Problem 37 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

AREA BETWEEN CURVES In Exercises 31 through 38 , sketch the indicated region \(R\) and find its area by integration. \(R\) is the region bounded by the curve \(y=2+x-x^{2}\) and the \(x\) axis.

Step-by-Step Solution

Answer

The area is \( \frac{16}{3} \).

Step 1 - Identify the Intersection Points

To find the area between the curve and the x-axis, first determine where the curve intersects the x-axis. To do this, set the function equal to zero: \[ 0 = 2 + x - x^2 \]. Solve this quadratic equation to find the values of x.

Step 2 - Solve the Quadratic Equation

Rearrange the quadratic equation to standard form: \[ x^2 - x - 2 = 0 \]. Factorize the equation: \[ (x-2)(x+1) = 0 \]. Thus, the solutions are: \[ x = 2 \] and \[ x = -1 \].

Step 3 - Set Up the Integral

Now, set up the definite integral to find the area between the curve and the x-axis from \( x = -1 \) to \( x = 2 \): \[ \text{Area} = \int_{-1}^{2} (2 + x - x^2) \, dx \].

Step 4 - Integrate the Function

Find the antiderivative of the integrand \( 2 + x - x^2 \). The integral is: \[ \int (2 + x - x^2) \, dx = 2x + \frac{1}{2}x^2 - \frac{1}{3}x^3 \].

Step 5 - Evaluate the Definite Integral

Evaluate the definite integral from \( x = -1 \) to \( x = 2 \): \[ [2x + \frac{1}{2}x^2 - \frac{1}{3}x^3]_{-1}^{2} \]. This equals: \[ (2(2) + \frac{1}{2}(2)^2 - \frac{1}{3}(2)^3) - (2(-1) + \frac{1}{2}(-1)^2 - \frac{1}{3}(-1)^3) \].

Step 6 - Simplify the Expression

Simplify the evaluated expression: \[ (4 + 2 - \frac{8}{3}) - (-2 + \frac{1}{2} + \frac{1}{3}) = (6 - \frac{8}{3}) - (-2 + \frac{1}{2} + \frac{1}{3}) = \frac{18}{3} - \frac{8}{3} - (-2 + \frac{1}{2} + \frac{1}{3}) = \frac{10}{3} - (-2 + \frac{4}{6}) = \frac{10}{3} - \left(- \frac{12}{6} \right) = \frac{10}{3} + \frac{12}{6} = \frac{10}{3} + 2 = \frac{16}{3} \].

Key Concepts

Integration

Definite integration, on the other hand, finds the net area under the curve within specified bounds. For example in the provided problem, we calculate the area between the curve and the \( x \)-axis from \( x = -1 \) to \( x = 2 \). This involves evaluating the antiderivative at these bounds and subtracting the results: \textbackslash begins[[ \int_{-1}^{2} (2 + x - x^2) dx = [2x + \frac{1}{2}x^2 - \frac{1}{3}x^3]_{-1}^{2} \text-backslash ends]].

Definite Integral

To evaluate, first find the antiderivative, which gives: \[ \int (2 + x - x^2)dx = 2x + \frac{1}{2}x^2 - \frac{1}{3}x^3 \]. Next, plug in the upper limit \( x = 2 \) and the lower limit \( x = -1 \) and subtract: \[ [2x + \frac{1}{2}x^2 - \frac{1}{3}x^3]_{-1}^{2} \].

Intersection Points

Rearrange and solve the quadratic equation: \[ x^2 - x - 2 = 0 \]. Then factorize: \[ (x-2)(x+1) = 0 \], yielding the roots \( x = 2 \) and \( x = -1 \). These are the points where the curve intersects the \( x \)-axis.

Quadratic Equation

One common method of solving a quadratic equation is by factoring. For \( x^2 - x - 2 = 0 \), we factorize into: \[ (x-2)(x+1) = 0 \], giving us solutions \( x = 2 \) and \( x = -1 \).
These solutions are also the x-intercepts of our original curve.