Problem 79
Question
Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. If \(f\) and \(g\) are continuous on \([a, b]\) and \(k\) is a constant, then $$ \int_{a}^{b}[k f(x)+g(x)] d x=k \int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x $$
Step-by-Step Solution
Verified Answer
The statement is true. This is because of the linearity property of integration, which states that for any constants \(c\) and \(d\), and continuous functions \(f\) and \(g\) on the interval \([a, b]\), we have:
$$
\int_{a}^{b}[c f(x) + d g(x)] dx = c \int_{a}^{b} f(x) dx + d \int_{a}^{b} g(x) dx
$$
Applying this property with \(c=k\) and \(d=1\), we obtain the given equation, which is valid for all continuous functions on the interval \([a,b]\) and any constant \(k\).
1Step 1: State the property
The statement uses the linearity of integration: for continuous functions \( f, g \) on \( [a,b] \) and constants \( c, d \):
\( \int_a^b [cf(x) + dg(x)]\,dx = c\int_a^b f(x)\,dx + d\int_a^b g(x)\,dx \).
\( \int_a^b [cf(x) + dg(x)]\,dx = c\int_a^b f(x)\,dx + d\int_a^b g(x)\,dx \).
2Step 2: Conclusion
The statement is true by the linearity property of the definite integral, which follows from the linearity of the Riemann sum.
Key Concepts
Continuous FunctionsDefinite IntegralProperties of Integrals
Continuous Functions
In the realm of calculus, continuous functions play a pivotal role. A function is described as continuous on an interval, like \[a, b\], if at every point within that interval, the function can be graphically drawn without lifting the pen from the paper—that is, it has no breaks, jumps, or sharp corners.
Why is continuity important? It ensures that within the interval \[a, b\], the function behaves nicely, which has profound implications on its integrability. Essentially, if a function is continuous on a closed interval, it guarantees that the definite integral over that interval is well-defined. This concept underpins numerous theorems in calculus, such as the Fundamental Theorem of Calculus.
Why is continuity important? It ensures that within the interval \[a, b\], the function behaves nicely, which has profound implications on its integrability. Essentially, if a function is continuous on a closed interval, it guarantees that the definite integral over that interval is well-defined. This concept underpins numerous theorems in calculus, such as the Fundamental Theorem of Calculus.
Visualizing Continuity
Imagine a smooth curve that flows along the graph without interruption—this is a graphic representation of a continuous function. It's this smoothness that facilitates the practical application of integration, as the area under the curve doesn't have any gaps or undefined regions.Definite Integral
The definite integral is a fundamental concept that quantifies the accumulation of quantities such as area, volume, and other quantities that can be accumulated. Given a continuous function \( f(x) \) and an interval \[a, b\], the definite integral of the function from \( a \) to \( b \) is denoted as \(\int_{a}^{b} f(x) dx\) and represents the net area under the function's curve between \( a \) and \( b \).
To understand it intuitively, imagine summing up an infinite number of infinitesimally thin rectangles under the curve of \( f(x) \)—the width of these rectangles approaches zero and their number approaches infinity. The definite integral is the limit of these sums as the rectangle widths shrink to zero, a process known as 'integration'.
To understand it intuitively, imagine summing up an infinite number of infinitesimally thin rectangles under the curve of \( f(x) \)—the width of these rectangles approaches zero and their number approaches infinity. The definite integral is the limit of these sums as the rectangle widths shrink to zero, a process known as 'integration'.
Definite Integral and Its Calculation
Calculating the definite integral involves finding the anti-derivative of the function, applying the limits of integration, and then subtracting the lower limit value from the upper limit value. This process is greatly facilitated when the function in question is continuous.Properties of Integrals
Understanding the properties of integrals is crucial for tackling integration problems efficiently. Two of these properties are particularly significant: linearity and additivity.
Linearity of integration establishes that the integral operator is linear, meaning that for continuous functions \( f \) and \( g \) and constants \( k \) and \( c \), the integral of the sum is the sum of the integrals, and the integral of a constant multiplied by a function is the constant multiplied by the integral of the function. In short, \(\int_{a}^{b}[k f(x) + c g(x)] dx = k \int_{a}^{b} f(x) dx + c \int_{a}^{b} g(x) dx\).
Additivity of integration, on the other hand, tells us that if we have two adjacent intervals, the integral over the entire interval is the sum of the integrals over the subintervals. Hence, for continuous functions and an intermediate point \( c \) in the interval \[a, b\], it holds that \(\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx\).
Linearity of integration establishes that the integral operator is linear, meaning that for continuous functions \( f \) and \( g \) and constants \( k \) and \( c \), the integral of the sum is the sum of the integrals, and the integral of a constant multiplied by a function is the constant multiplied by the integral of the function. In short, \(\int_{a}^{b}[k f(x) + c g(x)] dx = k \int_{a}^{b} f(x) dx + c \int_{a}^{b} g(x) dx\).
Additivity of integration, on the other hand, tells us that if we have two adjacent intervals, the integral over the entire interval is the sum of the integrals over the subintervals. Hence, for continuous functions and an intermediate point \( c \) in the interval \[a, b\], it holds that \(\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx\).
Applying Properties for Simplification
In solving integrals, exploiting these properties can immensely simplify the process. For instance, when faced with a complex integrand that can be broken down into simpler parts, using linearity can make the task more manageable. Similarly, additivity can be used to break an interval into parts to tackle an integral piece by piece if needed.Other exercises in this chapter
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