Problem 39 - College Algebra and Calculus: An Applied Approach

Question

$$ \text { Evaluate the definite integral. } $$$$ \int_{1}^{2} \frac{x^{3}-4 x^{2}-3 x+3}{x^{2}-3 x} d x $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{5}{2} - 6ln2 \)

Step 1: Breakdown the integrand

Start by dividing the numerator by the denominator to simplify the integrand. The goal is to rewrite the fraction as a sum of simpler fractions which can be integrated more easily. \[ \frac{x^{3}-4 x^{2}-3 x+3}{x^{2}-3 x} = x - 1 + \frac{6}{x - 3} \]

Step 2: Integrate each term individually

Next, integrate each term individually over the interval \[1, 2\]. This is possible due to the properties of integrals. The integral of a sum is equal to the sum of integrals. \[\int_{1}^{2} \left( x - 1 + \frac{6}{x - 3} \right) dx = \int_{1}^{2} xdx -\int_{1}^{2}dx + \int_{1}^{2}\frac{6}{x - 3}dx \]

Step 3: Calculate the antiderivatives and evaluate the definite integrals

Now it's time to find the antiderivatives for each individual integral and use the Fundamental Theorem of Calculus. This theorem states that to evaluate a definite integral, one should calculate the antiderivative at the top limit and subtract the antiderivative at the bottom limit. \[\int_{1}^{2}xdx = \frac{x^2}{2}|_{1}^{2} = \frac{4}{2} - \frac{1}{2}= \frac{3}{2} \] \[\int_{1}^{2}dx = (x)|_{1}^{2} = 2 - 1 = 1\] \[\int_{1}^{2}\frac{6}{x - 3}dx = 6ln|x-3||_{1}^{2} = 6ln|2-3| - 6ln|1-3| = 6ln|1| - 6ln|2| = 0 - 6ln2 = -6ln2\]

Step 4: Sum up all results

Finally, sum up the results of the three individual calculations to find the result of the original problem. \[ \frac{3}{2} + 1 - 6ln2 = \frac{5}{2} - 6ln2 \]

Key Concepts

Integration by Partial Fractions

Integration by partial fractions is a technique used to decompose a complex rational function into simpler fractions that can be individually integrated. In the given exercise, the integrand \(\frac{x^3 - 4x^2 - 3x + 3}{x^2 - 3x}\) is simplified by dividing the numerator by the denominator, leading to a sum of simpler terms \( x - 1 + \frac{6}{x - 3}\). Each term represents a partial fraction.

When faced with rational functions, partial fractions can significantly simplify the integration process by breaking down the integrand into elemental parts, each of which corresponds to a basic antiderivative. However, the method of partial fractions only applies to rational functions where the degree of the numerator is less than the degree of the denominator, or after polynomial long division has been used to ensure this condition is met, as demonstrated in the solution.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical principle that connects differentiation and integration, offering a way to evaluate definite integrals. The theorem posits that if \(F\) is an antiderivative of a function \(f\) on an interval \[a,b\], then the definite integral of \(f\) from \(a\) to \(b\) can be found by \(F(b) - F(a)\).

It serves as the backbone for calculating definite integrals, as seen in our exercise example. After the integration process is applied to each partial fraction, you apply the theorem to evaluate the integral at the upper limit, \(b = 2\), and subtract the value at the lower limit, \(a = 1\). This results in distinct numerical values for each term, which are then combined to give the final answer for the integral.

Antiderivative Calculation

An antiderivative of a function is another function whose derivative yields the original function. In the context of definite integrals, calculating antiderivatives is paramount to determine the integral's value. For the solution provided, antiderivatives for each term of the decomposed function were calculated. Specifically:

\( \int x dx = \frac{x^2}{2} \) and \( \int \frac{6}{x - 3} dx = 6ln|x - 3|\)

By evaluating these antiderivatives at the limits of integration, you can easily find the definite integral's value. This step is where the concrete calculations come into play, and having a firm grasp on finding antiderivatives is key to successfully solving integrals.

Integral Properties

Integral properties are rules that simplify computation and allow for the integration of complex functions. One of the most utilized properties is the linearity of integrals, enabling us to integrate terms independently and sum them up later. This property is evident in the exercise, where the integrand is separated into distinct terms:

\( \int (x - 1 + \frac{6}{x - 3}) dx = \int x dx - \int dx + \int \frac{6}{x - 3} dx \).

These properties streamline the evaluation process by permitting you to dissect a complex integral into manageable parts—treating each as a standalone integral—then integrate each part and sum the results. Recognizing and applying integral properties proficiently is fundamental in solving calculus problems.