Consider the following for the next two (02) items that follow : Let \(l=\int_a^b \frac{|x|}{x} d x, \) a < b
What is l equal to when a < b < 0 ?
a - b
Evaluating the Definite Integral of \( \frac{|x|}{x} \) when \( a < b < 0 \)
The problem asks us to evaluate the definite integral \( l=\int_a^b \frac{|x|}{x} d x \) under the specific condition that \( a < b < 0 \). This means both the lower limit \( a \) and the upper limit \( b \) of the integration are negative, and the entire interval \( [a, b] \) lies to the left of zero on the number line.
Understanding the Integrand \( \frac{|x|}{x} \)
The function inside the integral is \( \frac{|x|}{x} \). This function involves the absolute value of \( x \). Let's consider how the absolute value function behaves:
- If \( x > 0 \), then \( |x| = x \). In this case, \( \frac{|x|}{x} = \frac{x}{x} = 1 \).
- If \( x < 0 \), then \( |x| = -x \). In this case, \( \frac{|x|}{x} = \frac{-x}{x} = -1 \).
- If \( x = 0 \), the expression \( \frac{|x|}{x} \) is undefined because the denominator is zero.
The given condition for this specific problem is \( a < b < 0 \). This tells us that for every value of \( x \) within the interval of integration \( [a, b] \), \( x \) is strictly negative (\( x < 0 \)).
Therefore, for all \( x \) in the interval \( [a, b] \) when \( a < b < 0 \), the integrand simplifies to \( \frac{|x|}{x} = -1 \).
Evaluating the Definite Integral
Since the integrand is constantly \( -1 \) over the interval \( [a, b] \) under the given condition \( a < b < 0 \), the integral becomes:
\( l = \int_a^b \frac{|x|}{x} dx = \int_a^b (-1) dx \)
To evaluate this definite integral, we find the antiderivative of \( -1 \) and apply the Fundamental Theorem of Calculus. The antiderivative of \( -1 \) with respect to \( x \) is \( -x \) (plus a constant of integration, but that is not needed for definite integrals).
Applying the Fundamental Theorem of Calculus: \( \int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a) \) Here, \( f(x) = -1 \) and \( F(x) = -x \).
So, \( l = [-x]_a^b = -(b) - (-(a)) = -b + a = a - b \).
Conclusion
When \( a < b < 0 \), the value of the definite integral \( l = \int_a^b \frac{|x|}{x} dx \) is \( a - b \).
Revision Table: Key Concepts for Integral Evaluation
| Concept | Description |
|---|---|
| Definite Integral | Represents the accumulation of a quantity over an interval, often visualized as the signed area under the curve of a function. |
| Absolute Value Function | A function \( |x| \) that returns the non-negative value of \( x \). Defined as \( |x| = x \) for \( x \ge 0 \) and \( |x| = -x \) for \( x < 0 \). |
| Integrand | The function being integrated in an integral. In this case, it is \( \frac{|x|}{x} \). |
| Fundamental Theorem of Calculus | A theorem linking differentiation and integration, providing a method to calculate definite integrals using antiderivatives. |
Additional Information: Integrating Functions with Absolute Value
When evaluating integrals involving absolute value functions like \( |x| \), the key step is to split the integral into intervals where the expression inside the absolute value maintains a constant sign. For example, to integrate \( \int_a^b |x| dx \), you would typically split the integral at \( x=0 \) if \( a < 0 \) and \( b > 0 \), because the definition of \( |x| \) changes at \( 0 \).
In this specific problem, \( \int_a^b \frac{|x|}{x} dx \), the function \( \frac{|x|}{x} \) is equivalent to \( 1 \) for \( x > 0 \) and \( -1 \) for \( x < 0 \). The point where the function might change its definition is \( x=0 \).
Since the given condition \( a < b < 0 \) ensures that the entire interval of integration \( [a, b] \) is strictly less than zero, we don't need to split the integral. The integrand \( \frac{|x|}{x} \) is consistently \( -1 \) throughout the interval, simplifying the calculation significantly. If the interval included \( 0 \) (e.g., \( a < 0 < b \)), we would need to split the integral at \( 0 \), like \( \int_a^b \frac{|x|}{x} dx = \int_a^0 \frac{|x|}{x} dx + \int_0^b \frac{|x|}{x} dx \).
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