Problem 31 - Calculus and its applications

Question

Find the area of the region bounded by the graphs of the given equations. $$ f(x)=x^{2}-7 x+20, \quad g(x)=2 x+6 $$

Step-by-Step Solution

Answer

The area of the region between the curves is approximately 45.83 square units.

Step 1: Understand the Problem

We need to find the area between the two curves: the parabola described by the function \( f(x) = x^2 - 7x + 20 \) and the line described by the function \( g(x) = 2x + 6 \). To calculate the area between these curves, we must first find the points where they intersect.

Step 2: Find Points of Intersection

Set the two equations equal to find the intersection points: \( x^2 - 7x + 20 = 2x + 6 \). Rearrange to form a quadratic equation: \( x^2 - 9x + 14 = 0 \). Solve for \( x \) using the quadratic formula where \( a = 1 \), \( b = -9 \), and \( c = 14 \). The roots are \( x = \frac{9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot 14}}{2 \cdot 1} = 7, 2 \). So, the lines intersect at \( x = 2 \) and \( x = 7 \).

Step 3: Set Up the Integral for Area

The area between the curves from \( x = 2 \) to \( x = 7 \) can be found by integrating the difference of the functions: \( \int_{2}^{7} [(2x + 6) - (x^2 - 7x + 20)] \, dx \). Simplify the expression: \( \int_{2}^{7} (-x^2 + 9x - 14) \, dx \).

Step 4: Integrate the Function

Perform the integration: \( \int (-x^2 + 9x - 14) \, dx = -\frac{x^3}{3} + \frac{9x^2}{2} - 14x + C \). Evaluate from \( x = 2 \) to \( x = 7 \).

Step 5: Calculate the Definite Integral

Evaluate the antiderivative at the upper and lower bounds: \[ F(7) = -\frac{7^3}{3} + \frac{9 \times 7^2}{2} - 14 \times 7 \] \[ F(2) = -\frac{2^3}{3} + \frac{9 \times 2^2}{2} - 14 \times 2 \]. Calculate these to find: \[ F(7) = -\frac{343}{3} + \frac{441}{2} - 98 \] \[ F(2) = -\frac{8}{3} + \frac{36}{2} - 28 \]. Subtract these results to find the area.

Step 6: Simplify and Compute Final Area

Substitute and simplify the numbers: \[ F(7) - F(2) = \left(-\frac{343}{3} + \frac{441}{2} - 98 \right) - \left(-\frac{8}{3} + 18 - 28 \right) \]. Calculate each term and simplify: \[ = -114.33 + 220.5 - 98 + 2.67 - 18 \]. Add and subtract terms to find the area of approximately 45.83 square units.

Key Concepts

Intersection Points

When we're tasked with finding the area between two curves, the first step is to determine where these curves intersect. This helps us set the boundaries for our calculation. In this exercise, we have one curve as a parabola and the other as a straight line.
To find intersection points, we set the equations equal to each other:

• For the parabola, use: \( f(x) = x^2 - 7x + 20 \)
• For the line, use: \( g(x) = 2x + 6 \)
• Set them equal: \( x^2 - 7x + 20 = 2x + 6 \)

Rearranging gives a quadratic equation: \( x^2 - 9x + 14 = 0 \).
Solving this equation using the quadratic formula, we find that the roots of the equation — which are also the intersection points — are \( x = 2 \) and \( x = 7 \). These points define the segment over which we will integrate to find the area.

Definite Integral

The definite integral is a mathematical tool used to calculate the exact area under a curve between two points, in this case, between our intersection points. To find the area between our two curves, we integrate the difference of the functions over the interval from \( x = 2 \) to \( x = 7 \).
Here's how it's set up:

• Difference of functions: \( (2x + 6) - (x^2 - 7x + 20) \)
• Simplified: \( -x^2 + 9x - 14 \)
• Set up the integral: \( \int_{2}^{7} (-x^2 + 9x - 14) \, dx \)

This integral represents the total area between the curves from \( x = 2 \) to \( x = 7 \). It's like summing up tiny slices of area from one point to another, ensuring all parts are accounted for, to get a precise measurement.

Quadratic Equation

In the context of this problem, a quadratic equation helps us find where the curves intersect, which is pivotal to determining the area between curves.
A quadratic equation generally has the form \( ax^2 + bx + c = 0 \). In our calculation for intersection points, the equation becomes \( x^2 - 9x + 14 = 0 \).
To solve it, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is used. Here:

• \( a = 1 \)
• \( b = -9 \)
• \( c = 14 \)

By substituting these values into the formula,

• \( b^2 - 4ac = 81 - 56 = 25 \)
• This gives us roots: \( x = \frac{9 \pm 5}{2} \) which simplifies to \( x = 7 \) and \( x = 2 \)

Solving quadratic equations efficiently can unravel key features like roots or intersection points, integral for many calculus problems like finding areas between curves.