91Ó°ÊÓ

When dealing with improper integrals, the concept of limits of integration is crucial. The limits define where you start and stop when calculating an integral. Usually, these limits are finite numbers, like 0 to 5.
In this exercise, however, the lower limit is \(-\infty\), an infinite value, which makes the integral improper. To tackle this, imagine a workaround: replace the infinite limit with a variable, such as \(a\), which will later approach \(-\infty\). This transforms your integral into a manageable form. This approach changes our integral to:

• \( \int_{a}^{-2} \frac{3}{x^{3}} \, dx \)

This small adjustment allows us to apply the valuable tools of calculus to solve the problem.

The term "antiderivative" refers to the opposite of differentiation. It's essentially finding a function whose derivative is the given function. To solve our integral, we first need the antiderivative of \( \frac{3}{x^3} \).
To find this, we rewrite it in a form suitable for basic integration rules:

• The integrand \( \frac{3}{x^3} \) can be rewritten as \( 3x^{-3} \).

Once rewritten, apply the power rule for integration. The power rule states that the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) (as long as \( n eq -1\)):

• \( \int 3x^{-3} \, dx = 3 \cdot \frac{x^{-2}}{-2} = -\frac{3}{2x^2} + C \)

The "+ C" symbolizes that there are infinitely many functions that can have the same derivative. However, for definite integrals, we will focus on evaluation, eliminating \(C\) in the process.

When a limit relates to infinity, like in improper integrals, we define it as an infinite limit. Our original problem specifies a lower bound of \(-\infty\). Infinite limits can seem challenging, but we handle them using limits.The expression changes into taking the limit as a variable approaches infinity. For the given exercise:

• \( \lim_{a \to -\infty} \int_{a}^{-2} \frac{3}{x^{3}} \, dx \)

This simply means integrating, then determining the function's behavior as our variable approaches infinity. After obtaining the antiderivative, we apply it over \([-\infty, -2]\) by taking:

• \( [-\frac{3}{2x^2}]_a^{-2} \)

This requires expanding and simplifying to evaluate the limit:

• \( -\frac{3}{8} + \frac{3}{2a^2} \) approaches \( -\frac{3}{8} \) as \( 1/a^2 \) goes to 0.

Thus, our improper integral evaluates to \:

• -\frac{3}{8}

The infinite limit transforms a seemingly complex problem into solvable steps, using limits as a tool.