Problem 23 - Calculus Single Variable

Question

Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=x^{2}+x+1 \quad g(x)=2 x^{2}+3 x-7 $$

Step-by-Step Solution

Answer

The area between the curves is 36 square units.

Step 1: Find Points of Intersection

To find the area between the curves, we first need the points of intersection of the two functions. Solve the equation for when the functions are equal: \[ f(x) = g(x) \]Substitute the given functions: \[ x^2 + x + 1 = 2x^2 + 3x - 7 \]Rearrange terms to find the x-values: \[ 0 = 2x^2 + 3x - 7 - x^2 - x - 1 \]Simplify to: \[ 0 = x^2 + 2x - 8 \]Solve for \(x\) by factoring or using the quadratic formula: \[ x^2 + 2x - 8 = (x + 4)(x - 2) = 0 \]The solutions are \(x = -4\) and \(x = 2\).

Step 2: Define the Area Formula Between Two Functions

The area between two curves from \(x = a\) to \(x = b\) is given by:\[ \int_{a}^{b} (f(x) - g(x)) \, dx \]Since \(g(x)\) is above \(f(x)\) in this interval, the integral becomes:\[ \int_{-4}^{2} [(2x^2 + 3x - 7) - (x^2 + x + 1)] \, dx \]Simplify the expression inside the integral:\[ = \int_{-4}^{2} (x^2 + 2x - 8) \, dx \]

Step 3: Evaluate the Integral

Integrate the simplified function:\[ \int (x^2 + 2x - 8) \, dx \]\[ = \frac{x^3}{3} + x^2 - 8x \]Evaluate the definite integral from \(-4\) to \(2\):\[ \left[ \frac{2^3}{3} + 2^2 - 8(2) \right] - \left[ \frac{(-4)^3}{3} + (-4)^2 - 8(-4) \right] \]Simplify each part:For \(x = 2\):\[ \frac{8}{3} + 4 - 16 \]For \(x = -4\):\[ \frac{-64}{3} + 16 + 32 \]

Step 4: Calculate Each Part and Find Area

Calculate the numerical values from earlier:\[ \frac{8}{3} + 4 - 16 = \frac{8}{3} + \frac{12}{3} - \frac{48}{3} = \frac{-28}{3} \]\[ \frac{-64}{3} + 16 + 32 = \frac{-64}{3} + \frac{48}{3} + \frac{96}{3} = \frac{80}{3} \]Find the difference to determine the area:\[ \frac{80}{3} - \frac{-28}{3} = \frac{80}{3} + \frac{28}{3} = \frac{108}{3} = 36 \]

Step 5: Conclusion: Find the Area

The area between the curves, over the interval where they intersect, is 36 square units.

Key Concepts

Points of Intersection

Points of intersection are crucial when finding the area between two curves. They represent where the two curves meet on the Cartesian plane, and they are essential to determine the integration limits. Essentially, we start by setting the functions equal to each other to find these points. For the given exercise, we equate the functions:

• \( f(x) = x^2 + x + 1 \),
• \( g(x) = 2x^2 + 3x - 7 \).

Solving \( f(x) = g(x) \) means finding the values of \( x \) where these functions output the same \( y \)-value. Subtract \( f(x) \) from \( g(x) \), rearrange the equation, and simplify, resulting in a quadratic equation: \( x^2 + 2x - 8 = 0 \).
Then, we solve this quadratic equation either by factoring or using the quadratic formula. In this case, it factors nicely away: \( (x + 4)(x - 2) = 0 \). Solving this gives the intersection points \( x = -4 \) and \( x = 2 \), which are our boundaries for integration.

Definite Integral

The definite integral is used to find the area under a curve between two specific x-values, bounds which, in this exercise, are the points of intersection found earlier. When finding the area between two curves, we take the integral of the top function minus the bottom function over the segment they intersect. The basic formula to find the area is:

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

For your exercise, since \( g(x) \) is above \( f(x) \) over the interval \( x = -4 \) to \( x = 2 \), the area is given by:
\[ \int_{-4}^{2} ((2x^2 + 3x - 7) - (x^2 + x + 1)) \, dx \].
This setup makes sure that the area calculation remains positive, respecting the geometrical representation of the space caught between the two curves.

Quadratic Functions

Quadratic functions are polynomials of degree 2, taking the form \( ax^2 + bx + c \), where \( a eq 0 \). In this exercise, both \( f(x) \) and \( g(x) \) are quadratic functions:

• \( f(x) = x^2 + x + 1 \),
• \( g(x) = 2x^2 + 3x - 7 \).

The characteristic feature of quadratic functions is the parabolic shape of their graphs. The leading coefficient, \( a \), determines if the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)). Quadratics can be manipulated and understood for various intersections and area calculations, often leading to solutions via factoring or quadratic formulas.
The points of intersection, determined by solving the equation \( x^2 + 2x - 8 = 0 \), revealed where these curves touch each other, making it a setup for applying definite integrals.

Integration Techniques

Integration techniques are crucial for calculating areas between curves effectively. In this exercise, we perform basic polynomial integration because both functions involved result in a simplified polynomial expression when subtracted: \( x^2 + 2x - 8 \).
For polynomials, integration involves applying the power rule in reverse. For example, the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). Thus, we integrate each term separately:

• \( \int x^2 \, dx = \frac{x^3}{3} \)
• \( \int 2x \, dx = x^2 \)
• \( \int -8 \, dx = -8x \)

Evaluating these integrals within the bounds defined by the intersection points \( x = -4 \) and \( x = 2 \) requires substituting these limits to find the difference in resultant areas, eventually leading to the calculated area of 36 square units.