Chapter 5: Problem 315
Find the area under the graph of \(g(t)=\frac{t}{\left(1-t^{2}\right)^{a}}\) between \(t=0\) and \(t=x,\) where \(0 < x < 1\) and \(a > 0\) is fixed. Evaluate the limit as \(x \rightarrow 1\)
Short Answer
Expert verified
The integral evaluates to a finite limit if and only if \( a > 1 \).
Step by step solution
01
Setup the Integral
To calculate the area under the curve defined by the function \( g(t) = \frac{t}{(1-t^2)^a} \) from \( t = 0 \) to \( t = x \), we set up an integral:\[A(x) = \int_{0}^{x} \frac{t}{(1-t^2)^a} \, dt\]
02
Change of Variables
Consider the substitution \( u = 1 - t^2 \), which implies that \( du = -2t \, dt \). This substitution simplifies the integral. When \( t = 0, u = 1 \) and when \( t = x, u = 1 - x^2 \).After substitution, the integral becomes:\[A(x) = \int_{1}^{1-x^2} \frac{-1}{2} \cdot u^{-a} \, du = \frac{1}{2} \int_{1-x^2}^{1} u^{-a} \, du\]
03
Evaluate the Definite Integral
The integral \( \int u^{-a} \, du = \frac{u^{1-a}}{1-a} \), as long as \( a eq 1 \). We evaluate the definite integral:\[\frac{1}{2(1-a)} \left[ (1)^{1-a} - (1-x^2)^{1-a} \right] = \frac{1}{2(1-a)} \left[ 1 - (1-x^2)^{1-a} \right]\]
04
Evaluate the Limit as \(x \rightarrow 1\)
To find \( \lim_{x \to 1} A(x) \), evaluate:\[\lim_{x \to 1} \frac{1}{2(1-a)} \left[ 1 - (1-x^2)^{1-a} \right]\]As \( x \to 1 \), \( 1-x^2 \to 0 \). Since \( a > 0\), \((1-x^2)^{1-a} \) tends towards infinity if \( 1-a < 0 \) (or, equivalently, \( a > 1 \)), leading to a finite limit. However, if \( a \leq 1 \), the term results in a zero or undefined value, respectively.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The definite integral is a powerful tool used in calculus to calculate the total area under a curve over a specific interval. This process transforms the function over an interval into a single number, representing this area. For the problem at hand, we are interested in the area between the curve defined by the function \( g(t) = \frac{t}{(1-t^2)^a} \) and the x-axis, from \( t = 0 \) to \( t = x \). Here, the integral is defined as:
- \( A(x) = \int_{0}^{x} \frac{t}{(1-t^2)^a} \, dt \)
Change of Variables
The method of change of variables, or substitution, is a technique often employed in integration to simplify complex integrals. It involves introducing a new variable to make the integration process easier. In this exercise, we employ the substitution \( u = 1 - t^2 \). This substitution is strategic because it converts the given variable from \( t \) to \( u \), simplifying the expression.
- Differentially, \( du = -2t \, dt \), letting us rearrange the integral's differential component.
- By substituting, the integration limits also transform: when \( t = 0 \), \( u = 1 \); when \( t = x \), \( u = 1 - x^2 \).
- The original integral expression becomes:
- \( A(x) = \int_{1}^{1-x^2} \frac{-1}{2} \cdot u^{-a} \, du \)
Limit Evaluation
Limit evaluation is a fundamental aspect of calculus, often used to determine the behavior of a function as it approaches a particular value. In this exercise, after finding the integral, the final step is to evaluate the limit of the result as \( x \rightarrow 1 \).
- The expression obtained from evaluating the integral is:
- \( \lim_{x \to 1} \frac{1}{2(1-a)} \left[ 1 - (1-x^2)^{1-a} \right] \)
- As \( x \to 1 \), \( 1-x^2 \to 0 \).
- Special considerations arise depending on the value of \( a \):
- If \( a > 1 \), since \( (1-x^2)^{1-a} \rightarrow \infty \), the integral has a finite and meaningful limit.
- If \( a \leq 1 \), complicated outcomes can occur, such as results tending towards zero or being undefined.