Problem 18 - Calculus: Early Transcendentals

Question

Sketch the region enclosed by the given curves and find its area. \( y = \sqrt{x - 1} \) , \( x - y = 1 \)

Step-by-Step Solution

Answer

The area enclosed by the curves is \( \frac{1}{6} \).

Step 1: Express Both Curves in Terms of y

The given equations are: \( y = \sqrt{x - 1} \) and \( x - y = 1 \). First, solve \( x - y = 1 \) for \( y \).\[y = x - 1\]Now, both curves are expressed as functions of \( x \): \( y = \sqrt{x-1} \) and \( y = x-1 \).

Step 2: Find the Points of Intersection

Set the equations equal to each other to find the points of intersection:\[\sqrt{x - 1} = x - 1\]\(\Rightarrow x - 1 = (x - 1)^2\). Solve this quadratic equation: \[x^2 - 3x + 2 = 0\]\[(x - 1)(x - 2) = 0\]\( x = 1 \) or \( x = 2 \). Substitute these \( x \) values back to find \( y \): For \( x = 1 \), \( y = \sqrt{1 - 1} = 0 \). For \( x = 2 \), \( y = \sqrt{2 - 1} = 1 \). So, points are \((1, 0)\) and \((2, 1)\).

Step 3: Sketch the Curves

Plot the curve \( y = \sqrt{x - 1} \) which starts from \( x = 1 \) and opens rightwards, reaching \( (2, 1) \). Also, plot the line \( y = x - 1 \) which is a straight line passing through points \( (1, 0) \) and \( (2, 1) \). The region enclosed is bounded between \( x = 1 \) and \( x = 2 \) by these two equations.

Step 4: Calculate the Area Between Curves

The curves intersect at \( x = 1 \) and \( x = 2 \). To find the area between the curves, compute the integral of the difference of the functions from \( x = 1 \) to \( x = 2 \):\[\int_{1}^{2} ((x - 1) - \sqrt{x - 1})\, dx\]. Evaluate this integral to find the enclosed area. First, calculate:\[\int_{1}^{2} (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2} = \left( \frac{4}{2} - 2 \right) - \left( \frac{1}{2} - 1 \right) = 0.5\]Then, calculate:\[\int_{1}^{2} \sqrt{x - 1} \, dx = \left[ \frac{2}{3}(x-1)^{3/2} \right]_{1}^{2} = \frac{2}{3}\]Thus, the area enclosed by the curves is:\[0.5 - \frac{2}{3} = \frac{1}{6}\].

Step 5: Conclude the Calculation Process

Finish by stating the area of the region enclosed by the intersecting curves based on the previous calculations.

Key Concepts

Area Between Curves

The concept of finding the area between curves is a fundamental topic in calculus that deals with calculating the region enclosed by two curves on a graph. To find this area, one typically sets up an integral that represents the difference between the functions defining the curves. For our specific example, the curves we are looking at are represented by the equations:

• \( y = \sqrt{x-1} \)
• \( y = x-1 \)

The area between these curves is calculated using the integral of the difference:

• \[ \int_{1}^{2} ((x - 1) - \sqrt{x - 1}) \, dx \]

This particular integral will provide the net area between the curves from one point of intersection to the other. It is important to note that the upper curve in the integral should be subtracted by the lower curve, which ensures a positive result.

Points of Intersection

To determine where two curves intersect, we equate their respective equations and solve for the points they share in common. For the given exercise, the equations are:

• \( y = \sqrt{x - 1} \)
• \( y = x - 1 \)

We set them equal to find intersections:

• \( \sqrt{x - 1} = x - 1 \)

Solving this equation simplifies into:

• \( x - 1 = (x - 1)^2 \)

This is solved by re-arranging into quadratic form and factoring, which yields:

• \( (x - 1)(x - 2) = 0 \)

Thus, the solutions are \( x = 1 \) and \( x = 2 \). Substituting these values back into the original equations helps us find the corresponding y-values, giving us the points

• \((1, 0)\)
• \((2, 1)\)

These points mark where the curves cross each other.

Integral Calculus

Integral calculus, part of the broader scope of calculus, is instrumental in computing areas under curves and, by extension, the area between curves. In solving problems like these, we take advantage of definite integrals to find such areas efficiently.
When dealing with two functions, say \( f(x) \) and \( g(x) \), where \( f(x) \) is the upper function and \( g(x) \) is the lower over a certain interval [a, b], the area is given by:

• \[ \int_{a}^{b} (f(x) - g(x)) \, dx \]

The integration process involves evaluating this integral, taking into account any simplifications that make the calculation direct and accurate. For the example we are exploring, the definite integral set between the points of intersection \( x = 1 \) and \( x = 2 \) requires evaluating two separate integrals:

• \( \int_{1}^{2} (x-1) \, dx \)
• \( \int_{1}^{2} \sqrt{x-1} \, dx \)

These integrals combine to yield the area as the difference between the two calculated values.

Sketching Regions

Sketching the region enclosed by curves is crucial in visualizing and understanding the problem at hand. It helps to identify the bounds and behavior of each curve. In this exercise, we have:

• The curve \( y = \sqrt{x - 1} \), a rightward-opening parabola starting at \( x = 1 \).
• The line \( y = x - 1 \) which is a straight line intersecting the previous curve.

When sketching, plot the known points

• \((1, 0)\) and
• \((2, 1)\)

for clear visualization. Additionally, marking the intersections simplifies determining which function lies above the other between the established limits.
This visualization confirms the region's bounds and ensures the correct identification and integration of the area between these curves. It clarifies any potential ambiguity in setting up the integral setup while enhancing comprehension of the integral calculus involved.