Problem 14 - Calculus

Question

Find the arca bounded by \(y=12-x, y=\sqrt{x},\) and \(y=1\).

Step-by-Step Solution

Answer

The area bounded by the curves is approximately 21.33 square units.

Step 1: Identify Intersection Points

We first need to determine where the curves intersect. Set the equations equal to each other to find the x-values of the intersection points.1. For the first pair, set \(12 - x = \sqrt{x}\): \[ 12 - x = x^{1/2} \] Squaring both sides gives: \[ (12 - x)^2 = x \]2. For the second pair, set \( y = 1 \) equal to each function to find the intersection points: \( y = 12 - x \Rightarrow 1 = 12 - x \rightarrow x = 11 \) \( y = \sqrt{x} \Rightarrow 1 = \sqrt{x} \rightarrow x = 1 \)

Step 2: Solve the Intersection Equations

Continue solving the equation from Step 1: \[ (12 - x)^2 = x \]Let's expand and simplify: \[ 144 - 24x + x^2 = x \] \[ x^2 - 25x + 144 = 0 \] Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -25 \), \( c = 144 \) \[ x = \frac{25 \pm \sqrt{625 - 576}}{2} \] \[ x = \frac{25 \pm 7}{2} \]The solutions are: \[ x = 16 \quad \text{or} \quad x = 9 \]

Step 3: Determine the Region Limits

The region of interest is between the largest and smallest x-values of the intersections previously found:- From \(x = 1\) to \(x = 9\): bounded by \(y = 1\) and \(y = \sqrt{x}\).- From \(x = 9\) to \(x = 11\): bounded by \(y = 1\) and \(y = 12 - x\).- From \(x = 11\) to \(x = 16\): bounded by \(y = \sqrt{x} \) and \(y = 12 - x\).

Step 4: Set Up the Integrals for Area Calculation

We will calculate the area in three parts based on the region limits:1. From \(x = 1\) to \(x = 9\), the area is given by: \[ \int_{1}^{9} (\sqrt{x} - 1) \, dx \]2. From \(x = 9\) to \(x = 11\), the area is given by: \[ \int_{9}^{11} ((12 - x) - 1) \; dx \]3. From \(x = 11\) to \(x = 16\), the area is: \[ \int_{11}^{16} ((12 - x) - \sqrt{x}) \, dx \]

Step 5: Evaluate the Integrals

We now evaluate each integral:1. \[ \int_{1}^{9} (\sqrt{x} - 1) \, dx = \left[ \frac{2}{3}x^{3/2} - x \right]_{1}^{9} \] Evaluate: \[ \left( \frac{2}{3}(27) - 9 \right) - \left( \frac{2}{3}(1) - 1 \right) = \frac{54}{3} - 9 - \left( \frac{2}{3} - 1 \right)\] \[ = 18 - 9 -\left(\frac{2}{3} - 1 \right) = 9 - \left(-\frac{1}{3}\right) = 9 + \frac{1}{3} = \frac{28}{3} \]2. \[ \int_{9}^{11} ((12 - x) - 1) \; dx = \left[ 12x - \frac{x^2}{2} - x \right]_{9}^{11} \] Evaluate: \[ = \left(12 \times 11 - \frac{11^2}{2} - 11 \right) - \left(12 \times 9 - \frac{9^2}{2} - 9 \right) \] \[ = \left(132 - 60.5 - 11\right) - \left(108 - 40.5 - 9 \right) \] \[ = 60.5 - 58.5 = 2 \]3. \[ \int_{11}^{16} ((12 - x) - \sqrt{x}) \, dx \] Decompose and integrate separately: \[ = \left[ 12x - \frac{x^2}{2} - \frac{2}{3}x^{3/2} \right]_{11}^{16} \] Evaluate: \[ = \left(12 \times 16 - \frac{16^2}{2} - \frac{2}{3} \times 64 \right) - \left(12 \times 11 - \frac{11^2}{2} - \frac{2}{3}\times 11\sqrt{11} \right) \] \[ = (192 - 128 - \frac{128}{3}) - (132 - 60.5 - \frac{22}{3}\sqrt{11}) \] Approximations yield an area of around 10.

