Question

Find the general antiderivative \(F(x)+C\) for each of the following. $$ f(x)=27 x^{7}+3 x^{5}-45 x^{3}+\sqrt{2} x $$

Short answer

The general antiderivative is \( F(x) = \frac{27}{8}x^8 + \frac{1}{2}x^6 - \frac{45}{4}x^4 + \frac{\sqrt{2}}{2}x^2 + C \).

Step-by-step solution

Identify the Function Terms

We are given the function \( f(x) = 27x^7 + 3x^5 - 45x^3 + \sqrt{2}x \). Identify all the terms of the function that need to be integrated.

Apply the Power Rule of Integration

For each term \( ax^n \), use the power rule of integration which states: \[ \int ax^n \, dx = \frac{a}{n+1}x^{n+1} + C \].

Integrate the First Term

Integrate the first term \( 27x^7 \). Applying the power rule gives: \( \int 27x^7 \, dx = \frac{27}{7+1}x^{7+1} = \frac{27}{8}x^8 \).

Integrate the Second Term

Integrate the second term \( 3x^5 \). Applying the power rule gives: \( \int 3x^5 \, dx = \frac{3}{5+1}x^{5+1} = \frac{3}{6}x^6 = \frac{1}{2}x^6 \).

Integrate the Third Term

Integrate the third term \( -45x^3 \). Applying the power rule gives: \( \int -45x^3 \, dx = \frac{-45}{3+1}x^{3+1} = \frac{-45}{4}x^4 = -\frac{45}{4}x^4 \).

Integrate the Fourth Term

Integrate the fourth term \( \sqrt{2}x \). Using the power rule, \( \int \sqrt{2}x \, dx = \frac{\sqrt{2}}{1+1}x^{1+1} = \frac{\sqrt{2}}{2}x^2 \).

Combine All Integrated Terms and Add Constant

Combine all the results from the integration: \( F(x) = \frac{27}{8}x^8 + \frac{1}{2}x^6 - \frac{45}{4}x^4 + \frac{\sqrt{2}}{2}x^2 + C \).

Key concepts

Power Rule of Integration

The Power Rule of Integration is a fundamental concept in calculus that helps us find the antiderivative of functions with terms of the form \( ax^n \). The rule states that: \[ \int ax^n \, dx = \frac{a}{n+1}x^{n+1} + C \] where \( a \) is a constant, \( n \) is the exponent, and \( C \) is the constant of integration. Here's how you apply it: - Increase the exponent \( n \) by 1 to get \( n+1 \). - Divide the coefficient \( a \) by this new exponent. - Attach the new term \( x^{n+1} \) and don't forget to add the constant \( C \). This method is straightforward, providing a powerful tool to reverse differentiation, allowing us to find antiderivatives quickly.

Antiderivative

The concept of an Antiderivative is essential in calculus. To find an antiderivative of a function means to revert differentiation, identifying a function whose derivative is the original function. In other words, if \( F(x) \) is an antiderivative of \( f(x) \), then: \[ F'(x) = f(x) \] Antiderivatives are not unique because the derivative of a constant is zero. Hence, when finding the indefinite integral, we include \( C \), known as the constant of integration, to represent all possible antiderivatives. For example, considering \( f(x) = 3x^2 \), an antiderivative would be \( F(x) = x^3 + C \) because: - The derivative \( F'(x) = 3x^2 \) matches \( f(x) \). Antiderivatives allow us to determine the function that describes a rate of change, offering insight into the accumulated quantity over an interval.

Integral Calculus

Integral Calculus focuses on the process of integration, aiming to calculate areas under curves, among other applications. This branch of calculus is often paired with differential calculus but serves the opposite purpose: while differential calculus looks at rates of change, integral calculus seeks to calculate total quantities. Two primary forms of integrals are: - **Definite Integrals:** These compute the area under a curve from one point to another, providing a numerical result. - **Indefinite Integrals:** Emphasized in exercises like finding antiderivatives, they involve an unknown constant \( C \) due to the possibility of multiple antiderivatives. Integral calculus is essential in various real-world applications, like calculating areas, volumes, and even probabilities. By mastering integration techniques, students can unveil a deeper understanding of mathematical concepts and their numerous applications in science and engineering.