91Ó°ÊÓ

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler parts. This makes it easier to integrate or differentiate the expression.
To apply this, first factor the denominator of the given integral. For example, in the exercise, we noticed that the denominator \( x^4 - 16 \) can be factored as \( (x^2 - 4)(x^2 + 4) \).
Next, express the integrand as a sum of fractions, each with simpler denominators. In our case: \ \ \ \ \ \ \ \ \[ \frac{5x^3 - 4x}{(x^2 - 4)(x^2 + 4)} = \frac{Ax + B}{x^2 - 4} + \frac{Cx + D}{x^2 + 4} \]
Here, 'A', 'B', 'C', and 'D' are constants that we need to determine.
This method is useful because each fraction is easier to integrate independently.

Integration techniques are methods used to compute integrals. In our exercise, after applying partial fraction decomposition, we have two simpler integrals: \ \[ \frac{5x}{x^2 + 4} \text{ and } \frac{4}{x^2 + 4} \]
For the first integral, use substitution. Let \( u = x^2 + 4 \), then \( du = 2x dx \). This transforms the integral into a simpler form that's easier to evaluate.
For the second integral, use the standard formula: \ \[ \frac{1}{a} \tan^{-1} \frac{x}{a} \text{ where } a = 2 \].
These techniques make solving the integral straightforward and manageable. By splitting complex expressions into simpler parts, the overall computation is simplified.

Calculus is the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series.
In this exercise, we evaluated the definite integral: \ \[ \begin{align*} \frac{5}{2} \times \bigg[\frac{1}{2}(4^2 + 4 - (3^2 + 4))\bigg] - 2\bigg(\tan^{-1}(2) - \tan^{-1}(\frac{3}{2})\bigg) \. \ \frac{5}{2} \times \bigg[\frac{1}{2}(16 - 9)\bigg] - 2\bigg(\tan^{-1}(2) - \tan^{-1}(1.5)\bigg) \. \ \frac{5}{2} \times \bigg[\frac{7}{2}\bigg] - 2\bigg(\tan^{-1}(2) - \tan^{-1}(1.5)\bigg) \. \ \frac{35}{4} - 2\bigg(\tan^{-1}(2) - \tan^{-1}(1.5)\bigg) \. \ \frac{35}{4} - 2\bigg(1.107 - 0.9828\bigg) \. \ \frac{35}{4} - 2(0.1242) \. \ \frac{35}{4} - 0.2484 \. \ 8.75 - 0.2484 = 8.5016 \] \text{(Approximation)} \ Calculus allows us to work with the underlying principles of change and motion, providing tools necessary for a wide range of real-world applications. Understanding the definite integral helps us find the area under curves, along with its connections to physics, economics, and beyond.}