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Chapter 5: Problem 1

In Problems 1-40, find the general antiderivative of the given function. $$ f(x)=4 x^{2}-x $$

Short Answer

Expert verified
The general antiderivative is \( F(x) = \frac{4}{3}x^3 - \frac{1}{2}x^2 + C \).

Step by step solution

01

Recall the General Rule for Antiderivatives

To find the antiderivative of a function, we use the power rule for integration. For a function of the form \( f(x) = ax^n \), its antiderivative is given by \( F(x) = \frac{a}{n+1}x^{n+1} + C \), where \( C \) is the constant of integration.
02

Apply the Power Rule to Each Term

First, consider the term \( 4x^2 \). Using the power rule, its antiderivative is \( \frac{4}{2+1}x^{2+1} = \frac{4}{3}x^3 \).Next, examine the term \( -x \). This can be rewritten as \( -1x^1 \), and its antiderivative is \( \frac{-1}{1+1}x^{1+1} = -\frac{1}{2}x^2 \).
03

Combine the Antiderivatives

Now, add the antiderivatives of all terms together and include the constant of integration:\[ F(x) = \frac{4}{3}x^3 - \frac{1}{2}x^2 + C \]
04

Write the General Antiderivative

Finally, express the overall function: \( F(x) = \frac{4}{3}x^3 - \frac{1}{2}x^2 + C \), where \( C \) is any constant. This represents the general antiderivative of the function \( f(x) = 4x^2 - x \).

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Key Concepts

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

Power Rule for Integration
The power rule for integration is a foundational tool in calculus. It allows us to find antiderivatives of functions that take the form of a power of a variable. To apply the power rule to any function of the form \( ax^n \), follow these steps:
  • Increase the exponent, \( n \), by 1. This means you will use \( n+1 \) as the new exponent.
  • Divide the coefficient, \( a \), by this new exponent to find the new coefficient of your term.
After performing these steps, you'll have the antiderivative of that term in the form \( \frac{a}{n+1}x^{n+1} \).Using the power rule, you don't have to remember complicated integration techniques for simple polynomials. Just follow the rule, and you'll convert any polynomial term into its antiderivative. Remember, each term in a polynomial needs to be integrated separately before adding them back together.
Constant of Integration
When integrating, especially when finding general antiderivatives, it's crucial to include a constant of integration, typically denoted by \( C \). This constant accounts for the fact that indefinite integrals (antiderivatives) include all possible vertical shifts of a function.
  • Without the constant \( C \), you would miss identifying the complete family of antiderivatives for a given function.
  • The constant of integration can be any real number, meaning each different value of \( C \) represents a different antiderivative.
In practical problem-solving, this constant ensures that your solutions reflect every possible scenario of the original function's integral, capturing every nuance of how it might appear in the real world.
Calculus Problem-Solving
Solving calculus problems, such as finding antiderivatives, involves methodically working through a series of systematic steps to ensure accuracy. Here are some key aspects to focus on:
  • Start by understanding the problem: Identify the function and the need, such as finding an antiderivative.
  • Choose the right rule: For polynomial functions, use the power rule as discussed.
  • Step through the process: Integrate each term separately, being careful to adjust exponents and coefficients correctly.
  • Check your work: After finding the antiderivative, consider its derivative to see if it returns to the original function.
  • Practice makes perfect: The more problems you solve, the more comfortable you will become with the techniques.
Effective problem-solving in calculus isn't just about applying rules, but rather understanding the rules so well that you can apply them creatively in various scenarios. This skill comes with practice, patience, and persistence.

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Most popular questions from this chapter

This problem illustrates the fact that \(f^{\prime}(c)=0\) is not a sufficient condition for the existence of a local extremum of a differentiable function.] Show that the function \(f(x)=x^{3}\) has a horizontal tangent at \(x=0 ;\) that is, show that \(f^{\prime}(0)=0\), but \(f^{\prime}(x)\) does not change sign at \(x=0\) and, hence, \(f(x)\) does not have a local extremum at \(x=0\).

Determine whether each function has absolute maxima and minima and find their coordinates. For each function, find the intervals on which it is increasing and the intervals on which it is decreasing. $$ y=(2-x)^{2},-2 \leq x \leq 3 $$

The height \(y\) in feet of a tree as a function of the tree's age \(x\) in years is given by $$ y=121 e^{-17 / x} \quad \text { for } x>0 $$ (a) Determine (1) the rate of growth when \(x \rightarrow 0^{+}\) and (2) the limit of the height as \(x \rightarrow \infty\). (b) Find the age at which the growth rate is maximal. (c) Show that the height of the tree is an increasing function of age. At what age is the height increasing at an accelerating rate and at what age at a decelerating rate? (d) Sketch the graph of both the height and the rate of growth of the tree as functions of age.

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{\sin x} $$

Find the limits in Problems Be sure to check whether you can apply I'Hospital's rule before you evaluate the limit. $$ \lim _{x \rightarrow-\infty} x e^{x} $$
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