Problem 74 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Given that \(\int_{-1}^{2} f(x) d x=2\) and \(\int_{0}^{2} f(x) d x=3\), evaluate a. \(\int_{-1}^{0} f(x) d x\) b. \(\int_{0}^{2} f(x) d x-\int_{-1}^{0} f(x) d x\)

Step-by-Step Solution

Answer

a. \(\int_{-1}^{0} f(x) dx = -1\) b. \(\int_{0}^{2} f(x) dx - \int_{-1}^{0} f(x) dx = 4\)

Step 1: Divide the given integral at x=0

Using the property mentioned above, we have: \(\int_{-1}^{2} f(x) dx = \int_{-1}^{0} f(x) dx + \int_{0}^{2} f(x) dx\)

Step 2: Substitute given values

We know that \(\int_{0}^{2} f(x) dx = 3\) and \(\int_{-1}^{2} f(x) dx = 2\). Substituting these values, we get: \(2=\int_{-1}^{0} f(x) dx + 3\)

Step 3: Solve for the unknown integral

Now we will solve for \(\int_{-1}^{0} f(x) dx\): \(\int_{-1}^{0} f(x) dx = 2 - 3 = -1\) So, \(\int_{-1}^{0} f(x) dx = -1\). b. To find \(\int_{0}^{2} f(x) dx - \int_{-1}^{0} f(x) dx\), we will simply subtract the value of the first integral from the second integral.

Step 1: Subtract the values of the integrals

We know that \(\int_{0}^{2} f(x) dx = 3\) and \(\int_{-1}^{0} f(x) dx = -1\). Hence, \(\int_{0}^{2} f(x) dx - \int_{-1}^{0} f(x) dx = 3 - (-1)\)

Step 2: Simplify the expression

Simplifying the expression, we get: \(\int_{0}^{2} f(x) dx - \int_{-1}^{0} f(x) dx = 3 + 1 = 4\) So, \(\int_{0}^{2} f(x) dx - \int_{-1}^{0} f(x) dx = 4\).

Key Concepts

Definite Integrals

A definite integral can be thought of as the signed area under a curve, bounded by specific limits. Essentially, when you compute a definite integral, you're finding the total accumulation (like area) between the graph of a function and the x-axis over a certain interval. This is why definite integrals are so useful in problems related to total change or quantity.

Here's how they work:

• You have a function, often denoted by \(f(x)\), and you're interested in studying its behavior over a particular interval. Let's consider the interval [a, b].
• The integral \(\int_{a}^{b} f(x) \, dx\) will give you the net area between the curve and the x-axis from a to b. "Net" because areas above the x-axis are considered positive, while areas below are negative.
• This is distinguished from indefinite integrals, which lack specified limits and represent families of functions as opposed to explicit quantities.

In practice, definite integrals are evaluated by finding an antiderivative of the function, and then applying the Fundamental Theorem of Calculus, which simplifies calculations through substitution into this antiderivative. This theorem effectively turns the often complex task of area calculation into a much simpler one of function evaluation.

Properties of Integrals

Understanding the properties of integrals can simplify many calculations. These properties are derived from the fundamental nature of integration and are incredibly useful for breaking down complex integrals into simpler parts.

Some key properties include:

• Linearity: The integral of a sum is the sum of integrals. Mathematically, \(\int (a \cdot f(x) + b \cdot g(x)) \, dx = a \cdot \int f(x) \, dx + b \cdot \int g(x) \, dx\).
• Additivity over Intervals: This means that the integral over a large interval can be split into smaller ones. \(\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\) is one such important property, helping in handling complex intervals.
• Zero Width Intervals: If the two limits of an integral are the same, the integral is zero, i.e., \(\int_{a}^{a} f(x) \, dx = 0\).
• Order of Limits: Flipping the limits of integration changes the sign of the integral, \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\).

Having a firm grasp of these properties not only makes solving integrals easier but also helps in understanding the interaction and behavior of functions under integration.

Mathematical Problem Solving

When tackling calculus problems, especially those involving integrals, a structured problem-solving approach can greatly enhance clarity and efficiency. Here's a useful strategy to approach such problems:

• **Understand the Problem:** Begin by reading the problem carefully. Understand what is given, and what needs to be found. Sketching graphs might be helpful if dealing with function behavior.
• **Identify the Variables and Limits:** Pay close attention to the functions being integrated and the limits of integration. This is critical in defining what part of the function’s graph contributes to the integral.
• **Apply the Relevant Techniques:** Use properties of integrals to break down the problem. Subdivide intervals if needed or use known results to simplify what you're working with.
• **Calculate Carefully:** Once set up correctly, perform the calculations with precision, typically using tools like the Fundamental Theorem of Calculus to evaluate definite integrals.
• **Review Your Solution:** Check your answers for consistency and accuracy. Make sure they reflect the problem's context – especially the relationships involved and the direction of integration.

Developing these skills boosts not just speed, but also confidence in handling a variety of integral-related calculus problems.