Problem 12 - Calculus: Early Transcendentals

Question

Find the area bounded by the curves. \(y=x^{2}-2 x\) and \(y=x-2\)

Step-by-Step Solution

Answer

The area bounded by the curves is \(\frac{1}{6}\) square units.

Step 1: Set Equations Equal to Find Intersection Points

To find the points where the two curves intersect, set the equations equal to each other: \[ x^2 - 2x = x - 2 \]Simplify to get: \[ x^2 - 3x + 2 = 0 \]This equation can be solved by factoring.

Step 2: Factor the Quadratic Equation

The quadratic equation \(x^2 - 3x + 2 = 0\) can be factored into:\[ (x-1)(x-2) = 0 \]This gives solutions for \(x\) as 1 and 2. These points determine the interval over which we'll integrate.

Step 3: Determine Functions' Relative Positions

Between the intersection points \(x = 1\) and \(x = 2\), evaluate which function is on top by plugging in a midpoint, such as \(x = 1.5\):For \(y = x^2 - 2x\), at \(x = 1.5\), \(y = 1.5^2 - 2 \cdot 1.5 = -0.75\).For \(y = x - 2\), at \(x = 1.5\), \(y = 1.5 - 2 = -0.5\).Thus, \(y = x - 2\) is on top.

Step 4: Set Up the Integral for Area

Since \(y = x - 2\) is on top, the area \(A\) between the curves can be found by:\[ A = \int_{1}^{2} ((x-2) - (x^2 - 2x)) \; dx \]Simplify the expression inside the integral:\[ A = \int_{1}^{2} (-x^2 + 3x - 2) \; dx \]

Step 5: Integrate to Find the Area

Perform the integration: \[A = \int_{1}^{2} (-x^2 + 3x - 2) \; dx = \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_{1}^{2} \]Calculate this:\[= \left(-\frac{2^3}{3} + \frac{3(2)^2}{2} - 2(2)\right) - \left(-\frac{1^3}{3} + \frac{3(1)^2}{2} - 2(1)\right)\]\[= \left(-\frac{8}{3} + 6 - 4 \right) - \left(-\frac{1}{3} + \frac{3}{2} - 2 \right) \]\[= \left(-\frac{8}{3} + 2 \right) - \left(-\frac{1}{3} + \frac{3}{2} - 2 \right)\]\[= \left(\frac{-8}{3} + \frac{6}{3}\right) - \left(\frac{-1}{3} + \frac{4.5}{3} - \frac{6}{3}\right) \]\[= \left(-\frac{2}{3}\right) - \left(-\frac{1}{3} + \frac{1.5}{3}\right) \]\[= -\frac{2}{3} + \frac{5}{6}\]Simplify to get the final area:\[= \frac{5}{6} - \frac{4}{6} = \frac{1}{6}\]

Step 6: Conclude with the Calculated Area

The area bounded by the curves \(y = x^2 - 2x\) and \(y = x - 2\) between \(x = 1\) and \(x = 2\) is \(\frac{1}{6}\) square units.

Key Concepts

Intersection Points

To find the area between two curves, identifying the intersection points is a crucial step. Intersection points are where the curves meet, and they determine the boundaries for calculating the area. In our problem, we set the given equations equal: \[ x^2 - 2x = x - 2 \]. This is necessary because at these points, both functions have the same value of \(y\) for a particular \(x\). Solving for \(x\) gives the intersection points, which are critical for knowing the limits of integration for the area calculation. By rearranging and simplifying the equation to \(x^2 - 3x + 2 = 0\), we find that the curves intersect at \(x = 1\) and \(x = 2\). These values show us that the relevant section of the curves lies between these two points. Look out for these intersection points because without them, determining the limits for any integral that defines area would be a challenge.

Quadratic Factoring

Factoring a quadratic equation is an algebraic method used to solve equations in the form of \(ax^2 + bx + c = 0\). This process is key in our exercise as it allows us to find the \(x\) values where the curves intersect. Once we rearranged and simplified the intersection equation to \(x^2 - 3x + 2 = 0\), we factored it as \((x-1)(x-2) = 0\). Quadratic factoring turns the equation into a product of linear equations, which brings solutions when set to zero.

• First, identify factors of the constant term (here, 2) that add up to the coefficient of the linear term (here, -3).
• Next, rewrite the quadratic as a product of binomials, \((x-1)(x-2) = 0\).
• Finally, solve each binomial by setting them equal to zero, giving \(x = 1\) and \(x = 2\).

This method provides a simple technique to find roots of quadratic equations, allowing us to efficiently identify intersection points for further computations or analysis.

Integral Calculus

Integral calculus is used to calculate the area under curves. To find the area between two curves, we use the concept of integrating the difference between the functions over the interval defined by their intersection points.

• Firstly, check which function is on top between the intersection points by evaluating their values within the interval. In this exercise, between \(x = 1\) and \(x = 2\), the function \(y = x - 2\) lies above \(y = x^2 - 2x\).
• Set up the integral as the difference \((x-2) - (x^2 - 2x)\), representing the vertical distance between the two functions.
• This expression, once simplified to \(-x^2 + 3x - 2\), becomes the integrand.

The integration technique involves finding the anti-derivative of this expression over the interval \([1, 2]\). It calculates a definite measure - here, the area between the two curves. Understanding integral calculus is essential as it translates geometric concepts into solvable mathematical problems.

Definite Integral

The concept of a definite integral enables us to calculate an exact area between curves over specific intervals. A definite integral has limits, which in this case are the intersection points \(x = 1\) to \(x = 2\). The expression inside the integral, the integrand, is the simplified form of the difference between the two given functions.

• Set up the definite integral: \[ A = \int_{1}^{2} (-x^2 + 3x - 2) \, dx \]
• Perform the integration to find the anti-derivative of the polynomial expression.
• Evaluate this anti-derivative at the boundaries of the interval \(x = 1\) and \(x = 2\).
• Subtracting these values gives the exact area between the curves over this range.

In our problem, the calculation yielded an area of \(\frac{1}{6}\) square units. This showcases the power of definite integrals in providing precise quantities derived from continuous data, like curves on a graph.