Question

Given \(\int_{-2}^{2} g(x) d x=8\) and \(\int_{0}^{2} g(x) d x=3,\) find (a) \(\int_{-2}^{0} g(x) d x\) (b) \(\int_{2}^{-2} g(x) d x\) (c) \(\int_{0}^{-2} 5 g(x) d x\) (d) \(\int_{-2}^{2} 2 g(x) d x\)

Short answer

Question: Find the value of the following definite integrals: a) \(\int_{-2}^{0} g(x) d x\) b) \(\int_{2}^{-2} g(x) d x\) c) \(\int_{0}^{-2} 5 g(x) d x\) d) \(\int_{-2}^{2} 2 g(x) d x\) Answer: a) \(\int_{-2}^{0} g(x) d x = 5\) b) \(\int_{2}^{-2} g(x) d x = -8\) c) \(\int_{0}^{-2} 5 g(x) d x = -25\) d) \(\int_{-2}^{2} 2 g(x) d x = 16\)

Step-by-step solution

Use Definite Integral Property: \(\int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx = \int_{a}^{b} f(x)dx\)

Given that \(\int_{-2}^{2} g(x) d x=8\) and \(\int_{0}^{2} g(x) d x=3,\) we can replace \(a=-2, c=0, b=2\) in the property to find \(\int_{-2}^{0} g(x) d x.\) So, \(\int_{-2}^{0} g(x) d x + \int_{0}^{2} g(x) d x = 8\). Then, by substituting the known integral value, we get \(\int_{-2}^{0} g(x) d x = 8 - 3 = 5.\) b) To find \(\int_{2}^{-2} g(x) d x\)

Use Definite Integral Property: \(\int_{b}^{a} f(x)dx = - \int_{a}^{b} f(x)dx\)

Given that \(\int_{-2}^{2} g(x) d x=8\), we can replace \(a=-2, b=2\) in the property to find \(\int_{2}^{-2} g(x) d x\), which gives us \(\int_{2}^{-2} g(x) d x = - \int_{-2}^{2} g(x) d x = -8.\) c) To find \(\int_{0}^{-2} 5 g(x) d x\)

Use Definite Integral Property: \(\int_{a}^{b} kf(x)dx = k \times \int_{a}^{b} f(x)dx\)

First, we need to find \(\int_{0}^{-2} g(x) d x\). We can use the property \(\int_{b}^{a} f(x)dx = - \int_{a}^{b} f(x)dx\). So, \(\int_{0}^{-2} g(x) d x = - \int_{-2}^{0} g(x) d x\). As found earlier, \(\int_{-2}^{0} g(x) d x = 5\). Thus, \(\int_{0}^{-2} g(x) d x = -5\). Now, replacing \(a=0, b=-2, k=5, f(x)=g(x)\) in the property, we find \(\int_{0}^{-2} 5 g(x) d x = 5 \times \int_{0}^{-2} g(x) d x = 5 \times (-5) = -25.\) d) To find \(\int_{-2}^{2} 2 g(x) d x\)

Use Definite Integral Property: \(\int_{a}^{b} kf(x)dx = k \times \int_{a}^{b} f(x)dx\)

Given that \(\int_{-2}^{2} g(x) d x=8\), we can replace \(a=-2, b=2, k=2, f(x)=g(x)\) in the property to find \(\int_{-2}^{2} 2 g(x) d x\). Therefore, \(\int_{-2}^{2} 2 g(x) d x = 2 \times \int_{-2}^{2} g(x) d x = 2 \times 8 = 16.\)

Key concepts

Properties of Integrals

Definite integrals have some unique properties that make them very useful, especially when solving complex calculus problems. These properties help break down integrals into simpler parts and reassemble them in more comprehensible ways.
\(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \) is one such property, showing that changing the limits of integration reverses the sign of the integral value.
Another crucial property is \(\int_{a}^{b} kf(x) \, dx = k \times \int_{a}^{b} f(x) \, dx\). This states that a constant factor can be taken outside the integral, simplifying calculations when a function is multiplied by a constant.
Understanding these properties enhances skills in manipulating integrals for easier computation.

Integral Calculus

Integral calculus is a branch of calculus focused on the concept of integrals. It is the counterpart to differential calculus, which focuses on derivations. Integral calculus is crucial because it deals with accumulation of quantities, such as areas under curves and total values from rates of change.
The definite integral is a fundamental concept in integral calculus, used to find the exact area under a graph. It is expressed as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration.
Another key aspect of integral calculus is solving differential equations, which often involves reversing differentiation processes.

• Definite Integrals: Summarize changes over intervals.
• Indefinite Integrals: Integrals without bounds, related to the antiderivative.
• Fundamental Theorem of Calculus: Links differentiation and integration.

By developing an understanding of these areas, students gain powerful tools for solving real-world problems.

Calculus AB

Calculus AB is frequently part of high school and introductory college math courses, providing students with an understanding of foundational calculus principles. The course covers both differential and integral calculus.
In Calculus AB, students learn to solve problems using the derivative for rates of change and the integral for total accumulation or net change.
Students explore the properties and methods for computing both definite and indefinite integrals. These basics form a critical foundation for further study in calculus-intensive fields.
Mastery of Calculus AB topics helps students prepare for more advanced studies in mathematics, science, and engineering.

• Conceptual Understanding: Develop reasoning about changes and accumulation.
• Problem Solving: Apply calculus concepts to new situations.

Building a strong understanding in Calculus AB lays the groundwork for success in complex problems and advanced math courses.

AP Calculus Exam

The AP Calculus Exam tests high school students' grasp of the material covered in AP Calculus courses, including both Calculus AB and BC. This exam is widely recognized for college credit and placement in advanced coursework.
Integral concepts, particularly the properties and applications of definite integrals, are key components of the exam. Students must be adept at making sense of integrals in different contexts and applying the fundamental theorem of calculus effectively.
The exam requires a solid internalization of calculus principles and the ability to solve practical problems using integration techniques.

Exam Format: Consists of multiple-choice and free-response questions.

Scoring: Combines both parts to determine competency levels, often resulting in college credit or course advancement.

Being well-prepared for the AP Calculus Exam can significantly influence students' academic paths, facilitating higher college placement and advanced learning opportunities.