Problem 9
Question
Suppose that f and g are integrable and that\begin{equation} \int_{1}^{2} f(x) d x=-4, \quad \int_{1}^{5} f(x) d x=6, \int_{1}^{5} g(x) d x=8 \end{equation} Use the rules in Table 5.6 to find $$ \begin{array}{ll}{\text { a. } \int_{2}^{2} g(x) d x} & {\text { b. } \int_{5}^{1} g(x) d x} \\ {\text { c. } \int_{1}^{2} 3 f(x) d x} & {\text { d. } \int_{2}^{5} f(x) d x} \\ {\text { e. } \int_{1}^{5}[f(x)-g(x)] d x} & {\text { f. } \int_{1}^{5}[4 f(x)-g(x)] d x}\end{array} $$
Step-by-Step Solution
Verified Answer
a. 0; b. -8; c. -12; d. 10; e. -2; f. 16.
1Step 1: Calculate \( \int_{2}^{2} g(x) dx \)
Since the limits of integration are the same (2 to 2), the value of the integral is zero because there is no interval over which to integrate: \[\int_{2}^{2} g(x) dx = 0\]
2Step 2: Calculate \( \int_{5}^{1} g(x) dx \)
To evaluate an integral with reversed limits, change the limits and multiply by -1:\[\int_{5}^{1} g(x) dx = -\int_{1}^{5} g(x) dx = -8\]
3Step 3: Calculate \( \int_{1}^{2} 3f(x) dx \)
By the constant multiple rule, you can factor constants out of the integral:\[\int_{1}^{2} 3f(x) dx = 3 \cdot \int_{1}^{2} f(x) dx = 3 \cdot (-4) = -12\]
4Step 4: Calculate \( \int_{2}^{5} f(x) dx \)
Use the property of integral interval addition:\[\int_{1}^{5} f(x) dx = \int_{1}^{2} f(x) dx + \int_{2}^{5} f(x) dx\]Substitute known values and solve for the desired integral:\[6 = (-4) + \int_{2}^{5} f(x) dx \quad \Rightarrow \quad \int_{2}^{5} f(x) dx = 6 + 4 = 10\]
5Step 5: Calculate \( \int_{1}^{5} [f(x) - g(x)] dx \)
Use the linearity of integration, which allows you to integrate terms separately:\[\int_{1}^{5} [f(x) - g(x)] dx = \int_{1}^{5} f(x) dx - \int_{1}^{5} g(x) dx\]Substitute the known values:\[6 - 8 = -2\]
6Step 6: Calculate \( \int_{1}^{5} [4f(x) - g(x)] dx \)
Separate into two integrals and factor constants:\[\int_{1}^{5} [4f(x) - g(x)] dx = 4 \cdot \int_{1}^{5} f(x) dx - \int_{1}^{5} g(x) dx\]Substitute known values:\[4(6) - 8 = 24 - 8 = 16\]
Key Concepts
Integrable FunctionsConstant Multiple RuleLinearity of IntegrationProperties of Definite Integrals
Integrable Functions
An integrable function is one that can be mathematically represented by a definite integral over a certain interval. This means we can compute areas under the curve of this function.
For example, if you have a function like \( f(x) \), it is considered integrable if its integral over an interval \([a, b]\), such as \( \int_{a}^{b} f(x)\, dx \), exists.
This is a key concept because it allows us to compute areas, volumes, and more complex phenomena across mathematics and applied fields.
For example, if you have a function like \( f(x) \), it is considered integrable if its integral over an interval \([a, b]\), such as \( \int_{a}^{b} f(x)\, dx \), exists.
This is a key concept because it allows us to compute areas, volumes, and more complex phenomena across mathematics and applied fields.
- To be integrable, a function may have certain properties such as being bounded or continuous; however, there are exceptions.
- Most functions encountered in basic calculus courses, such as polynomials or trigonometric functions, are integrable over their domains.
Constant Multiple Rule
The constant multiple rule of integration is a handy tool when working with integrals. It allows you to 'factor out' constant multipliers outside of an integral.
Suppose you have an integral that looks like this: \( \int_{a}^{b} c \cdot f(x) \, dx \), where \(c\) is a constant.
According to the constant multiple rule, you can rewrite it as \( c \cdot \int_{a}^{b} f(x) \, dx \).
This simplification is powerful because it allows us to easily deal with constants, focusing instead on the function inside.
Suppose you have an integral that looks like this: \( \int_{a}^{b} c \cdot f(x) \, dx \), where \(c\) is a constant.
According to the constant multiple rule, you can rewrite it as \( c \cdot \int_{a}^{b} f(x) \, dx \).
This simplification is powerful because it allows us to easily deal with constants, focusing instead on the function inside.
- Remember, this rule holds for any constant, whether it is positive or negative.
- For example, in our exercise, we simplified \( \int_{1}^{2} 3f(x) \, dx \) to \( 3 \cdot \int_{1}^{2} f(x) \, dx \), making the calculation straightforward.
Linearity of Integration
Linearity of integration is a useful property that lets you break down integrals of sums or differences into simpler parts.
This property means that the integral of a sum is the sum of the integrals and the integral of a difference is the difference of the integrals:
\( \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \).
Similarly, for differences:
\( \int_{a}^{b} [f(x) - g(x)] \, dx = \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx \).
This property is very helpful in dissecting complex expressions and solving them piece by piece.
This property means that the integral of a sum is the sum of the integrals and the integral of a difference is the difference of the integrals:
\( \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \).
Similarly, for differences:
\( \int_{a}^{b} [f(x) - g(x)] \, dx = \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx \).
This property is very helpful in dissecting complex expressions and solving them piece by piece.
- In our exercise, this property was crucial for steps 5 and 6 where we had to deal with combinations of \(f(x)\) and \(g(x)\).
- It simplifies handling complex integrals by breaking them into more manageable parts.
Properties of Definite Integrals
Definite integrals have certain properties that can help simplify calculations and understand the behavior of functions over intervals.
Some essential properties include:
In the provided exercise, we used the zero width property and additivity to solve specific integrals.
Some essential properties include:
- Zero Width Property: \( \int_{a}^{a} f(x) \, dx = 0 \) implies that integrating over an interval with no width results in zero.
- Reversal Property: If you reverse the limits of an integral, the integral's value will be the negative of the original: \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).
- Additivity of Intervals: Integrals over adjacent intervals can be added: \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \).
In the provided exercise, we used the zero width property and additivity to solve specific integrals.
Other exercises in this chapter
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