Problem 8 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. $$\int_{0}^{1} \frac{x-1}{x^{2}+3 x+2} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(\ln\left(\frac{27}{4}\right)\).

Step 1: Factor the Denominator

The denominator of the integrand is quadratic and can be factored. The expression \(x^2 + 3x + 2\) factors as \((x+1)(x+2)\). This gives us a starting point for decomposing the fraction.

Step 2: Setup Partial Fraction Decomposition

Express \(\frac{x-1}{(x+1)(x+2)}\) as partial fractions: \(\frac{A}{x+1} + \frac{B}{x+2}\). Clear the denominators: \(x-1 = A(x+2) + B(x+1)\).

Step 3: Solve for Constants A and B

Expand and equate coefficients: \(x-1 = (A+B)x + (2A+B)\). Setting coefficients equal, we get two equations: \(A + B = 1\) and \(2A + B = -1\). Solve these to find \(A = -2\) and \(B = 3\).

Step 4: Rewrite the Integral

Substitute back into the integral using the values of \(A\) and \(B\). The integral becomes \(\int_{0}^{1} \left(\frac{-2}{x+1} + \frac{3}{x+2}\right) dx\).

Step 5: Integrate Each Term Separately

Now, separate the integrals: \(-2\int_{0}^{1} \frac{1}{x+1} dx + 3\int_{0}^{1} \frac{1}{x+2} dx\). These integrals are standard and their results involve the natural logarithm.

Step 6: Evaluate Each Integral

Calculate \(-2\left[\ln|x+1|\right]_{0}^{1}\) and \(3\left[\ln|x+2|\right]_{0}^{1}\). This results in: \(-2[\ln 2 - \ln 1] + 3[\ln 3 - \ln 2]\).

Step 7: Simplify the Expression

Simplify using the property of logarithms \(\ln a - \ln b = \ln\frac{a}{b}\). Thus, the expression simplifies to \(-2\ln 2 + 3\ln 3 - 3\ln 2\). Combine terms: \(\ln\left(\frac{27}{4}\right)\).

Step 8: Final Answer

Putting everything together, the value of the integral is \(\ln\left(\frac{27}{4}\right)\).

Key Concepts

Partial Fractions

Partial fraction decomposition is a key technique in integral calculus when dealing with rational functions. A rational function is the ratio of two polynomials. For complicated functions, especially those with polynomials in the denominator, it can make integration easier by breaking the fraction into simpler parts.

Consider a rational function, like the one we had in the exercise: \[\frac{x-1}{(x+1)(x+2)}\]To decompose it, we first factor the denominator, as was done in "Step 1" of the original solution. Factoring helps us recognize the simpler fractions we can express the original fraction into.

• Factor the denominator if possible.
• Decompose the fraction into a sum of simpler fractions. Example: \[ \frac{x-1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \]
• Find the constants (A and B) by clearing the denominators and equating coefficients.

These steps simplify the function, and with partial fractions, the original problem can be approached in a way that individual pieces become easier to integrate separately, as seen in the exercise steps.

Definite Integrals

Definite integrals are a core component of calculus, specifically in measuring the accumulated quantity and understanding changes across an interval. Simply put, a definite integral calculates the area under a curve between two bounds. For our function, that meant evaluating:

\[\int_{0}^{1} \left(\frac{-2}{x+1} + \frac{3}{x+2}\right) dx\]To do so:

• Identify the bounds which, in this case, were from 0 to 1.
• Perform the integration of simpler fractions individually over the given limits.
• Calculations yield a specific value representing the area under the curve within these bounds.

The arithmetic involved definitely simplifies each integration process, particularly since the partial fraction decomposition results in manageable terms. As shown, the integral evaluates over specified limits instead of focusing on an indefinite scope. This effectively 'bounds' the area calculation.

Logarithmic Functions

Logarithmic functions often appear when integrating expressions involving fractional terms as seen in the simple parts of partial fraction decomposition.

Integrating terms like:\[\int \frac{1}{x+c} dx\]Commonly results in natural logarithms. This is because the integral of \(\frac{1}{x}\) is \(\ln |x| + C\), where \(C\) is the integration constant. For definite integrals:

• Evaluate the logarithm over the specified bounds. For our problem, this meant computing \([-2\ln(x+1) + 3\ln(x+2)]\) from 0 to 1.
• Apply logarithmic properties like \(\ln a - \ln b = \ln\frac{a}{b}\) to simplify the result further.
• Combine terms efficiently to achieve a more compact representation, which in the solution was \(\ln\frac{27}{4}\).

These functions play a vital role in establishing the relationship between growth patterns, decay, and spread across different scales in integral calculus. Understanding logarithmic integration helps make the complex simple with powerful results, as demonstrated in the exercise solution.