Problem 32 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{0}^{2}(x-4)(x-1) d x$$

Step-by-Step Solution

Answer

Evaluating the definite integral, we obtain: \[ \int_{0}^{2}(x-4)(x-1) dx = \frac{2}{3}. \]

Step 1: Expand the expression

First, let's expand the expression inside the integral: \((x-4)(x-1)\). Using the distributive property, we obtain: \((x-4)(x-1) = x(x-1) -4(x-1) = x^2 - x - 4x + 4 = x^2 -5x + 4\)

Step 2: Evaluate the definite integral

Now we can evaluate the definite integral of the expanded expression: \[ \int_{0}^{2}(x^2 - 5x + 4) dx. \] We find the antiderivative of \(x^2 - 5x +4\), which gives us: \[ \int (x^2 - 5x + 4) dx = \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x + C. \]

Step 3: Apply the fundamental theorem of calculus

Now we apply the fundamental theorem of calculus to find the definite integral: \[ \int_{0}^{2}(x^2 - 5x + 4) dx = \left[\frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x\right]_0^2. \]

Step 4: Calculate the result

We now substitute the limits of integration into the antiderivative expression and calculate the result: \[ \begin{aligned} \left[\frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x\right]_0^2 &= \left(\frac{1}{3}(2)^3 - \frac{5}{2}(2)^2 + 4(2)\right) - \left(\frac{1}{3}(0)^3 - \frac{5}{2}(0)^2 + 4(0)\right) \\ &= \left(\frac{8}{3} - \frac{20}{2} + 8\right) \\ &= \left(\frac{8}{3} - 10 + 8\right) \\ &= \left(\frac{2}{3}\right). \end{aligned} \] So the result is: \[ \int_{0}^{2}(x-4)(x-1) dx = \frac{2}{3}. \]

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation and integration, two central concepts in calculus. It establishes a connection between the antiderivative of a function and the distinct evaluation of a definite integral.
In simple terms, it states that if you have a continuous function that has an antiderivative, you can evaluate an integral across a given interval by looking at the values of this antiderivative at the bounds of that interval.

• The first part of the theorem tells us that an integral can be reserved by a function's derivative.
• The second part allows us to evaluate the integral based on the antiderivative. For a function \( f(x) \) over an interval \([a, b]\), the definite integral is \( F(b) - F(a) \), where \( F \) is the antiderivative.

Polynomial Integration

Polynomial integration involves finding the antiderivative for polynomial expressions. These are expressions formed from variables, coefficients, and the operations of addition, subtraction, and multiplication, but no divisions by variables.
Integrating a polynomial is straightforward when you know the rule: every term \( ax^n \) in the polynomial becomes \( \frac{a}{n+1}x^{n+1} \) when integrated.
This means raising the power of your term by one and dividing by the new power. Considered a fundamental process in calculus, mastering polynomial integration allows you to evaluate many problems involving motion, area, and volume among others.

Antiderivative

An antiderivative is a function that reverses the process of differentiation. If you have a function \( f(x) \), an antiderivative is another function \( F(x) \) such that \( F'(x) = f(x) \). In essence, finding an antiderivative means determining a function that, when differentiated, will give you the original function.
This process is essential for solving definite integrals, as it helps you evaluate the integral across an interval using the Fundamental Theorem of Calculus.
The concept of an antiderivative is central to understanding integration, as it serves as the primary tool to unlock the area under a curve.

Distributive Property

The distributive property is a key algebraic identity that allows for expressions to be rewritten, often simplifying the integration process for composite expressions. It states that multiplying a single term by a sum is the same as multiplying each term of the sum individually by the single term. If you have an expression like \((x-4)(x-1)\), the distributive property allows you to expand this out:

• First, multiply each term in \( (x-1) \) by \( x \) producing \( x^2 - x \).
• Then, multiply each term in \( (x-1) \) by \( -4 \) yielding \( -4x + 4 \).
• Finally, combine the like terms to simplify the expression to \( x^2 - 5x + 4 \).

This method is crucial for making polynomials easier to integrate, as it breaks a complex function into simpler components that can be managed piece by piece.