Problem 12 - 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. $$ \int_{0}^{1}[2 f(s)+g(s)] d s $$

Step-by-Step Solution

Answer

The value of the integral \( \int_{0}^{1}[2 f(s)+g(s)] \, ds \) is 3.

Step 1: Distribute the Integral

Start by applying the linearity property of integrals. This property allows you to separate the integral of a sum into the sum of integrals:\[\int_{0}^{1} [2f(s) + g(s)] \, ds = \int_{0}^{1} 2f(s) \, ds + \int_{0}^{1} g(s) \, ds\]

Step 2: Factor Out Constants

Using the linearity property again, pull the constant (in this case, 2) outside of the first integral:\[\int_{0}^{1} 2f(s) \, ds = 2 \int_{0}^{1} f(s) \, ds\]

Step 3: Substitute Known Values

Substitute the given values of the definite integrals into the expression obtained in Step 2:\[2 \int_{0}^{1} f(s) \, ds + \int_{0}^{1} g(s) \, ds = 2 \cdot 2 + (-1)\]From the given, \(\int_{0}^{1} f(s) \, ds = 2\) and \(\int_{0}^{1} g(s) \, ds = -1\).

Step 4: Calculate the Expression

Carry out the arithmetic computation to find the value of the integral:\[4 + (-1) = 3\]

Key Concepts

Linearity of Integrals

When dealing with integrals, the concept of linearity simplifies many calculations. The linearity property states that the integral of a sum is equal to the sum of the integrals. It also allows us to factor out constants outside the integral.
To illustrate, consider the integral:

• \[\int_{0}^{1} [2f(s) + g(s)] \, ds \]

Applying the linearity property, we split this into:

• \[\int_{0}^{1} 2f(s) \ ds + \int_{0}^{1} g(s) \ ds\]

This means we evaluate each part separately.
In more detail:

• For the term involving the function \(f(s)\), you can remove the constant "2" out of the integral:\[\int_{0}^{1} 2f(s) \ ds = 2 \int_{0}^{1} f(s) \ ds\]
• Similarly, the integral of \(g(s)\) remains the same:\[\int_{0}^{1} g(s) \ ds\]

This basic linearity is critical for simplifying complex integrals.

Interval Additivity

Another helpful property of integrals is interval additivity. This means that an integral over a whole interval can be broken down into a sum of integrals over sub-intervals. This property is perfect when you need to calculate integrals defined piecewise.
Imagine you have a situation where a function spans several intervals. For instance, if:

• \[\int_{0}^{2} g(x) \, dx = \int_{0}^{1} g(x) \, dx + \int_{1}^{2} g(x) \, dx\]

This property lets you dissect an integral into manageable parts, making the calculations easier.
In the given problem, while we're focused on the interval from 0 to 1 for both functions, it's useful to know that definitions of integrals over larger intervals can be broken down if needed. By knowing parts of the interval, you apply interval additivity to calculate or verify the integral over the entire range.

Properties of Integrals

Understanding properties of integrals is fundamental to solving integral problems efficiently. Besides linearity and interval additivity, another important property is symmetry, although it's not used here, it's good to know.
Here, let's focus on the properties being applied:

• **Linearity:** As demonstrated, it enables simplification of integrals involving sums and allows factoring out constants.
• **Interval Additivity:** Helps break down integrals over large intervals into smaller, more calculable segments.
• **Substitution of Known Values:** Once you simplify the integral expressions using these properties, substitute any given values to easily finalize your computations.

In our example, substituting the given values of \[\int_{0}^{1} f(s) \, ds = 2\]and \[\int_{0}^{1} g(s) \, ds = -1\],streamlines the process to find:\[2 \cdot 2 + (-1) = 3\]By keeping these properties in mind, you can elegantly work through integral problems without unnecessary complexity.