steps include:

• Plotting each equation on the Cartesian plane.
• Observing where curves meet, which informs intersection points and integration limits.
• Summarizing which areas lie above or below each curve across specified intervals.

In the exercise, sketch the graphs of equations like \(y = \sqrt{x} \) and \(y = x - 4\), alongside boundary \(x = 0\), forming a full picture of the region to be integrated. Graphing helps confirm the step-by-step solution, ensuring clarity in problem-solving and accuracy in further computation.