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1-20 Find the most general antiderivative of the function. (Check your answer by differentiation.) \(g(x)=\frac{5-4 x^{3}+2 x^{6}}{x^{6}}\)

Short Answer

Expert verified
The most general antiderivative is \( G(x) = -x^{-5} + 2x^{-2} + 2x + C \).

Step by step solution

01

Simplify the Function

Start by simplifying the function: \( g(x) = \frac{5-4x^{3}+2x^{6}}{x^{6}} \).Divide each term by \( x^6 \):\[ g(x) = \frac{5}{x^6} - \frac{4x^3}{x^6} + \frac{2x^6}{x^6} = 5x^{-6} - 4x^{-3} + 2. \]
02

Integrate Each Term

Find the antiderivative of each term:- For \( 5x^{-6} \), use \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \): \[ \int 5x^{-6} \, dx = 5 \cdot \frac{x^{-5}}{-5} = -x^{-5}. \]- For \( -4x^{-3} \), apply the power rule: \[ \int -4x^{-3} \, dx = -4 \cdot \frac{x^{-2}}{-2} = 2x^{-2}. \]- For \( 2 \), integrate as: \[ \int 2 \, dx = 2x. \]
03

Combine Antiderivatives

Combine the antiderivatives obtained from each term, and don't forget the constant of integration, \( C \):\[ G(x) = -x^{-5} + 2x^{-2} + 2x + C. \]
04

Differentiate to Check

Differentiate \( G(x) = -x^{-5} + 2x^{-2} + 2x + C \) to verify:\[ G'(x) = 5x^{-6} - 4x^{-3} + 2, \] which matches the original function \( g(x) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus that involves finding the antiderivative of a function. An antiderivative is essentially the opposite of differentiation. It represents the reverse process, where we reconstruct a function from its derivative. Whenever you integrate a function, you aim to find a function (or family of functions) whose derivative is the original function.
When dealing with integration, it is important to remember a few key concepts:
  • The power rule is a primary technique for integrating polynomial terms. This states that for any term in the form of \( x^n \), its antiderivative is \( \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \). The \( + C \) is crucial because it represents the constant of integration, acknowledging that indefinite integrals represent a family of functions.
  • Integration of constant terms, such as with \( 2 \), is straightforward because the antiderivative of any constant \( a \) is \( ax + C \).
  • When presented with a composite function, such as a sum or difference of separate functions, you integrate each component individually and then combine the results.
By understanding and applying these rules, you can solve integration problems, just like in the given exercise.
Calculus
Calculus is an area of mathematics that focuses on change and motion. It is divided into two primary branches: differentiation and integration. These two branches are interconnected and often serve opposite functions.
  • Differentiation deals with rates of change and slopes of curves. It is used to find derivatives, which describe how a function changes at any given point.
  • Integration, on the other hand, is about accumulation and areas under curves. It involves finding the antiderivatives of functions.
When learning calculus, it's essential to understand how differentiation and integration complement each other. For example:
  • If differentiation is used to find the velocity of an object at a specific time, integration could be used to find the total distance traveled over a time interval.
  • In the context of the exercise, you found an antiderivative (an integrated function), then checked it by differentiating, which brings you full circle to the original function.
This interplay between the two provides a powerful mathematical framework for solving a wide variety of problems in physics, engineering, and beyond.
Differentiation
Differentiation is another cornerstone of calculus, concentrating on how a function changes at a specific point. Essentially, it helps you calculate the derivative of a function, which is the rate at which the function’s value is changing at any given point. In mathematical terms, the derivative \( f'(x) \) of a function \( f(x) \) provides the slope of the tangent line at any point \( x \).
In the context of solving the problem, differentiation serves another important purpose:
  • Once an antiderivative is calculated through integration, differentiation allows you to check your work. By differentiating the antiderivative, you should obtain the original function if all calculations are correct.
  • This step is imperative in ensuring that the integration process did not involve any errors, and it affirms that the constant of integration \( C \) doesn't affect the derivative since the derivative of a constant is zero.
Grasping differentiation deeply can open doors to understanding other advanced concepts, like optimization and curve sketching. The process follows certain rules, like the power rule, which conveniently complements the power rule in integration, making the learning curve for both more intuitive.

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Most popular questions from this chapter

If an electrostatic field \(E\) acts on a liquid or a gaseous polar dielectric, the net dipole moment P per unit volume is $$\mathrm{P}(\mathrm{E})=\frac{\mathrm{e}^{\mathrm{E}}+\mathrm{e}^{-\mathrm{E}}}{\mathrm{e}^{\mathrm{E}}-\mathrm{e}^{-\mathrm{E}}}-\frac{1}{\mathrm{E}}$$ Show that\(\lim _{\mathrm{E} \rightarrow 0^{+}} \mathrm{P}(\mathrm{E})=0.\)

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