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 in Problems. $$ \int_{0}^{1}[2 f(s)+g(s)] d s $$

Short answer

The integral evaluates to 3.

Step-by-step solution

Apply Linearity Property

We start by applying the linearity property of integrals, which states that the integral of a sum is the sum of the integrals and that constants can be factored out of the integral. So we express the given integral \( \int_{0}^{1}[2f(s) + g(s)] ds \) as two separate integrals: \[\int_{0}^{1}[2f(s) + g(s)] ds = \int_{0}^{1}2f(s) ds + \int_{0}^{1}g(s) ds.\]

Factor Out Constant from First Integral

In the first integral \( \int_{0}^{1}2f(s) ds \), factor out the constant 2 as follows:\[\int_{0}^{1}2f(s) ds = 2\int_{0}^{1}f(s) ds.\]

Substitute Known Integral Values

Now substitute the known integral values from the problem statement. From the problem, \( \int_{0}^{1}f(x) dx = 2 \) and \( \int_{0}^{1}g(x) dx = -1 \). So:\[2\int_{0}^{1}f(s) ds = 2 \times 2 = 4,\]and\[\int_{0}^{1}g(s) ds = -1.\]

Add the Results

Finally, add the results from the two integrals obtained in Step 3:\[4 + (-1) = 3.\] Thus, the value of the integral \( \int_{0}^{1}[2f(s) + g(s)] ds \) is 3.

Key concepts

Linearity of Integrals

When tackling definite integrals, the linearity property is an essential tool. This property allows us to handle more complex integrals by breaking them down into simpler parts. Here's how it works:

• The integral of a sum of functions is equal to the sum of their integrals. So, for any functions \( f(x) \) and \( g(x) \), \( \int f(x) + g(x) \, dx = \int f(x) \, dx + \int g(x) \, dx \).
• Furthermore, if a function is multiplied by a constant, the constant can be taken outside the integral. For example, \( \int c \cdot f(x) \, dx = c \int f(x) \, dx \).

This property made it possible for us to simplify the integral \( \int_{0}^{1}[2f(s) + g(s)] ds \). We separated it into two integrals: one with \( 2f(s) \) and another with \( g(s) \). The constant \( 2 \) was moved outside of the first integral with ease, showcasing the power of linearity.

Interval Additivity

Interval Additivity is one of those intuitive yet powerful properties of definite integrals. This principle states that if a region can be divided into smaller parts, then the integral over the entire region can be broken into integrals over its parts. Formally,

• For a function \( f(x) \) over the interval \([a, b] \), \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \), where \( c \) is any point between \( a \) and \( b \).

In our original problem, this property could allow us, for example, to break down an integral over a larger range into smaller, known integral values, helping us to determine the final result efficiently. It shows us how flexible an integral's range can be, as we saw with the given values.

Properties of Integrals

Properties of integrals include a wide array of rules that assist in solving problems involving integration. Apart from linearity and interval additivity, other crucial properties include:

• Non-negativity: If \( f(x) \geq 0 \) on interval \([a, b] \), then \( \int_{a}^{b} f(x) \, dx \geq 0 \).
• Order of Limits: If you reverse the limits of integration, the sign of the integral changes. Thus, \( \int_{a}^{b} f(x) \, dx = - \int_{b}^{a} f(x) \, dx \).
• Zero Width: An integral over an interval where the start and end are the same is zero, \( \int_{a}^{a} f(x) \, dx = 0 \).

Understanding these basic properties can simplify complex integral calculus processes into manageable parts, aiding in precise mathematical solutions.

Integral Calculus

Integral Calculus is a vast field concerning the accumulation of quantities and the spaces under curves. It's closely related to differential calculus, together forming the two main branches of calculus. Let's delve into some of its major aims:

• Finding Areas: It helps calculate the area under curves, which is fundamental in physics, engineering, and probability.
• Solving Differential Equations: Many real-world phenomena are modeled with differential equations, and integral calculus often provides solutions to these equations.
• Accumulation Function: The definite integral concept helps in understanding accumulated change, which is essential in fields like economics.

Mastering integral calculus empowers students to apply mathematics to real-world problems, making it a cornerstone of mathematical education and practical application.