Problem 47 - Applied Calculus

Question

Find the area bounded by the given curves. \(y=6 x^{2}-10 x-8\) and \(y=3 x^{2}+8 x-23\)

Step-by-Step Solution

Answer

The area bounded by the curves is 32 square units.

Step 1: Set the equations equal to each other

To find the points of intersection, set the two equations equal to each other: \(6x^2 - 10x - 8 = 3x^2 + 8x - 23\). This forms a new equation: \(3x^2 - 18x + 15 = 0\).

Step 2: Simplify and solve the quadratic equation

Divide the entire equation by 3 to simplify: \(x^2 - 6x + 5 = 0\). Now factor the quadratic equation to solve for \(x\): \((x - 5)(x - 1) = 0\). Thus, \(x = 5\) and \(x = 1\) are the points of intersection.

Step 3: Evaluate the definite integral

The area between two curves from \(x = a\) to \(x = b\) is given by the integral \(\int_a^b (f(x) - g(x)) \,dx\), where \(f(x)\) is the top function and \(g(x)\) is the bottom function. Identify the top and bottom functions between \(x = 1\) and \(x = 5\). Here, \(y = 6x^2 - 10x - 8\) is above \(y = 3x^2 + 8x - 23\). Set up the integral: \(\int_1^5 ((6x^2 - 10x - 8) - (3x^2 + 8x - 23)) \,dx\). Simplifying inside the integral gives: \(\int_1^5 (3x^2 - 18x + 15) \,dx\).

Step 4: Integrate the polynomial

Compute the integral: \(\int_1^5 (3x^2 - 18x + 15) \,dx = \left[ x^3 - 9x^2 + 15x \right]_1^5\).

Step 5: Evaluate the definite integral

Substitute the limits into the integrated function: Evaluate at the upper limit (5): \((5)^3 - 9(5)^2 + 15(5) = 125 - 225 + 75 = -25\). Evaluate at the lower limit (1): \((1)^3 - 9(1)^2 + 15(1) = 1 - 9 + 15 = 7\). Subtract the results: \(-25 - 7 = -32\). The absolute value of this area is \(32\), since area cannot be negative.

Key Concepts

Definite Integral

The concept of a definite integral is crucial in finding the area between curves. Unlike an indefinite integral, which represents a family of antiderivatives, a definite integral has specific limits and results in a specific numerical value. When calculating the area between two curves, the definite integral helps by allowing us to subtract the area below the lower curve from the area below the upper curve, giving the precise area sandwiched between them.

To use the definite integral for finding the area, follow these steps:

• Identify the upper and lower bounds (the points where the curves intersect).
• Determine which function is above and which is below within those bounds.
• Set up the integral with these bounds, subtracting the lower function from the upper function.
• Evaluate the integral (and remember to take the absolute value if necessary, as area is always positive).

Utilizing a definite integral provides a structured way to compute areas when dealing with complex functions and intersections.

Quadratic Equations

Quadratic equations often arise when solving problems related to curves, such as finding their points of intersection. A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Solving these equations is essential for determining key values like intersection points, which in turn help define the limits for definite integrals.

Here's how we generally solve a quadratic equation:

• Use factoring, if possible, to simplify the equation and find its roots.
• Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) if the equation does not factor easily.
• Consider completing the square as another alternative method.

In the provided exercise, the equation was simplified to \((x - 5)(x - 1) = 0\), illustrating the factoring method. By solving this, we found that \(x = 5\) and \(x = 1\) are where the curves intersect, essential for setting up the integral.

Polynomial Integration

Polynomial integration is the process of integrating polynomial functions, which are expressions like \(ax^n + bx^{n-1} + cx^{n-2} + \ldots + d\). Integrating these functions involves raising the degree of each term and dividing by the new exponent. This technique is fundamental when dealing with functions represented by polynomials, especially in the context of finding areas between curves.

When performing polynomial integration:

• Increase the power of each term by one.
• Divide each new term by its new exponent.

For example, integrating \(3x^2\) becomes \(x^3\), \(-18x\) becomes \(-9x^2\), and a constant like \(15\) becomes \(15x\). Each integrated term contributes to the final expression used to find the area between curves when evaluated with definite integrals.

In the exercise, this method resulted in the integral expression \(\left[ x^3 - 9x^2 + 15x \right]_1^5\), which was evaluated at the specified limits to give the area.