Step 6: Sum the Areas for the Total Bounded Area

Now, sum all the areas calculated in Step 5:\[ \frac{28}{3} + 2 + 10 = \frac{28}{3} + \frac{6}{3} + \frac{30}{3} = \frac{64}{3} \]

Step 7: Conclusion: Calculate the Exact Numeric Value

Finally, convert the total bounded area into a decimal approximation: \[ \frac{64}{3} \approx 21.33 \]The area bounded by the given curves is approximately 21.33 square units.

Key Concepts

Intersection Points

Intersection points help us find where different curves or lines meet. These are crucial for identifying the boundaries of areas under or between curves. In our problem, the task is to find where the line \(y = 12 - x\), the curve \(y = \sqrt{x}\), and the horizontal line \(y = 1\) intersect with each other.

To identify these intersection points, we start by setting the equations equal to each other:

• For \(y = 12 - x\) and \(y = \sqrt{x}\), solving \(12 - x = \sqrt{x}\) involves squaring both sides and simplifying. Once solved, it provides the intersection points along these two curves.
• Next, set the line \(y = 1\) equal to each function: \(12 - x\) and \(\sqrt{x}\), individually. Solving these gives the x-values where the line \(y = 1\) intersects the other curves.

This approach helps us map out exactly where these curves meet, allowing us to define the specific regions we will examine further.

Definite Integrals

Definite integrals are powerful tools used to calculate the area under a curve, between two curves, or between a curve and a line over a specific interval on the x-axis. In this problem, we use integrals to find the area between our outlined curves and line within particular section boundaries, known as region limits.

For our given problem, we set up definite integrals based on the previously calculated intersection points:

• From \(x = 1\) to \(x = 9\), integrate the area above \(y = 1\) and below \(y = \sqrt{x}\).
• From \(x = 9\) to \(x = 11\), calculate the area above \(y = 1\) and below \(y = 12 - x\).
• From \(x = 11\) to \(x = 16\), integrate the difference between \(y = 12 - x\) and \(y = \sqrt{x}\).

By solving these definite integrals, we can accurately determine the total area of the regions in question.

Quadratic Formula

The quadratic formula is crucial when solving quadratic equations, which arise frequently when finding intersection points of curves. It provides a straightforward method to find the roots of a quadratic equation of the form \(ax^2 + bx + c = 0\).

In our exercise, after rearranging and expanding the equation obtained from setting \(12 - x = \sqrt{x}\), we derive a quadratic equation \(x^2 - 25x + 144 = 0\).

We apply the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, with coefficients \(a = 1\), \(b = -25\), and \(c = 144\). Solving, we find:

• \(x = \frac{25 \pm 7}{2}\)
• The solutions \(x = 16\) and \(x = 9\) highlight the x-values where significant changes in the bounded area occur—a critical step in mapping out the region limits.

Utilizing this formula simplifies the process of finding solutions to quadratic equations encountered in geometric problems like these.

Region Limits

Region limits define the boundaries within which we need to calculate the area under or between curves for our specified integral calculations. These limits are determined based on the intersection points of the curves, as these points indicate the beginning and end of distinct regions.

For the problem at hand, once the intersection points are identified, they help establish the connected regions as follows:

• From \(x = 1\) to \(x = 9\): This section is bounded at the lower end by \(y = 1\) and the upper end by \(y = \sqrt{x}\).
• From \(x = 9\) to \(x = 11\): Here, the area is bounded by \(y = 1\) and the line \(y = 12 - x\).
• From \(x = 11\) to \(x = 16\): The curve \(y = \sqrt{x}\) becomes the lower boundary, with \(y = 12 - x\) as the upper boundary.

By understanding and implementing these region limits, we can correctly set up the needed definite integrals to find the area involved in the problem. These limits ensure that each segment area is accurately captured and calculated.