Problem 30 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The indefinite integral of the function $$6\sqrt[3]{x}$$ is $$\left(\frac{9}{2}\right)x^{\frac{4}{3}} + C.$$

Step 1: Apply the power rule of integration

To apply the power rule for integration, we need to identify the power of x in the given function, which is $$\frac{1}{3}.$$ So, we will integrate $$6x^{\frac{1}{3}}$$ with respect to x. $$\int 6x^{\frac{1}{3}} dx = 6\int x^{\frac{1}{3}} dx$$ Now, applying the power rule for integration: $$6\int x^{\frac{1}{3}} dx = 6\left(\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right) + C$$

Step 2: Simplify the result

Now, we will simplify the result of integration: $$6\left(\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right) + C = \frac{6}{\frac{4}{3}}x^{\frac{4}{3}} + C$$ Multiplying the fraction by its reciprocal: $$\frac{6}{\frac{4}{3}}x^{\frac{4}{3}} + C = \frac{6\cdot\frac{3}{4}}{1}x^{\frac{4}{3}} + C = \left(\frac{9}{2}\right)x^{\frac{4}{3}} + C$$ So, the indefinite integral of $$6\sqrt[3]{x}$$ is $$\left(\frac{9}{2}\right)x^{\frac{4}{3}} + C.$$

Step 3: Check the result by differentiation

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