91Ó°ÊÓ

The concept of a definite integral is crucial in calculus. It represents the accumulation of quantities, like areas under curves, computed over an interval. When we talk about the definite integral of a function, we are looking for the net area between the curve of the function and the x-axis from a starting point (lower limit) to an endpoint (upper limit). This area can be positive or negative, depending on whether the function is above or below the x-axis.

The objective in this exercise involves evaluating the definite integral of a function from 1 to \( \sqrt{3} \). This involves calculating the integral of our decomposed partial fractions over that interval. By using the result of the integration, we achieve a solution that provides the exact value of the area—or accumulated quantity—given by the function over the specified range.

An integrand is the function that you want to integrate. It is located inside the integral sign, and it tells you what function to accumulate over the given limits. In this particular problem, the integrand is \( \frac{3t^2 + t + 4}{t^3 + t} \).

Notice that the task is to express this integrand as a sum of partial fractions. By doing this, we simplify the integration process. Instead of dealing with a complex rational expression, we decompose it into simpler fractions. These are easier to integrate, especially when each type can be handled with known integration techniques.

• The numerator, \( 3t^2 + t + 4 \), is a polynomial that we aim to express in terms of simpler fractions.
• The denominator, \( t^3 + t \), gets factored into \( t(t^2 + 1) \) as part of the decomposition process.

This results in a more manageable expression for integration.

Coefficient comparison is a technique used to determine the unknown constants in an equation in algebra and calculus. During partial fraction decomposition, it becomes necessary to find specific values for coefficients to ensure equality of both sides of the equation.

In the exercise, once we assume a partial fraction decomposition setup, we equate it with the original integrand. We multiply out both sides to remove fractions and compare the coefficients of like terms:

• Compare powers of \( t \) to create equations from coefficients: \( A + B = 3 \), \( C = 1 \), \( A = 4 \).
• These simple equations are then solved to find \( A = 4 \), \( B = -1 \), and \( C = 1 \).

This technique ultimately helps rewrite our integrand as a sum of simple fractions, ready for integration.

Trigonometric integration involves integrating functions that incorporate trigonometric identities or expressions. These types of integrals often appear in calculus and are solved using specific strategies and formulas.

In the exercise, part of the integration involves the function \( \frac{1}{t^2+1} \). The integral of this function over any interval leads to an arctangent expression due to its relationship to the derivative of \( \tan^{-1}(t) \).

• Recognize the integral: \( \int \frac{1}{t^2+1} \, dt = \tan^{-1}(t) + C \).
• The interval \( [1, \sqrt{3}] \) results in an exact expression \( \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \), yielding values based on known angle measures: \( \frac{\pi}{3} - \frac{\pi}{4} \).

Here, the use of trigonometric integration aids in converting a seemingly complex integral into a recognizable trigonometric identity that is easier to evaluate.