91Ó°ÊÓ

Integration is a fundamental concept in calculus. It allows us to find the antiderivative of a function, essentially working as the reverse process of differentiation. In our exercise, the function \( f(x) = 2x^2 + x - 1 \) is subjected to integration to find its most general antiderivative. When integrating, each term of the function is integrated separately. This means calculating indefinite integrals for each part independently:

• \( \int 2x^2 \, dx = \frac{2}{3}x^3 \)
• \( \int x \, dx = \frac{1}{2}x^2 \)
• \( \int (-1) \, dx = -x \)

Integrating constant terms, like \(-1\) in this function, involves simply multiplying by \(x\). Each integral gives an antiderivative for its respective term. The general antiderivative is a combination of these individual components including a constant of integration, with a general expression given by:\[ F(x) = \frac{2}{3}x^3 + \frac{1}{2}x^2 - x + C \]

Polynomial expansion simplifies complex expressions through multiplication. It helps us rewrite products of polynomials as a sum or difference of terms. Initially, the function \((x + 1)(2x - 1)\) appears as a product of two binomials. To make integration straightforward, we expand them into a single polynomial expression.

• First, distribute each term from the first binomial \((x + 1)\) to each term in the second binomial \((2x - 1)\).
• Multiply the inner and outer terms: \(x(2x - 1) + 1(2x - 1)\).
• Add similar terms to simplify: \(2x^2 - x + 2x - 1 = 2x^2 + x - 1\).

After expanding, we find a simplified polynomial \(2x^2 + x - 1\), making it easier to apply integration techniques. Polynomial expansion reduces computational complexity and reveals the individual parts of the polynomial, making calculus operations clearer.

Differentiation is used to verify our calculated antiderivative. It involves finding the derivative of a function, undoing integration, to check if the original function is obtained. After finding the antiderivative \(F(x) = \frac{2}{3}x^3 + \frac{1}{2}x^2 - x + C\), we differentiate \(F(x)\) to ensure accuracy:

• Differentiate each term separately.
• \(\frac{d}{dx}\left(\frac{2}{3}x^3\right) = 2x^2\)
• \(\frac{d}{dx}\left(\frac{1}{2}x^2\right) = x\)
• \(\frac{d}{dx}(-x) = -1\)
• \(\frac{d}{dx}(C) = 0\)

Adding these derivatives confirms that \( F'(x) = 2x^2 + x - 1 \), successfully reproducing the original function \( f(x) \). Differentiation remains a crucial step to validate the accuracy of integration.

The constant of integration, denoted as \(C\), arises in the process of finding antiderivatives. It acknowledges that there are infinite functions differing by a constant which share the same derivative. During the integration of our example function, each term yields its own constant:

• \( \int 2x^2 \, dx = \frac{2}{3}x^3 + C_1 \)
• \( \int x \, dx = \frac{1}{2}x^2 + C_2 \)
• \( \int (-1) \, dx = -x + C_3 \)

Each constant \( C_i \) combines into a single constant \( C = C_1 + C_2 + C_3 \). This collective constant represents the family of curves parallel to the antiderivative function's graph. Since the derivative of a constant is zero, the actual value of \(C\) does not affect the outcome when differentiating back. The constant of integration exemplifies the indefinite nature of antiderivatives, capturing all possible vertical shifts in the function.