Problem 90 - Calculus Volume 1

Question

Suppose that \(\int_{0}^{4} f(x) d x=5\) and \(\int_{0}^{2} f(x) d x=-3,\) and \(\int_{0}^{4} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=2 .\) In the following exercises, compute the integrals. \(\int_{0}^{2}(f(x)-g(x)) d x\)

Step-by-Step Solution

Answer

\( \int_{0}^{2}(f(x)-g(x)) \, dx = -5 \)

Step 1: Understand the Problem

You are given the integrals of functions \(f(x)\) and \(g(x)\) over the intervals [0, 4] and [0, 2]. You need to find the integral of the function \(f(x) - g(x)\) over the interval [0, 2].

Step 2: Use the Property of Linearity

The integral of a difference of two functions is the difference of their integrals. This means \( \int_{0}^{2} (f(x) - g(x)) \, dx = \int_{0}^{2} f(x) \, dx - \int_{0}^{2} g(x) \, dx \).

Step 3: Substitute the Given Values

Substitute the given values from the problem: \( \int_{0}^{2} f(x) \, dx = -3 \) and \( \int_{0}^{2} g(x) \, dx = 2 \). So, the integral becomes: \( \int_{0}^{2} (f(x) - g(x)) \, dx = -3 - 2 \).

Step 4: Simplify the Expression

Simplify the expression to find the result: \( -3 - 2 = -5 \). Thus, \( \int_{0}^{2} (f(x) - g(x)) \, dx = -5 \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, representing the accumulated area under a curve for a given interval. When we calculate the definite integral of a function over a specific interval, like from 0 to 2, we are essentially finding the net "area" between the function's graph and the x-axis between these bounds.
This net area accounts for the space above and below the axis, which means parts below the x-axis are considered negative. So, when you see an integral like \( \int_{0}^{2} f(x) \, dx = -3 \), it implies the function has more area below the x-axis from 0 to 2.

• The limits of integration (0 and 2 in this example) define the boundaries of this calculation.
• Definite integrals provide a way to calculate total displacement, not just distance traveled, in physical problems.

Understanding this concept allows you to see integrals as powerful tools to aggregate quantities over intervals.

Linearity of Integration

Linearity is one of the key properties that makes integration such a versatile tool in calculus. It states that the integral of a sum (or difference) of functions is the sum (or difference) of their integrals. This rule simplifies calculations significantly.
For example, given \( \int_{0}^{2} (f(x) - g(x)) \, dx \), we can split it into two separate integrations: \( \int_{0}^{2} f(x) \, dx - \int_{0}^{2} g(x) \, dx \).

• This property allows you to tackle complex functions by breaking them down into simpler parts.
• Linearity ensures that constants can be factored out of the integral, such as \( c \int f(x) \, dx = \int c \, f(x) \, dx \).

In practical terms, this means you can handle each part of the expression individually and reassemble the pieces after computing their integrals. It's a bit like solving a complex puzzle by handling manageable sections first.

Integral Properties

Several integral properties are essential when solving calculus problems, and knowing them can make the process more straightforward. Each property provides a toolset for handling and simplifying integral computations. Let's explore a few of these helpful properties:
- **Additivity over intervals**: If you know an integral over a larger interval, you can break it down into smaller segments. For instance, \( \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx \).
- **Zero-width interval**: The integral over a zero-width interval is zero, \( \int_{a}^{a} f(x) \, dx = 0 \), since no area exists under a point.

• This property comes in handy when verifying calculations, like ensuring continuity between function segments.
• Each property acts as a guide for dissecting problems and strategically applying mathematical rules to simplify expressions.

Mastering these integral properties helps you tackle integrals more efficiently by deploying the right tool for each specific scenario.