Problem 12
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 9-16. $$ \int_{0}^{1}[2 f(s)+g(s)] d s $$
Step-by-Step Solution
Verified Answer
The integral evaluates to 3.
1Step 1: Use Linearity of Integrals
The integral \( \int_{0}^{1} [2f(s) + g(s)] \, ds \) can be decomposed using the linearity property of integrals: \( \int_{0}^{1} [2f(s) + g(s)] \, ds = \int_{0}^{1} 2f(s) \, ds + \int_{0}^{1} g(s) \, ds \).
2Step 2: Factor Out Constants
Factors outside the integrals can be moved outside the integral. Therefore, \( \int_{0}^{1} 2f(s) \, ds = 2 \int_{0}^{1} f(s) \, ds \).
3Step 3: Substitute Known Values
Substitute the known integral values: \( \int_{0}^{1} f(x) \, dx = 2 \) and \( \int_{0}^{1} g(x) \, dx = -1 \). Hence, \( 2 \int_{0}^{1} f(s) \, ds = 2 \times 2 = 4 \) and the second integral remains \( \int_{0}^{1} g(s) \, ds = -1 \).
4Step 4: Combine the Results
Combine the results by adding the evaluated integrals: \( 4 + (-1) = 3 \).
Key Concepts
Properties of Definite IntegralsLinearity of IntegralsFactor Out ConstantsAdditivity of Integrals
Properties of Definite Integrals
Definite integrals have properties that make them powerful tools for calculus. One key property is their ability to measure the accumulation of quantities. For example, when considering speed over time, the definite integral calculates the total distance traveled. This property helps in solving problems involving continuous functions within a given interval.
Moreover, definite integrals possess two important characteristics: bounds and area interpretation. The integral is evaluated between two points, known as bounds, which limits the scope of calculation to a specific interval. The evaluation of a definite integral can be visualized as the area under the curve of the function over this interval.
Understanding these concepts helps students solve complex calculus problems by providing a clear framework for how to handle continuous data.
Moreover, definite integrals possess two important characteristics: bounds and area interpretation. The integral is evaluated between two points, known as bounds, which limits the scope of calculation to a specific interval. The evaluation of a definite integral can be visualized as the area under the curve of the function over this interval.
Understanding these concepts helps students solve complex calculus problems by providing a clear framework for how to handle continuous data.
Linearity of Integrals
The linearity of integrals is a crucial concept in calculus. It states that the integral of a sum is the sum of the integrals. This means for functions like \( f(x) \) and \( g(x) \), \( \int_a^b [f(x) + g(x)] \, dx \) is equal to \( \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \). By rewriting integrals in this way, we simplify calculations.
This property provides flexibility, allowing students to break down complex functions into simpler parts. It helps manage functions containing multiple terms by addressing each term separately before combining them back together.
Linearity also indicates that a constant multiplied by a function can be factored out of the integral, further simplifying the evaluation.
This property provides flexibility, allowing students to break down complex functions into simpler parts. It helps manage functions containing multiple terms by addressing each term separately before combining them back together.
Linearity also indicates that a constant multiplied by a function can be factored out of the integral, further simplifying the evaluation.
Factor Out Constants
A powerful property of integrals is the ability to factor out constants. If you have a constant multiplying a function, such as \( c \cdot f(x) \), you can take the constant \( c \) out of the integral. So, \( \int_a^b c \cdot f(x) \, dx \) becomes \( c \cdot \int_a^b f(x) \, dx \).
This simplifies the integration process by reducing the function to a form that's easier to evaluate. By performing this step, you separate the constant from the variable part of the function, which makes calculations more straightforward.
This action is beneficial when you're dealing with functions that are scaled versions of simpler functions, as it saves time and effort during problem-solving.
This simplifies the integration process by reducing the function to a form that's easier to evaluate. By performing this step, you separate the constant from the variable part of the function, which makes calculations more straightforward.
This action is beneficial when you're dealing with functions that are scaled versions of simpler functions, as it saves time and effort during problem-solving.
Additivity of Integrals
Additivity allows us to tackle complicated integrals by breaking them into manageable parts. The additivity property of integrals states that the integral over a larger interval can be divided into smaller, adjacent intervals. For functions defined from \( a \) to \( c \), this means \( \int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx \) if \( a < b < c \).
This property is particularly useful when dealing with piecewise functions or where different rules or behaviors exist over different sections of the domain. By applying interval additivity, you isolate and solve parts of a problem incrementally, which simplifies the process and enhances understanding.
In summary, additivity makes handling integrations over complex intervals much more manageable by using a step-by-step approach.
This property is particularly useful when dealing with piecewise functions or where different rules or behaviors exist over different sections of the domain. By applying interval additivity, you isolate and solve parts of a problem incrementally, which simplifies the process and enhances understanding.
In summary, additivity makes handling integrations over complex intervals much more manageable by using a step-by-step approach.
Other exercises in this chapter
Problem 12
In Problems 1-14, use the Second Fundamental Theorem of Calculus to evaluate each definite integral. \(\int_{\pi / 6}^{\pi / 2} 2 \sin t d t\)
View solution Problem 12
\(g(x)=\tan x \sec ^{2} x ; \quad[0, \pi / 4]\)
View solution Problem 13
In Problems 11-16, evaluate the definite integrals using the definition, as in Examples 3 and \(4 .\) \(\int_{-2}^{1}(2 x+\pi) d x\)
View solution Problem 13
Write the indicated sum in sigma notation. $$ a_{1}+a_{3}+a_{5}+a_{7}+\cdots+a_{99} $$
View solution