91Ó°ÊÓ

Polynomial long division is a method used to divide polynomials, simplifying complex expressions into simpler components. It's similar to the long division process used with numbers, but instead, you work with variable expressions. In the exercise, we perform polynomial long division on the expression \(\frac{2x^5 + x^4 - 3x^3 + 2x^2 + 8x + 1}{x^2 + 1}\).Here are the steps followed:

• Write both the dividend (numerator) and divisor (denominator) in expanded form, considering all power terms, including those with zero coefficients.
• Begin dividing the leading term of the dividend by the leading term of the divisor. This gives the first term of the quotient.
• Multiply the entire divisor by this term and subtract this from the original dividend.
• Repeat the process with the new dividend obtained after each subtraction, continuing until the degree of the remaining dividend is less than that of the divisor.

After performing these steps, you end up with the simplified expression \(2x^3 + x^2 - 3x + 8 + \frac{1}{x^2+1}\), which can now be integrated more easily.

The antiderivative, or indefinite integral, is the reverse process of differentiation, which helps us find a function whose derivative equals the given function. Determining the antiderivative is crucial for solving definite integrals, like in our problem, where the integrand is simplified after polynomial long division.For the simplified terms resulting from our division, we find the antiderivatives separately:

• The antiderivative of \(2x^3\) is \(\frac{1}{2}x^4\).
• The antiderivative of \(x^2\) is \(\frac{1}{3}x^3\).
• The antiderivative of \(-3x\) is \(-\frac{3}{2}x^2\).
• The antiderivative of the constant \(8\) is \(8x\).
• The antiderivative of \(\frac{1}{x^2+1}\) is \(\arctan(x)\).

These computations allow us to proceed to integrate over the specified limits of \(-1\) to \(1\). Understanding antiderivatives is key because it translates our activity from operations on terms to finding actual areas under curves.

Once we have the antiderivatives, the next step is to evaluate the definite integral. A definite integral \(\int_{a}^{b} f(x) \, dx\) represents the signed area under the curve of \(f(x)\), from \(x = a\) to \(x = b\).For our expression, \(\int_{-1}^{1} (2x^3 + x^2 - 3x + 8 + \frac{1}{x^2+1}) dx\), we follow these steps:

• Split the integral into the sum of simpler integrals: \(\int_{-1}^{1} 2x^3 \, dx + \int_{-1}^{1} x^2 \, dx - \int_{-1}^{1} 3x \, dx + \int_{-1}^{1} 8 \, dx + \int_{-1}^{1} \frac{1}{x^2+1} \, dx\).
• Compute each antiderivative over the limits \(-1\) to \(1\).
• Add the results of these individual integrals to obtain the total evaluated integral.

This systematic approach helps in accurately determining the values, factoring in both the algebraic expressions and trigonometric terms, leading to a precise evaluation of the given integral problem.

The Fundamental Theorem of Calculus is pivotal as it bridges the worlds of differentiation and integration, showing that they are inverse processes. It has two main parts:1. **First Part**: If a function \(f\) is continuous on \([a, b]\) and \(F\) is its antiderivative, then the integral of \(f\) from \(a\) to \(b\) is \(F(b) - F(a)\).2. **Second Part**: If \(F\) is an antiderivative of \(f\) on \([a, b]\), then \( \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x)\).In our exercise, we apply the first part by using the antiderivative found for each term to evaluate their values at \(b = 1\) and \(a = -1\), then subtract:

• Apply \(F(b) - F(a)\) to each antiderivative calculated.
• Account for terms that might cancel each other due to symmetrical properties over the interval.

This theorem confirms that the sum of these operations equals the definite integral of the initial function. It simplifies complex problems by providing a method to find exact values, efficiently linking different calculus operations.