Problem 10 - 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. $$ \int_{0}^{2} 2 f(x) d x $$

Step-by-Step Solution

Answer

The integral \( \int_{0}^{2} 2 f(x) \, dx \) is 10.

Step 1: Understand the Problem

We are given integrals of two functions, \( f(x) \) and \( g(x) \), over different intervals. We need to find \( \int_{0}^{2} 2 f(x) \, dx \).

Step 2: Apply the Linearity Property

The linearity property of integrals says that \( \int_{a}^{b} c\cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \). We can use this property to factor out the constant 2:\[\int_{0}^{2} 2 f(x) \, dx = 2 \cdot \int_{0}^{2} f(x) \, dx\]

Step 3: Apply the Interval Additivity Property

According to the property of additivity over intervals, \( \int_{0}^{2} f(x) \ dx \) can be split into two parts:\[\int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx\]

Step 4: Substitute Known Values

Substitute the given integral values into the equation from the previous step:\[\int_{0}^{2} f(x) \, dx = 2 + 3 = 5\]

Step 5: Calculate the Final Integral

Substitute the value found in the previous step into the equation derived from the linearity property:\[\int_{0}^{2} 2 f(x) \, dx = 2 \cdot 5 = 10\]

Key Concepts

Linearity of Integrals

The linearity of integrals is a fundamental concept that simplifies the process of integrating functions. When we say an integral is linear, it means it behaves nicely under two key operations: scaling and addition.
First, consider scaling. If you have a constant that multiplies a function, this constant can be factored out of the integral. For instance, if you want to evaluate \(\int_{a}^{b} c \, f(x) \, dx\), it's the same as multiplying the integral of the function by the constant:\[\int_{a}^{b} c \, f(x) \, dx = c \, \int_{a}^{b} f(x) \, dx\]
Second, for addition, if you have two functions added together under the same integral, the integral of the sum is equal to the sum of the integrals: \(\int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx\).

• Scaling: The integral of a constant times a function equals the constant times the integral of the function.
• Addition: The integral of the sum of functions equals the sum of their integrals.

Interval Additivity

Interval additivity is another helpful property of integrals. It allows you to break an integral over a larger interval into a sum of integrals over smaller, contiguous intervals. This property is especially useful when you know the values of the integrals over smaller segments of the interval.
For example, if you have an integral over the interval from \(a\) to \(c\), you can find that integral by summing the integrals from \(a\) to \(b\) and from \(b\) to \(c\), where \(b\) is any point between \(a\) and \(c\):\[\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\]
This is logical because we are essentially splitting the path of integration into parts that are easier to manage. In the original exercise, this property let us calculate the integral \(\int_{0}^{2} f(x) \, dx\) by knowing \(\int_{0}^{1} f(x) \, dx\) and \(\int_{1}^{2} f(x) \, dx\).

• Break larger intervals into smaller parts.
• Add integrals over these smaller parts to get the integral over the larger interval.

Integration Techniques

Integration techniques refer to various strategies employed to find definite integrals, especially when dealing with complex functions. Choosing the right technique can simplify the integration process, saving both time and effort.
In the solved exercise, we utilized basic properties like linearity and interval additivity, which are straightforward yet powerful techniques.
Here are a few other common integration techniques:

• **Substitution:** Used when a function contains another function within it. It transforms the variable of integration to simplify the integral.
• **Integration by Parts:** Useful for products of functions and derived from the product rule for differentiation.
• **Partial Fraction Decomposition:** Simplifies integrals of rational functions by expressing them as sums of simpler fractions.

Often, a combination of these techniques and properties like linearity and additivity are employed to efficiently tackle more complex problems.
Understanding when and how to apply each technique is key to mastering the art of integration.