Problem 11 - 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. $$ \int_{0}^{2}[2 f(x)+g(x)] d x $$

Step-by-Step Solution

Answer

7.

Step 1: Use given information

\(\int_0^1 f(x)\,dx = 2\), \(\int_1^2 f(x)\,dx = 3\), so \(\int_0^2 f(x)\,dx = 2 + 3 = 5\).
\(\int_0^1 g(x)\,dx = -1\), \(\int_0^2 g(x)\,dx = 4\), so \(\int_1^2 g(x)\,dx = 4 - (-1) = 5\).

Step 2: Compute the requested integral

\(\int_0^2 [f(x) + g(x)]\,dx = \int_0^2 f(x)\,dx + \int_0^2 g(x)\,dx = 5 + 4 = 9\).
Other combinations can be computed similarly using linearity and additivity.

Key Concepts

Linearity of Integration

The linearity of integration is a rule in calculus that helps simplify the process of finding definite integrals. It shows how we can break down and handle complex integrals easily. This property states that the integral of a sum is the sum of the integrals, and we can also pull out constant multipliers from the integral.

For example, if you have an integral of the form \( \int_{a}^{b} [c \cdot u(x) + v(x)] dx \), you can simplify it by breaking it apart as follows:

• \( c \int_{a}^{b} u(x) dx \) - Here, \( c \) is a constant multiplier, so we can factor it out.
• \( + \int_{a}^{b} v(x) dx \) - This is how we manage the sum.

By applying this property, integration becomes significantly more manageable, allowing us to solve larger and more complex integrals by simply splitting them into parts that are easier to evaluate.

Integration Properties

Integration properties are rules that govern how we compute integrals, ensuring that calculations stay consistent and are straightforward to follow. Key properties include:

• Additivity: The integral over a single interval can be split into the sum of integrals over sub-intervals.
• Linearity: As discussed earlier, the integral of a sum is the sum of the integrals. Constants can be factored out of the integral.
• Zero Width Interval: An integral over an interval of zero width is always zero.

These properties are useful as they allow us to manipulate integrals in order to solve them more effectively. In the problem above, these properties enable us to write the integral as a combination of different known parts, making calculations more straightforward.

Interval Additivity

Interval additivity is a fundamental concept that states that the value of an integral over an entire interval equals the sum of the integrals over any subdivisions of that interval. This property is incredibly useful when dealing with definite integrals across different intervals.

For example, as given: \[ \int_{0}^{2} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{2} f(x) dx \]This allows us to use known values for \( \int_{0}^{1} f(x) dx \) and \( \int_{1}^{2} f(x) dx \) to find the integral over the combined interval \( [0,2] \). In the initial problem, interval additivity helps us handle functions defined over multiple intervals, by summing their individual integrals. This simplification cuts down complex problems into smaller, approachable ones.

Calculus Problem Solving

Calculus problem solving often involves combining various calculus concepts to find unknown values and approach complex problems logically and systematically. When solving an integral, we need to utilize integration properties, linearity, and interval additivity effectively.

To solve integrals like the one in the problem, we:

• Identify known integrals using past computations or given values.
• Apply linearity to separate and simplify complex expressions.
• Use interval additivity to connect different intervals, using known values to compute unknown integrals.

This methodical approach helps in understanding each part of the problem and how they connect. By breaking it down step-by-step, calculus problems become less daunting and more intuitive.