Problem 22 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=\sqrt{x-1}, y=3-x, y=0$$

Step-by-Step Solution

Answer

Total area bounded by curves is 15.5 square units.

Step 1: Find the Intersection Points

First, determine where the curves intersect by equating the functions. Set \(y = \sqrt{x-1}\) and \(y = 3-x\), and solve for \(x\). We have \(\sqrt{x-1} = 3-x\). Square both sides to get \(x - 1 = (3-x)^2\). Simplifying gives \(x^2 - 7x + 10 = 0\). Solving this quadratic equation, we get the solutions \(x = 2\) and \(x = 5\). Thus, the curves intersect at these points.

Step 2: Determine the Integration Intervals

The regions of interest are between \(y = 0\), \(y = \sqrt{x-1}\), and \(y = 3-x\). The three intervals are from the x-intercept of \(y=\sqrt{x-1}\) (which is \(x=1\)), to \(x=2\), and then from \(x=2\) to \(x=5\). For these intervals, the functions that bound the areas differ.

Step 3: Integrate to Find Area from x=1 to x=2

From \(x=1\) to \(x=2\), the area is bounded above by \(y = \sqrt{x-1}\) and below by \(y = 0\). Calculate the integral: \[A_1 = \int_{1}^{2} \sqrt{x-1} \, dx.\] Use substitution, let \(u = x-1\), then \(du = dx\) and \(\int \sqrt{u} \, du = \frac{2}{3}u^{3/2} + C\). Evaluate from 0 to 1 to get:\[A_1 = \frac{2}{3}(1^{3/2} - 0^{3/2}) = \frac{2}{3}.\]

Step 4: Integrate to Find Area from x=2 to x=5

From \(x=2\) to \(x=5\), the area is bounded above by \(y=3-x\) and below by \(y=\sqrt{x-1}\). Calculate this area using the integral: \[A_2 = \int_{2}^{5} ((3-x) - \sqrt{x-1}) \, dx.\] Split into two integrals: \(\int_{2}^{5} (3-x) \, dx - \int_{2}^{5} \sqrt{x-1} \, dx\). For the first, \(\int (3-x) \, dx = 3x - \frac{x^2}{2}\). Evaluate from 2 to 5:\[= [3(5) - \frac{5^2}{2}] - [3(2) - \frac{2^2}{2}] = 4.5.\] For the second, repeat Step 3 techniques for \(\int \sqrt{x-1} \, dx\) on this interval, resulting in \(-\frac{12}{3}\). Combine these to obtain \[A_2 = 4.5 - (-12/3) = 4.5 + 4 = 8.5.\]

Step 5: Calculate Total Area

Sum the two separate area calculations to find the total bounded area: \[\text{Total Area} = A_1 + A_2 = \frac{2}{3} + 8.5 = \frac{2}{3} + \frac{17}{2} = \frac{42}{6} + \frac{51}{6} = \frac{93}{6} = 15.5.\]

Key Concepts

Intersection Points

In any problem involving finding the area between curves, identifying the intersection points is essential. These points reveal where the curves cross each other, and they will serve as the limits for our definite integrals. In our example, we considered the curves given by the equations:

• \(y = \sqrt{x-1}\)
• \(y = 3-x\)

To find the intersection points, we equated the two equations: \[\sqrt{x-1} = 3-x\] By solving this equation:

• Square both sides to eliminate the square root, resulting in \(x - 1 = (3-x)^2\).
• This expands and simplifies to \(x^2 - 7x + 10 = 0\).

The quadratic formula or factoring gives solutions at \(x = 2\) and \(x = 5\). These are the critical points where the functions intersect and define the intervals for integration.

Definite Integrals

Definite integrals are powerful tools in calculus used to find the accumulated quantities such as areas under curves. This is critical for problems involving finding bounded areas. After finding the intersection points, we set up definite integrals that describe the area between the functions over specific intervals. Consider the integration steps in our exercise:

• For the first interval from \(x = 1\) to \(x = 2\), we integrate the function \(y = \sqrt{x-1}\). This was expressed as:\[\int_{1}^{2} \sqrt{x-1} \, dx\]


• For the second interval from \(x = 2\) to \(x = 5\), the area is between the curves \(y = 3-x\) and \(y = \sqrt{x-1}\). Thus, the definite integral becomes:\[\int_{2}^{5} (3-x - \sqrt{x-1}) \, dx\].

Each definite integral provides the exact area below the curve and above the x-axis (or between two curves) within the specified limits, contributing to finding the total area of the region.

Area Between Curves

Calculating the area between two or more curves provides the quantitative measure of space enclosed by the curves. It's a common application of integration and highlights the importance of both definite integrals and intersection points. By breaking down the problem into intervals determined by the intersection points, we can focus on calculating the area for each section where one curve is consistently above the other.

• For \(x = 1\) to \(x = 2\), the region is simply beneath the curve \(y = \sqrt{x-1}\) and above the x-axis.


• From \( x = 2 \) to \( x = 5 \), it gets more intricate, as \(y = 3-x\) is uppermost while \(y = \sqrt{x-1}\) is below. Here, integrals are used to find the difference in height between the two functions.

The combined result from each interval gives the total area. In our solution, this process is carried out by adding up the areas from each calculated interval, leading to a precise calculation of the area bounded by the curves.