Question

\(\quad(\mathrm{dy} / \mathrm{dx})=\left(2 \mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}-3 \mathrm{x}+2\right) .\) Find \(\mathrm{y}=\mathrm{F}(\mathrm{x})\).

Short answer

The function y = F(x) is given by \( y = \frac{2}{5}x^5 - \frac{3}{2}x^4 + \frac{5}{3}x^3 - \frac{3}{2}x^2 + 2x + C \), where C is the integration constant.

Step-by-step solution

Multiply the two expressions

Given that (dy/dx) = (2x^2 + 1)(x^2 - 3x + 2), we will first multiply the expressions: (2x^2 + 1)(x^2 - 3x + 2) = 2x^4 - 6x^3 + 4x^2 + x^2 - 3x + 2 = 2x^4 - 6x^3 + 5x^2 - 3x + 2 Now the expression (dy/dx) becomes: dy/dx = 2x^4 - 6x^3 + 5x^2 - 3x + 2

Integrate the expression with respect to x

Now we need to integrate the expression: \( ∫(2x^4 - 6x^3 + 5x^2 - 3x + 2)dx = ∫(2x^4)dx - ∫(6x^3)dx + ∫(5x^2)dx - ∫(3x)dx + ∫(2)dx \)

Integrate each term individually

Integrate each term: \( ∫(2x^4)dx = \frac{1}{5}(2x^5) \\ ∫(6x^3)dx = \frac{1}{4}(6x^4) \\ ∫(5x^2)dx = \frac{1}{3}(5x^3) \\ ∫(3x)dx = \frac{1}{2}(3x^2) \\ ∫(2)dx = 2x \)

Combine the integrals and add integration constant

Now that we have integrated each term individually, it's time to combine them: \( y = \frac{1}{5}(2x^5) - \frac{1}{4}(6x^4) + \frac{1}{3}(5x^3) - \frac{1}{2}(3x^2) + 2x + C \) Where C is the integration constant.

Simplification

Let's simplify the expression: \( y = \frac{2}{5}x^5 - \frac{3}{2}x^4 + \frac{5}{3}x^3 - \frac{3}{2}x^2 + 2x + C \) Now we have found the function y = F(x), with y = \( \frac{2}{5}x^5 - \frac{3}{2}x^4 + \frac{5}{3}x^3 - \frac{3}{2}x^2 + 2x + C \).

Key concepts

Antiderivative

The concept of an antiderivative is a fundamental building block in calculus and is synonymous with the process of finding the original function when given its derivative. An antiderivative of a function, say, f(x), is another function, F(x), such that the derivative of F with respect to x gives us f(x), or formally, if \( \frac{dF}{dx} = f(x) \). In other words, a function F is an antiderivative of f if the differential of F equals f. The process of finding an antiderivative is known as antidifferentiation or integration.

In the textbook exercise provided, the antiderivative of the polynomial \( \frac{dy}{dx} = (2x^2 + 1)(x^2 - 3x + 2) \) is sought. The calculation of the antiderivative involves reversing the differentiation rules and is a crucial step in determining the original function y = F(x) from its derivative function.

Indefinite Integral

An indefinite integral, often referred to just as an integral, is the antiderivative of a function with respect to a given variable, usually indicated without limits of integration. It represents a family of functions rather than a specific value. That's why the result of an indefinite integral includes a constant of integration, commonly denoted as C. This constant exists because when we take the derivative of a constant, the result is zero, so when we are performing the reverse process of differentiation - integrating - we must acknowledge that there might have been a constant that 'disappeared' during differentiation.

When we tackle the integrals of polynomial functions, like the one in our exercise, we employ the power rule in reverse. This rule gives us a straightforward method to perform the integration of power functions of x by incrementing the exponent by one and dividing the term by the new exponent. Don't forget to add the constant of integration, C, at the end, as seen in the step-by-step solution.

Integration by Term

Integration by term is a technique used when integrating a polynomial or any function that can be broken down into simpler parts. Doing this allows us to apply integration rules to each part separately, simplifying the overall process considerably. The process of integrating by term is based on the linearity of integration, which allows us to integrate each term of a polynomial separately and then combine the results.

As illustrated in the solution steps, each monomial in the polynomial \(2x^4 - 6x^3 + 5x^2 - 3x + 2 \) is integrated separately, and the results are added together to create the integrated function plus the constant of integration C. This task would be quite complex without breaking it down term-wise; thus, understanding how to integrate by term is invaluable in handling polynomial integrals.

Polynomial Functions

Polynomial functions are mathematical expressions that consist of variables and constants combined using only addition, subtraction, multiplication, and non-negative integer exponentiation. The general form of a polynomial in a single variable x is \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \), where \( a_n, a_{n-1}, ..., a_1, a_0 \) are constants and \( n \) is a non-negative integer termed the degree of the polynomial.

The significance of polynomial functions in calculus is immense as they have clear rules for differentiation and integration, making them a great starting point for students learning the concepts. Moreover, polynomials are smooth, continuous functions, and their simplicity allows for many analytical computations—an excellent example being the polynomial provided in the exercise, which we are integrating to find F(x).