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Chapter 4: Problem 2

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

Short Answer

Expert verified
The antiderivative is \( F(x) = \frac{x^3}{6} - x^2 + 6x + C \).

Step by step solution

01

Identify the Function

The given function is \( f(x) = \frac{1}{2}x^2 - 2x + 6 \). We need to find its most general antiderivative.
02

Antiderivative of Each Term

To find the antiderivative, integrate each term separately.1. For \( \frac{1}{2}x^2 \), the antiderivative is \( \frac{1}{2} \cdot \frac{x^3}{3} = \frac{x^3}{6} \).2. For \(-2x\), the antiderivative is \(-2 \cdot \frac{x^2}{2} = -x^2 \).3. For \(6\), the antiderivative is \(6x \).
03

Combine the Antiderivatives

Combine all the antiderivatives and include the constant of integration \( C \). Thus, the most general antiderivative \( F(x) \) is:\[ F(x) = \frac{x^3}{6} - x^2 + 6x + C \]
04

Differentiation Check

Differentiate \( F(x) \) to verify:\[ \frac{d}{dx}\left(\frac{x^3}{6} - x^2 + 6x + C\right) = \frac{1}{2}x^2 - 2x + 6 \].This matches the original function \( f(x) \), confirming our solution 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 like finding the opposite of differentiation. When we integrate a function, we are essentially looking for what the original function might have been before it was differentiated. For the function given, \( f(x) = \frac{1}{2} x^2 - 2x + 6 \), we need to find its antiderivative, which is also known as its indefinite integral. This process involves reversing the rules of differentiation.To find the antiderivative, you integrate each term separately:
  • For \( \frac{1}{2}x^2 \), you use the power rule in reverse: add 1 to the exponent (2 becoming 3) and divide by this new exponent, making the antiderivative \( \frac{x^3}{6} \).
  • For \(-2x\), the process is similar, resulting in \(-x^2\).
  • And for a constant like 6, the antiderivative becomes 6 times \( x \).
The sum of these results gives you the most general antiderivative, with a constant \( C \) added to account for any constant term that could have been in the original function before differentiation.
Differentiation
Differentiation is the process of finding a derivative, basically determining the rate at which a function is changing. We use differentiation to confirm our integration was done correctly. By differentiating the antiderivative, we hope to obtain the original function. Here's how it looks in our example.Starting with the antiderivative \( F(x) = \frac{x^3}{6} - x^2 + 6x + C \), we differentiate:
  • When differentiating \( \frac{x^3}{6} \), you multiply the exponent down (3 becomes 3x) and subtract 1 from the exponent (becoming \( x^2 \)), resulting in \( \frac{1}{2}x^2 \).
  • For \(-x^2\), it simplifies to \(-2x\).
  • Differentiating 6x gives you 6.
  • The constant \( C \) becomes 0 because constants vanish in differentiation.
You should end up with \( \frac{1}{2}x^2 - 2x + 6 \), which is the original function. This shows that the integration step was done correctly.
Calculus Problem-Solving
When solving calculus problems, like finding the most general antiderivative, it's important to apply both integration and differentiation effectively.Here are some steps to help guide you through similar problems:
  • Understand the function: Break it down into manageable terms. In this case, recognizing each term as a polynomial or a constant makes applying rules of integration straightforward.
  • Perform integration carefully: Use the power rule where necessary, integrate constants by multiplying them by \( x \), and remember to always add the constant of integration \( C \).
  • Verify with differentiation: Once your antiderivative is found, differentiate it to ensure it simplifies back to the original function. This serves as a great check that you took the correct steps.
  • Build understanding with practice: Like most math skills, proficiency in calculus comes from working through numerous problems and learning from each solution process.
By maintaining a structured approach and checking your work through differentiation, you build a strong understanding of these core calculus concepts.

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Most popular questions from this chapter

A manufacturer has been selling 1000 flat-screen TVs a week at \(\$ 450\) each. A market survey indicates that for each \(\$ 10\) rebate offered to the buyer, the number of TVs sold will increase by 100 per week. (a) Find the demand function. (b) How large a rebate should the company offer the buyer in order to maximize its revenue? (c) If its weekly cost function is \(C ( x ) = 68,000 + 150 x\) , how should the manufacturer set the size of the rebate in order to maximize its profit?

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