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Chapter 4: Problem 30

Determine the following indefinite integrals. Check your work by differentiation. $$\int 6 \sqrt[3]{x} d x$$

Short Answer

Expert verified
Question: Find the indefinite integral of the given function and verify the result by differentiation: $$\int 6 \sqrt[3]{x} dx$$. Answer: The indefinite integral of the given function is $$\frac{9}{2}x^{\frac{4}{3}} + C$$.

Step by step solution

01

Identify the power of x in the integrand

In the given function, we can rewrite the cube root of x as $$x^{\frac{1}{3}}$$. So the function becomes $$\int 6x^{\frac{1}{3}}dx$$. Now, we can identify the power of x in the integrand, which is $$\frac{1}{3}$$.
02

Apply the power rule for integration

Using the power rule for integration, we will integrate the function with respect to x: $$\int 6x^{\frac{1}{3}} dx = 6\int x^{\frac{1}{3}} dx = 6\left(\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right) + C$$ Simplify the expression: $$= 6\left(\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right) + C = \frac{18}{4}x^{\frac{4}{3}} + C = \frac{9}{2}x^{\frac{4}{3}} + C$$ The indefinite integral of the given function is $$\frac{9}{2}x^{\frac{4}{3}} + C$$.
03

Check the result by differentiation

To check our work, we will differentiate the result with respect to x and see if we get back the original function. Differentiate $$\frac{9}{2}x^{\frac{4}{3}}$$ with respect to x: $$\frac{d}{dx}\left(\frac{9}{2}x^{\frac{4}{3}}\right) = \frac{9}{2}(\frac{4}{3})x^{\frac{4}{3}-1} = 6x^{\frac{1}{3}}$$ The result of differentiation matches the original function, which confirms that the indefinite integral is correct and is $$\frac{9}{2}x^{\frac{4}{3}} + C$$.

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