What is the integral of (x^2)/6 with respect to x?
Understand the Problem
The question is asking for the integral of the function (x^2)/6 with respect to x. We need to find the antiderivative by using the rules of integration.
Answer
$$ \int \frac{x^2}{6} \, dx = \frac{x^3}{18} + C $$
Answer for screen readers
The integral of ( \frac{x^2}{6} ) with respect to ( x ) is: $$ \int \frac{x^2}{6} , dx = \frac{x^3}{18} + C $$
Steps to Solve
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Set up the integral We need to integrate the function ( \frac{x^2}{6} ). This can be written as: $$ \int \frac{x^2}{6} , dx $$
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Factor out the constant Since ( \frac{1}{6} ) is a constant, we can factor it out of the integral: $$ \frac{1}{6} \int x^2 , dx $$
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Use the power rule for integration Now, we apply the power rule for integration, which states that: $$ \int x^n , dx = \frac{x^{n+1}}{n+1} + C $$ For our case, ( n = 2 ): $$ \int x^2 , dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C $$
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Apply the result of the power rule Substituting back into our equation, we have: $$ \frac{1}{6} \left( \frac{x^3}{3} + C \right) $$
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Simplify the expression Finally, we can simplify this expression: $$ \frac{x^3}{18} + \frac{1}{6}C $$
Since ( \frac{1}{6}C ) is just a constant, we can replace it with a new constant, ( C' ): $$ \int \frac{x^2}{6} , dx = \frac{x^3}{18} + C' $$
The integral of ( \frac{x^2}{6} ) with respect to ( x ) is: $$ \int \frac{x^2}{6} , dx = \frac{x^3}{18} + C $$
More Information
The result represents the family of antiderivatives for the function ( \frac{x^2}{6} ). The constant ( C ) accounts for all possible vertical shifts of the antiderivative graph.
Tips
- Forgetting to add the constant ( C ) at the end of the integration, which is essential in indefinite integrals.
- Not applying the power rule correctly by miscalculating the exponent or the denominator.
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