Problem 9 - Calculus

Question

Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=4\). Use properties of definite integrals (linearity, interval additivity, and so on) to calculate each of the integrals. $$ \int_{1}^{2} 2 f(x) d x $$

Step-by-Step Solution

Answer

The integral \( \int_{1}^{2} 2f(x) \, dx \) is 6.

Step 1: Understanding the Problem

We need to calculate \( \int_{1}^{2} 2f(x) \, dx \) using given integral values and properties of definite integrals. We know \( \int_{1}^{2} f(x) \, dx = 3 \).

Step 2: Apply Linearity Property

The linearity property of integrals allows us to factor out constants. Therefore, \( \int_{1}^{2} 2f(x) \, dx = 2 \cdot \int_{1}^{2} f(x) \, dx \).

Step 3: Substitute Known Values

From the problem, we know that \( \int_{1}^{2} f(x) \, dx = 3 \). Substitute this value into the expression: \( 2 \cdot \int_{1}^{2} f(x) \, dx = 2 \cdot 3 \).

Step 4: Calculate the Integral

Perform the multiplication to find the value of the integral: \( 2 \cdot 3 = 6 \).

Key Concepts

Linearity Property

The linearity property of definite integrals is a fundamental concept that greatly simplifies integral calculations. It allows us to manipulate integrals in ways that are often very straightforward. Let's break down how this property works.

Linearity in integrals means two main things:

• The integral of a sum is the sum of the integrals. That means for any two functions \(f(x)\) and \(g(x)\), \(\int (f(x) + g(x)) \ dx = \int f(x) \ dx + \int g(x) \ dx\).
• More relevant to the problem at hand, we can factor constants out of the integral. For instance, \(\int a \cdot f(x) \ dx = a \cdot \int f(x) \ dx\) where \(a\) is a constant.

In our exercise, this property has been applied to move the constant factor of 2 outside the integral. This makes the problem much easier to solve because we only need to multiply the result of the integral \(\int_{1}^{2} f(x) \ dx\) by 2, based on the known value of this integral.

Interval Additivity

The interval additivity property of integrals is a clever tool that relates to splitting or combining intervals when working with integrals. This property is especially useful when dealing with pieces of functions over different segments.

It's important to remember the additivity formula:

• For a given interval \([a, c]\), if there's a point \(b\) in between, then \(\int_{a}^{c} f(x) \ dx = \int_{a}^{b} f(x) \ dx + \int_{b}^{c} f(x) \ dx\).

Essentially, this means you can add up the integrals over smaller, contiguous parts to find the integral over the entire interval. Although this property wasn't directly used in the solution for \(\int_{1}^{2} 2f(x) \ dx\), it could be a helpful step in addressing more complex problems where you have to find integrals over combined intervals. It's like understanding how small pieces add up to the whole.

Integral Calculations

When performing integral calculations, using definite integrals involves finding the area under the curve of a function between specific bounds. You employ various properties, like linearity and interval additivity, to simplify these calculations.

For the exercise example, we knew \(\int_{1}^{2} f(x) \ dx = 3\), which represents the area under \(f(x)\) from 1 to 2. The solution required multiplying this known value by 2 due to the constant factor of 2 in \(\int_{1}^{2} 2f(x) \ dx\).

Performing the calculation:

• First, recognize the integral with a constant factor (linearity): \( \int_{1}^{2} 2f(x) \ dx = 2 \cdot \int_{1}^{2} f(x) \ dx \).
• Substitute the known integral value: \(2 \cdot 3\).
• Finally, solve by straightforward multiplication: \(6\).

Understanding these steps showcases how the properties of integrals simplify what could otherwise become a complex computation. They streamline the process into a few reliable calculations, allowing you to efficiently solve integral problems in your coursework or exams.