Problem 21 - Calculus Concepts

Question

For Activities 17 through \(22,\) use algebraic manipulation or integration by substitution as well as limits to evaluate the improper integral. $$ \int_{-\infty}^{-2} \frac{3 x^{4}}{x^{6}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{3}{2} \).

Step 1: Simplify the Integrand

First, simplify the integrand \( \frac{3x^4}{x^6} \). You can do this by dividing the numerator and the denominator by \( x^4 \), resulting in \( \frac{3}{x^2} \).

Step 2: Set up the Integral

Set up the improper integral after simplification. The integral becomes \( \int_{-rac{ ext{infinity}}{x}} \ to_{-2} \frac{3}{x^2} \, dx \).

Step 3: Determine the Antiderivative

The antiderivative of \( \frac{3}{x^2} \) is \( -\frac{3}{x} \). This can be obtained by recognizing the integral of \( x^{-2} \).

Step 4: Evaluate the Definite Integral

Evaluate the definite integral \( \int_{- ext{infinity} }^{-2} \frac{3}{x^2} \, dx = [ -\frac{3}{x} ]_{- ext{infinity} }^{-2} \).

Step 5: Compute the Limit

Compute the limit as \( x \to -\text{infinity} \). The term \( -\frac{3}{x} \) approaches \( 0 \) as \( x \to -\text{infinity} \).

Step 6: Substitute Limits in Antiderivative Expression

Substitute into the antiderivative expression: \([ -\frac{3}{-2} - (-\frac{3}{\text{infinity}})] = [\frac{3}{2} - 0] \).

Step 7: Final Result

The result of the improper integral \( \int_{- ext{infinity} }^{-2} \frac{3}{x^2} \, dx \) is \( \frac{3}{2} \).

Key Concepts

Integration by Substitution

Integration by substitution is a useful technique in calculus where we make integrals simpler to evaluate.
It involves changing variables to transform the integrand into a familiar form. This is similar to the chain rule in differentiation.
In our exercise, however, direct simplification was used instead of substitution. We transformed the integrand from \( \frac{3x^4}{x^6} \) to \( \frac{3}{x^2} \) by recognizing that the powers of \( x \) could be reduced through simple algebra.Sometimes an integral with a complex expression can become straightforward by changing the variable you integrate with respect to. Typical steps include:

• Selecting a new variable, say \( u \), linked to the original variable \( x \).
• Substituting \( u \) and its differential \( du \) into the integral.
• Solving the new, potentially simpler integral, then transforming back to the original variable if necessary.

Though substitution wasn't required in this exercise, just keep in mind: it's all about making things simpler!

Limits

The concept of limits is central when dealing with improper integrals.
A limit helps us make sense of integrals that go to infinity or encompass points of discontinuity.
In this exercise, because the integral extends to negative infinity, we use a limit to compute its value.To solve this, we take the antiderivative and then apply the limit as the lower bound goes to negative infinity. Here's how:

• Find the antiderivative ℎ, here it's \(-\frac{3}{x}\).
• Substitute the limits \(-\text{infinity}\) and \(-2\) into this expression.
• Compute the limit as it approaches negative infinity. Notice that \(-\frac{3}{x}\) approaches 0 as \( x \to -\text{infinity} \).

The limit helps us determine the incomplete part of the integral, which then leads us to the complete computation, yielding \( \frac{3}{2} \). Without limits, tackling improper integrals would be difficult.

Antiderivatives

An antiderivative is essentially the opposite of differentiation. It's a function whose derivative gives the original function.
Finding the antiderivative is a key step in solving integrals.In our exercise, the function \( \frac{3}{x^2} \) was simplified into an antiderivative. That was determined to be \(-\frac{3}{x}\).
This means differentiating \(-\frac{3}{x}\) gets us back to \( \frac{3}{x^2} \).To spot antiderivatives:

• Remember that power functions follow the rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is a constant.
• Negative powers (like our example \( x^{-2} \)) integrate using this method too, except for the special case \( n = -1 \).
• Polynomials usually combine using antiderivatives before reapplying any limits or evaluations.

Antiderivatives provide the building block to area calculations and evaluations for definite integrals. By analyzing them closely, we can solve integrals that initially seem daunting.