Problem 36 - College Algebra and Calculus: An Applied Approach

Question

$$ \text { Evaluate the definite integral. } $$ $$ \int_{0}^{1} \frac{x^{2}-x}{x^{2}+x+1} d x $$

Step-by-Step Solution

Answer

The value of the given integral is 0.

Step 1: Apply Partial Fraction Decomposition

We can write the fraction as two separate fractions using partial fraction decomposition. To do this, we need to equate \(\frac{x^{2}-x}{x^{2}+x+1}\) to the format \(\frac{A}{x-a} + \frac{B}{x-b}\) where A, a, B, and b are constants that we need to find. Comparing coefficients and solving the corresponding system of equations, we eventually get A = 1/2, a = -1/2, B = -1/2, b = -1/2.

Step 2: Evaluate the integrals separately

Now we split the original integral into two definite integrals: \(\int_{0}^{1} \frac{1/2}{x+1/2} dx - \int_{0}^{1} \frac{1/2}{x+1/2} dx\). These integrals can be easily evaluated, as they've been simplified to basic log forms.

Step 3: Apply Fundamental Theorem of Calculus

The second fundamental theorem of calculus states that if f is a continuous real-valued function defined on a closed interval [a, b] and F is an antiderivative of f on [a, b], then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). So, for the integrals we get, \(\frac{1}{2}\ln|x+1/2|\Big|_0^1 - \frac{1}{2}\ln|x+1/2|\Big|_0^1\). Remember, the symbol '|' enclosed by limits is used to denote that difference is to be taken between the value at upper limit and the value at lower limit.

Step 4: Calculate definite integral values

Substitute the upper and lower limits in the expressions. We have \(\frac{1}{2}\ln|1+1/2| - \frac{1}{2}\ln|0+1/2| - \frac{1}{2}\ln|1+1/2| + \frac{1}{2}\ln|0+1/2|\). When we simplify this, we get 0.

Step 5: Conclusion

The value of the definite integral \(\int_{0}^{1} \frac{x^{2}-x}{x^{2}+x+1} dx\) is 0.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a powerful technique used to simplify complex rational expressions. This method involves breaking down a fraction into simpler components, making it easier to integrate or differentiate.
The process generally involves:

• Expressing the rational function as a sum of simpler fractions.
• For our example, we started with \( \frac{x^{2}-x}{x^{2}+x+1} \) and wanted to express it in a form like \( \frac{A}{x-a} + \frac{B}{x-b} \).
• After equating to this format, we solve for constants \( A \), \( a \), \( B \), and \( b \) by comparing coefficients.

In our exercise, we solved the system to find \( A = \frac{1}{2} \), \( a = -\frac{1}{2} \), and \( B = -\frac{1}{2} \), \( b = -\frac{1}{2} \). This decomposition helps us separate the integrals and solve them more easily.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration. It tells us that if a function is continuous over an interval, the integral over that interval can be evaluated using its antiderivative.
Here's how it works:

• If \( F \) is an antiderivative of function \( f \) on an interval \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
• This approach allows us to evaluate definite integrals by calculating the difference between antiderivative values at the upper and lower bounds.

In practice, once we transformed the integral using partial fractions, we applied this theorem to compute: \( \frac{1}{2}\ln|x+1/2|\Big|_0^1 - \frac{1}{2}\ln|x+1/2|\Big|_0^1 \). This use of the theorem helps simplify and calculate the integral accurately.

Logarithmic Integration

Logarithmic integration involves integrating expressions that result in logarithmic functions. It's particularly useful when dealing with functions in the form \( \frac{1}{x+c} \).

• When you integrate \( \frac{1}{x+c} \), the result is \( \ln|x+c| + C \), where \( C \) is the integration constant.
• This method simplifies the evaluation by transforming complex expressions into logarithmic forms.

In the original problem, once the expression was decomposed, the integration of each part led to terms like \( \frac{1}{2}\ln|x+1/2| \). This made the evaluation straightforward. By evaluating these at the specified bounds, we determined the integral's value. Logarithmic integration thus acts as a key tool in tackling integrals involving rational functions.