Problem 86 - Calculus and its applications

Question

Solve each integral. Each can be found using rules developed in this section, but some algebra may be required. $$ \int \frac{t^{3}+8}{t+2} d t $$

Step-by-Step Solution

Answer

The integral is \( \frac{t^3}{3} - t^2 + 4t + C \).

Step 1: Understand the Problem

We need to evaluate the integral \( \int \frac{t^{3}+8}{t+2} \, dt \). This is a rational function that can be simplified by algebraic long division.

Step 2: Perform Algebraic Long Division

We divide \( t^3 + 8 \) by \( t + 2 \). Start by dividing the first term of the numerator by the first term of the denominator: \( t^3 / t = t^2 \). Multiply \( t^2 \) by \( t + 2 \) and subtract from the original polynomial, getting a new polynomial \( t^3 + 8 - (t^3 + 2t^2) = -2t^2 + 8 \). Repeat this process on \( -2t^2 + 8 \). The final result of the division is \( t^2 - 2t + 4 + \frac{0}{t+2} \).

Step 3: Rewrite the Integral

The integral \( \int \frac{t^{3}+8}{t+2} \, dt \) can now be rewritten using the result from division: \( \int (t^2 - 2t + 4) \, dt \).

Step 4: Integrate Term by Term

Integrate each term separately: \( \int t^2 \, dt = \frac{t^3}{3} \), \( \int -2t \, dt = -t^2 \), and \( \int 4 \, dt = 4t \). Combine these results for the final integral.

Step 5: Write Down the Solution

Combine all integrated terms: \( \frac{t^3}{3} - t^2 + 4t + C \), where \( C \) is the constant of integration. This is the final evaluated integral.

Key Concepts

Algebraic Long Division

Algebraic long division is a method used to divide polynomials, similar to how you divide numbers. This technique is especially useful when simplifying rational functions.
To apply this, treat the division like a standard quotient problem: divide each term of the polynomial and subtract step by step.
For example, to divide the polynomial \( t^3 + 8 \) by \( t + 2 \), you start by dividing the first term in the numerator \( t^3 \) by the first term in the denominator \( t \), which gives \( t^2 \).

• Multiply \( t^2 \) by \( t + 2 \) resulting in \( t^3 + 2t^2 \).
• Subtract this from \( t^3 + 8 \), resulting in \( -2t^2 + 8 \).
• Repeat this process for each subsequent term.

The operation continues until all terms have been divided. The result in this exercise is \( t^2 - 2t + 4 \). This means the polynomial \( t^3 + 8 \) divided by \( t + 2 \) simplifies into \( t^2 - 2t + 4 \) with no remainder.

Rational Functions

Rational functions are expressions that represent the ratio of two polynomials. They take the form \( \frac{P(t)}{Q(t)} \), where \( P(t)\) and \( Q(t)\) are polynomials.

• In our exercise, \( \frac{t^3 + 8}{t + 2} \) is a rational function.
• Key point: Simplifying rational functions often involves using techniques like algebraic long division.

Understanding the form and behavior of rational functions is crucial when dealing with calculus operations like integration.
When integrating, simplifying the expression helps in breaking it down into manageable parts. Converting the expression to separated polynomials through division allows us to integrate each term easily. This simplifies the process of finding an antiderivative for the function.

Antiderivatives

An antiderivative represents a function whose derivative is the original function. In terms of integrals, finding an antiderivative means solving an indefinite integral.

• Consider the integral \( \int (t^2 - 2t + 4) \, dt \).
• This leads to finding an antiderivative for each polynomial term separately.

The process involves:

• Applying basic integration rules to each term.
• For \( t^2 \), the antiderivative is \( \frac{t^3}{3} \).
• For \(-2t\), it's \(-t^2\).
• For \(4\), it's \(4t\).

After finding these antiderivatives, combine them to form \( \frac{t^3}{3} - t^2 + 4t + C \), where \( C \) is the constant of integration. This is the final solution for the integral, representing the family of functions whose derivative is the original rational function.