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Find the general antiderivative \(F(x)+C\) for each of the following. $$ f(x)=\frac{4 x^{6}+3 x^{4}}{x^{3}} $$

Short Answer

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

Step by step solution

01

Simplify the Function

To find the antiderivative, we first simplify the given function. The expression \(\frac{4x^6 + 3x^4}{x^3}\) can be separated into two fractions: \(\frac{4x^6}{x^3} + \frac{3x^4}{x^3}\). Simplifying each term, we get \(4x^3 + 3x\).
02

Identify the Antiderivative Formula

For each term in the simplified function, we can apply the antiderivative (integration) rule for power functions. This rule states that the antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\).
03

Compute the Antiderivative of Each Term

Applying the antiderivative rule to the term \(4x^3\), we get \(\frac{4x^{3+1}}{3+1} = \frac{4x^4}{4} = x^4\). For the term \(3x^1\), the antiderivative is \(\frac{3x^{1+1}}{1+1} = \frac{3x^2}{2}\).
04

Write the General Antiderivative

Combine the antiderivatives from Step 3 and add the constant of integration \(C\). The general antiderivative \(F(x)\) is \(x^4 + \frac{3}{2}x^2 + C\).

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Key Concepts

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

Antiderivative
An antiderivative is a function whose derivative equals the given function. If you differentiate an antiderivative, you return to the original function. Antidifferentiation, or finding the antiderivative, is like performing the reverse process of differentiation.
In practical terms, when you're given a function, say \(f(x)\), finding the antiderivative means identifying a function \(F(x)\) such that \(F'(x) = f(x)\). The notation for the antiderivative of a function \(f(x)\) includes a constant \(C\), which represents an arbitrary constant. This is because the derivative of a constant is zero, so any constant could be included without affecting differentiation results.
Understanding the concept of the constant \(C\) is crucial. While solving integration problems, always remember to add \(C\) to your final result. It's what ensures the generality of the antiderivative.
Power Rule
The Power Rule for integration is a fundamental part of finding antiderivatives, especially for polynomial functions. If you're comfortable with differentiation, you'll notice the similarities. The Power Rule states:
  • If \(f(x) = x^n\) then the antiderivative is \(F(x) = \frac{x^{n+1}}{n+1} + C\), n ≠ -1.
This rule simplifies the integration of any function where the variable is raised to a power. Applying it involves a simple adjustment to the exponent and dividing by the new exponent.
To see the Power Rule in action, consider the term \(4x^3\). When applying the Power Rule, increase the exponent by one to get \(4x^4\). Then divide by this new exponent, resulting in \(\frac{4x^4}{4}\), simplifying to \(x^4\). As straightforward as it sounds, practicing with diverse examples helps reinforce this concept.
Remember, integrating constant terms is just as straightforward. For example, with \(3x^1\), integrate by increasing the exponent to two and dividing by the new exponent, leading to \(\frac{3x^2}{2}\). Keep practicing to make the Power Rule second nature.
Simplifying Expressions
Simplifying expressions is a key step before integrating, especially with fractions or complex polynomials. Consider the original function \(\frac{4x^6 + 3x^4}{x^3}\). The complexity in direct integration can often be overwhelming, so simplifying first can save time and reduce errors.
In this exercise, start by splitting the overall fraction into individual fractions: \(\frac{4x^6}{x^3} + \frac{3x^4}{x^3}\). By simplifying each fraction separately, we reduce them to more manageable terms, \(4x^3 + 3x\).
Why do we simplify? Because straightforward terms make applying the Power Rule easier. Once reduced, you avoid potential confusion and can focus on integrating each straightforward component.
Here's a tip: Look for opportunities within expressions to cancel terms or reduce exponents before you start integrating. It breaks down what otherwise might involve more elaborate calculations, leading to correct results with less effort.

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