Question

Perform the indicated integrations. \(\int_{-2}^{-3} \frac{\sqrt{t^{2}}-1}{t^{3}} d t\)

Short answer

The value of the integral is approximately -0.0371.

Step-by-step solution

Adjust the Integrals Limits

First, recognize that the limits are inverted in the integral. An integral can be simplified by reversing the limits of integration, which also changes the sign of the integral:\[\int_{-2}^{-3} \frac{\sqrt{t^{2}}-1}{t^{3}} d t = -\int_{-3}^{-2} \frac{\sqrt{t^{2}}-1}{t^{3}} d t.\]

Simplify the Integrand

Next, simplify the integrand \( \frac{\sqrt{t^{2}}-1}{t^{3}} \). Since \( \sqrt{t^2} = |t| \), in the interval \([-3, -2]\) where \(t\) is negative, \( |t| = -t \). Thus, the integrand simplifies to:\[\frac{-t-1}{t^3} = \frac{-t}{t^3} - \frac{1}{t^3} = -\frac{1}{t^2} - \frac{1}{t^3}.\]

Integrate the Simplified Expression

Integrate each term separately within the new limits:\[\int_{-3}^{-2} \left( -\frac{1}{t^2} - \frac{1}{t^3} \right) dt = \int_{-3}^{-2} -t^{-2} dt - \int_{-3}^{-2} t^{-3} dt.\]

Evaluate the Integrals

First, evaluate \( \int -t^{-2} \, dt \):\[-t^{-2} \Rightarrow \int -t^{-2} \, dt = \frac{t^{-1}}{1} = -\frac{1}{t}.\]Next, evaluate \( \int t^{-3} \, dt \):\[t^{-3} \Rightarrow \int t^{-3} \, dt = \frac{t^{-2}}{-2} = -\frac{1}{2t^2}.\]

Apply the Evaluated Integral to the Limits

Sum the integrated terms from the evaluation:\[\left[-\frac{1}{t} - \frac{1}{2t^2} \right]_{-3}^{-2}\]Substitute the limits:\[-\left(\frac{1}{-2} + \frac{1}{2(-2)^2}\right) + \left(\frac{1}{-3} + \frac{1}{2(-3)^2}\right)\]This computes to:\[-\left(-\frac{1}{2} - \frac{1}{8}\right) + \left(-\frac{1}{3} - \frac{1}{18}\right) \approx 0.3333 - 0.3704 \approx -0.0371.\]

Simplify the Final Result

Calculate the difference between the computed expressions:\[0.3333 - 0.3704 = -0.0371.\]Thus, the value of the integral is approximately \(-0.0371\).

Key concepts

Integral Calculus

Integral calculus centers on the process of integrating functions, which is a fundamental concept in mathematics that helps calculate areas under curves, among other things. The goal is to find the integral of a function, which essentially provides the accumulated value, area or total change over an interval. Definite integration, as tackled in this exercise, involves integrating a function within specific limits.
To solve a definite integral, like in this exercise, you first work with the limits of integration. This shows you the range over which you're accumulating values. The general process involves finding the antiderivative of the function, then evaluating it at the upper and lower limits and finding the difference.
Integral calculus helps in various fields such as physics for calculating displacement, chemistry for finding reaction rates, and economics for determining consumer and producer surplus.

Integrating Rational Functions

Rational functions are expressions that are the ratio of two polynomials, like the integrand in our problem \( \frac{\sqrt{t^2} - 1}{t^3} \).
Integration of rational functions requires simplification to manageable forms, allowing for straightforward antiderivative calculations. Simplifying the integrand is essential: here, the absolute value nature of\(|t|\) was considered due to the square root, and since \(t\) in this interval is negative, \(\sqrt{t^2} = -t\).
Breaking the function into simpler terms, \( -\frac{1}{t^2} \) and \( -\frac{1}{t^3} \), allows us to integrate each component separately with straightforward rules. Understanding these foundational principles in calculus is significant because it equips students to handle more complex functions and prepares them for advanced calculus challenges.

Reverse Limits of Integration

Sometimes, integrals are presented with limits which are in reverse order, like from -2 to -3. When you encounter this, it’s important to remember that swapping the limits changes the sign of the integral. This property of definite integrals can be represented as: \[\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx.\]
Understanding this rule helps streamline problem-solving and is especially useful in proofs and calculations involving definite integrals. The initial step in solving our problem was to reverse the limits, turning a negative into a positive context for easier computation.
This concept emphasizes a crucial part of integration, where attention to the correct order of operations significantly impacts the final result. It illustrates one of many useful calculus strategies that aids in simplifying complex problems.