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Magazine Chapter 5: Problem 314
Find the area under the graph of \(f(t)=\frac{t}{\left(1+t^{2}\right)^{a}}\) between \(t=0\) and \(t=x\) where \(a>0\) and \(a \neq 1\) is fixed, and evaluate the limit as \(x \rightarrow \infty\) .
Short Answer
Expert verified
The area's limit as \( x \rightarrow \infty \) is \( \frac{1}{2(a-1)} \).
Step by step solution
01
Set Up the Integral
To find the area under the curve of the function, set up the definite integral of \( f(t) = \frac{t}{(1+t^2)^a} \) from \( t = 0 \) to \( t = x \). The area \( A \) is given by the integral: \[A(x) = \int_{0}^{x} \frac{t}{(1+t^2)^a} \, dt.\]
02
Use Substitution Method
A useful substitution for this integral is \( u = 1 + t^2 \). Then \( du = 2t \, dt \), or equivalently \( t \, dt = \frac{1}{2} \, du \). The bounds change from \( t = 0 \), which gives \( u = 1 \), to \( t = x \), which gives \( u = 1 + x^2 \). Substituting these values, the integral becomes: \[A(x) = \frac{1}{2} \int_{1}^{1+x^2} u^{-a} \, du.\]
03
Integrate with New Variable
The integral \( \int u^{-a} \, du \) can be evaluated to: \[A(x) = \frac{1}{2} \cdot \left[ \frac{u^{1-a}}{1-a} \right]_{1}^{1+x^2}.\]
04
Substitute Back the Limits
Substitute the limits back into the antiderivative: \[A(x) = \frac{1}{2(1-a)} \left[ (1 + x^2)^{1-a} - 1^{1-a} \right].\]
05
Evaluate Limit as x Approaches Infinity
Evaluate the limit of \( A(x) \) as \( x \) approaches infinity: \[\lim_{x \to \infty} A(x) = \frac{1}{2(1-a)} \left[ \lim_{x \to \infty} (1 + x^2)^{1-a} - 1 \right].\] Since \( a > 1 \), as \( x \to \infty \), \( (1 + x^2)^{1-a} \to 0 \). Therefore, \[\lim_{x \to \infty} A(x) = -\frac{1}{2(1-a)}.\]
06
Final Answer
Simplifying \( -\frac{1}{2(1-a)} \) can be rewritten as \( \frac{1}{2(a-1)} \). Therefore, the area approaches \( \frac{1}{2(a-1)} \) as \( x \to \infty \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral involves finding the area under a curve represented by a function over a specific interval.
It is an essential idea in calculus, particularly in understanding how to accumulate quantities.
In this exercise, the function defined is \( f(t) = \frac{t}{(1+t^2)^a} \), and we're interested in the area under this curve from \( t = 0 \) to \( t = x \).
By setting up a definite integral from 0 to \( x \), we can mathematically describe this area as \( A(x) = \int_{0}^{x} \frac{t}{(1+t^2)^a} \, dt \).
The definite integral essentially serves as a tool that not only helps in calculating this area but also offers insights into the changing behavior of the function over a given limit.
It is an essential idea in calculus, particularly in understanding how to accumulate quantities.
In this exercise, the function defined is \( f(t) = \frac{t}{(1+t^2)^a} \), and we're interested in the area under this curve from \( t = 0 \) to \( t = x \).
By setting up a definite integral from 0 to \( x \), we can mathematically describe this area as \( A(x) = \int_{0}^{x} \frac{t}{(1+t^2)^a} \, dt \).
The definite integral essentially serves as a tool that not only helps in calculating this area but also offers insights into the changing behavior of the function over a given limit.
Substitution Method
The substitution method is a powerful technique in integration used to simplify difficult integrals.
In this exercise, we used the substitution method to make the integration manageable.
The substitution \( u = 1 + t^2 \) transforms the integral by changing its variable of integration.
It turns the differential \( dt \) into \( du = 2t \, dt \) or \( t \, dt = \frac{1}{2} \, du \).
The substitution also changes the limits of integration: when \( t = 0 \), \( u = 1 \); and when \( t = x \), \( u = 1 + x^2 \).
This transformation simplifies the integral to \( A(x) = \frac{1}{2} \int_{1}^{1+x^2} u^{-a} \, du \), making it easier to evaluate.
In this exercise, we used the substitution method to make the integration manageable.
The substitution \( u = 1 + t^2 \) transforms the integral by changing its variable of integration.
It turns the differential \( dt \) into \( du = 2t \, dt \) or \( t \, dt = \frac{1}{2} \, du \).
The substitution also changes the limits of integration: when \( t = 0 \), \( u = 1 \); and when \( t = x \), \( u = 1 + x^2 \).
This transformation simplifies the integral to \( A(x) = \frac{1}{2} \int_{1}^{1+x^2} u^{-a} \, du \), making it easier to evaluate.
Limit Evaluation
Evaluating the limit is a crucial step in determining how a function behaves as it approaches a certain point, often infinity.
In this situation, finding the limit \( \lim_{x \to \infty} A(x) \) allows us to assess the behavior of the area under the curve as \( x \) progresses without bound.
The specific expression involves examining \[ \lim_{x \to \infty} \frac{1}{2(1-a)} \left[ (1 + x^2)^{1-a} - 1 \right]. \]
Given that \( a > 1 \), the term \((1 + x^2)^{1-a}\) approaches zero as \( x \to \infty \), due to the negative exponent, diminishing the influence of \( x \).
This results in the simplification of the limit, eventually leading to a clear and defined outcome of \( \frac{1}{2(a-1)} \).
In this situation, finding the limit \( \lim_{x \to \infty} A(x) \) allows us to assess the behavior of the area under the curve as \( x \) progresses without bound.
The specific expression involves examining \[ \lim_{x \to \infty} \frac{1}{2(1-a)} \left[ (1 + x^2)^{1-a} - 1 \right]. \]
Given that \( a > 1 \), the term \((1 + x^2)^{1-a}\) approaches zero as \( x \to \infty \), due to the negative exponent, diminishing the influence of \( x \).
This results in the simplification of the limit, eventually leading to a clear and defined outcome of \( \frac{1}{2(a-1)} \).
Area Under Curve
Finding the area under a curve is a fundamental application of integration, often used to calculate the total accumulated quantity of a given function.
In our problem, the task involves determining the area below the graph of \( f(t) = \frac{t}{(1+t^2)^a} \) from \( t = 0 \) to \( x \).
By integrating the function over this interval, we effectively sum up all the infinitesimally small areas beneath the curve.
This gives us \( A(x) \) as a function of \( x \), representing the accumulated area.
Once the limit of this area is calculated as \( x \to \infty \), it reveals an intriguing aspect of the function's infinite behavior.
The exercise finds that the area approaches \( \frac{1}{2(a-1)} \), offering insights into how the total area stabilizes despite the curve extending towards infinity.
In our problem, the task involves determining the area below the graph of \( f(t) = \frac{t}{(1+t^2)^a} \) from \( t = 0 \) to \( x \).
By integrating the function over this interval, we effectively sum up all the infinitesimally small areas beneath the curve.
This gives us \( A(x) \) as a function of \( x \), representing the accumulated area.
Once the limit of this area is calculated as \( x \to \infty \), it reveals an intriguing aspect of the function's infinite behavior.
The exercise finds that the area approaches \( \frac{1}{2(a-1)} \), offering insights into how the total area stabilizes despite the curve extending towards infinity.
