Problem 22 - Thomas Calculus

Question

In Exercises \(21-28,\) express the integrands as a sum of partial fractions and evaluate the integrals. $$ \int_{1}^{\sqrt{3}} \frac{3 t^{2}+t+4}{t^{3}+t} d t $$

Step-by-Step Solution

Answer

The evaluated integral is \(2 \ln(3) - \frac{1}{2}\ln(2) + \frac{\pi}{12}\).

Step 1: Factor the Denominator

First, factor the denominator \(t^3 + t\). Notice that we can factor \(t\) out of the expression, which gives us \(t(t^2 + 1)\). This means the integrand is \(\frac{3t^2 + t + 4}{t(t^2 + 1)}\).

Step 2: Set Up the Partial Fraction Decomposition

For partial fractions, express the integrand as \(\frac{A}{t} + \frac{Bt + C}{t^2 + 1}\). This assumes the form for a linear factor \(\frac{A}{t}\) and an irreducible quadratic factor \(\frac{Bt + C}{t^2 + 1}\).

Step 3: Solve for Coefficients

Writing \(\frac{3t^2+t+4}{t(t^2+1)} = \frac{A}{t} + \frac{Bt+C}{t^2+1}\), multiply through by the common denominator \(t(t^2 + 1)\) to get \(3t^2 + t + 4 = A(t^2 + 1) + (Bt + C)t\). Expanding gives \(3t^2+t+4 = At^2 + A + Bt^2 + Ct\). Collect like terms: \((A+B)t^2 + Ct + A = 3t^2 + t + 4\). Matching coefficients, solve: \(A + B = 3\), \(C = 1\), and \(A = 4\).

Step 4: Calculate Coefficients

From the equations in the previous step, solve \(A = 4\) and substitute into \(A + B = 3\) to get \(4 + B = 3\), so \(B = -1\). Thus, our coefficients are \(A = 4\), \(B = -1\), and \(C = 1\).

Step 5: Substitute and Integrate

Substitute back to get \(\frac{4}{t} + \frac{-t + 1}{t^2 + 1}\). Split the integral: \(\int_1^{\sqrt{3}} \left( \frac{4}{t} \right) dt + \int_1^{\sqrt{3}} \left( \frac{-t}{t^2+1} \right) dt + \int_1^{\sqrt{3}} \left( \frac{1}{t^2+1} \right) dt\).

Step 6: Evaluate Each Integral

The integral \(\int \frac{4}{t} dt\) evaluates to \(4 \ln |t|\). The integral \(\int \frac{-t}{t^2 + 1} dt\) evaluates to \(-\frac{1}{2} \ln |t^2 + 1|\). The integral \(\int \frac{1}{t^2 + 1} dt\) evaluates to \(\tan^{-1}(t)\).

Step 7: Evaluate Definite Integrals

Using the evaluations, solve: - \(4 \ln |t|\) from \(1\) to \(\sqrt{3}\): \(4 \ln(\sqrt{3}) - 4 \ln(1) = 2 \ln(3)\).- \(-\frac{1}{2} \ln |t^2 + 1|\) from \(1\) to \(\sqrt{3}\): \(-\frac{1}{2} [\ln(4) - \ln(2)] = -\frac{1}{2}(\ln 2)\).- \(\tan^{-1}(t)\) from \(1\) to \(\sqrt{3}\): \(\frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}\).

Step 8: Sum Up the Results

Add the results of each integral evaluation:2 ln(3) - 1/2 ln(2) + \(\pi/12\). Simplifying: \(2 \ln(3) - \frac{1}{2}\ln(2) + \frac{\pi}{12}\).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focusing on finding the integral of functions. Integrals allow us to determine the area under a curve, compute volumes, and solve numerous real-world problems. The integral of a function can be thought of as the inverse process of differentiation. It involves two main types: indefinite integrals, which provide a function that represents the accumulation of quantities, and definite integrals, which calculate the accumulation over a specific interval.

When discussing integrals, it's essential to understand some foundational components:

• The integrand is the function you are integrating.
• The limits of integration define the interval over which you'll find the area or accumulated quantity.
• The integration process often involves algebraic manipulation, such as partial fraction decomposition, to simplify complex integrands.

Integral calculus plays a crucial role in mathematical problem solving, especially when combined with techniques like substitution and integration by parts.

Definite Integrals

Definite integrals calculate the exact accumulation of a quantity over a specified interval. Unlike indefinite integrals, which include a constant of integration, definite integrals result in a numeric value representing the total accumulation. This is often visualized as the area under the curve of the function within the bounds of integration.

In practical terms, here’s how definite integrals work:

• The notation \( \int_{a}^{b} f(x) \, dx \) tells us that we're finding the accumulation of \( f(x) \) from \( x = a \) to \( x = b \).
• The Fundamental Theorem of Calculus connects differentiation and integration. Specifically, it allows us to compute a definite integral using the antiderivative of the function.
• After finding the antiderivative, you substitute the upper and lower limits of integration to calculate the definite integral. This involves plugging in the upper limit, evaluating, then subtracting the result of plugging in the lower limit.

Understanding definite integrals involves a grasp of integration techniques and application of the Fundamental Theorem, making it a cornerstone in mathematical analysis and solving concrete problems.

Mathematical Problem Solving

Mathematical problem solving often demands a multi-step approach, particularly when dealing with integrals that require simplification before integrating. In the provided exercise, partial fraction decomposition is the key technique used.

Here's a breakdown of this problem-solving method:

• **Factor the Denominator:** Identify any factors within the denominator of the integrand function to simplify it.
• **Set Up Partial Fractions:** Express the simplified integrand as a sum of simpler fractions, called partial fractions. Each fraction corresponds to a factor of the original denominator.
• **Solve for Coefficients:** Determine the coefficients of each partial fraction by equating and solving the resulting polynomial equation through comparing coefficients.
• **Integrate Each Fraction:** Once decomposed, integrate each of the partial fractions separately. The simplicity gained through this process makes complex integrals more manageable.
• **Evaluate the Definite Integral:** Substitute the bounds into the evaluated antiderivatives to find the definite integral's value.

This comprehensive strategy involves both algebraic and calculus skills, offering clear pathways to solving otherwise daunting integrals. Mastering these steps showcases the powerful intersection of algebra and calculus in mathematical problem solving.