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Chapter 4: Problem 13

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

Short Answer

Expert verified
The most general antiderivative is \(F(u) = \frac{1}{3}u^3 - 6u^{-1/2} + C\).

Step by step solution

01

Simplify the Function

First, we can simplify the function by dividing each term in the numerator by the denominator. We rewrite the given function: \[ f(u) = \frac{u^4}{u^2} + \frac{3 \sqrt{u}}{u^2} \]Simplifying, we get: \[ f(u) = u^2 + \frac{3}{u^{3/2}} \]
02

Write in Terms of Powers

Rewrite the function in terms of power functions to prepare it for integration. We have:\[ f(u) = u^2 + 3u^{-3/2} \]
03

Integrate the Function

Find the antiderivative by integrating each term separately:\[ \int f(u) \, du = \int (u^2) \, du + \int (3u^{-3/2}) \, du \]Calculating the integrals separately, we have:\[ \int (u^2) \, du = \frac{1}{3}u^3 \]\[ \int (3u^{-3/2}) \, du = 3 \cdot \left(\frac{u^{-1/2}}{-1/2}\right) = -6u^{-1/2} \]
04

Find the General Antiderivative

Combine the integrated terms to find the most general antiderivative:\[ F(u) = \frac{1}{3}u^3 - 6u^{-1/2} + C \]where \(C\) is the constant of integration.
05

Verify the Solution by Differentiation

Differentiate the antiderivative to ensure it matches the original function:\(F(u) = \frac{1}{3}u^3 - 6u^{-1/2} + C\)Taking the derivative:\[ F'(u) = \frac{3}{3}u^2 + 6 \cdot \left(\frac{-1}{2}\right)u^{-3/2} = u^2 + 3u^{-3/2} \]This simplifies back to \(f(u) = u^2 + \frac{3}{u^{3/2}}\), confirming our antiderivative is correct.

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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, serving as the reverse process of differentiation. Essentially, it accumulates quantities to find the total or collective value of something. When we talk about integration, terms like antiderivatives or indefinite integrals often come into play.
  • To integrate a function, you perform the opposite of differentiation, going from a given function to the function from which it was derived.
  • In the given exercise, integrating means finding the most general form of the antiderivative of the function.
  • The process requires recognizing each term as a power function, making them easier to integrate.
Integration is always followed by a constant of integration, denoted usually by "C," indicating that many antiderivatives exist that differ only by a constant value. This constant is vital as it represents an entire family of functions.
Differentiation
Differentiation is the calculus process of finding the derivative. The derivative represents the rate at which a function is changing at any given point.
  • In essence, while integration finds the total accumulation, differentiation measures how values change at specific instances.
  • In the step-by-step solution, after finding the antiderivative through integration, differentiation is employed to check that the original function is accurately recovered.
  • By differentiating the calculated antiderivative, we ensure it matches the original function, thus verifying the correctness of the solution.
Differentiation is like tracing back your steps in calculating an antiderivative to confirm all operations were correctly handled.
Calculus
Calculus is the branch of mathematics that primarily deals with change and motion. It is divided into two main parts: integration and differentiation. Both are essential tools used to analyze functions and their behaviors.
  • Integration, as covered, accumulates values, while differentiation finds how those values change.
  • Calculus enables us to solve complex problems such as finding areas under curves or predicting future trends based on current rates.
  • In everyday terms, calculus helps understand how things evolve over time or how they are constructed piece by piece.
The exercise above is a perfect instance of practical application, opting to find an antiderivative using calculus principles, simplifying, and ensuring accuracy through differentiation.
Power Functions
Power functions are functions of the form \(x^n\), where \(n\) is any real number. They are fundamental in calculus for integration and differentiation.
  • Understanding how to maneuver power functions is crucial for both differentiation and integration.
  • In the exercise, simplifying the function into components like \(u^2\) and \(3u^{-3/2}\) makes it straightforward to apply rules for integration.
  • Each term's power in a function significantly determines how easy or complex it is to integrate or differentiate the function.
By manipulating power functions, you can simplify complex expressions, making calculus operations easier, as seamlessly done in the provided solution to reach the final antiderivative.

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Most popular questions from this chapter

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