Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \sqrt{x}\left(2 x^{6}-4 \sqrt[3]{x}\right) d x$$

Short answer

Question: Determine the indefinite integral of the function $$\sqrt{x}(2 x^{6}-4 \sqrt[3]{x}).$$ Solution: The indefinite integral of the function is $$\int \sqrt{x}(2 x^{6}-4 \sqrt[3]{x}) dx = \frac{4}{15}x^{\frac{15}{2}} - \frac{24}{11}x^{\frac{11}{6}} + C$$.

Step-by-step solution

Rewrite the function as powers of x

First, rewrite the function in terms of powers of x to simplify the integration process: $$\int \sqrt{x}(2 x^{6}-4 \sqrt[3]{x}) dx = \int (2x^{6}-4x^{\frac{1}{3}})x^{\frac{1}{2}} dx $$

Distribute the x terms

Distribute the \(x^{\frac{1}{2}}\) to both terms inside the parentheses: $$\int (2x^{6}-4x^{\frac{1}{3}})x^{\frac{1}{2}} dx = \int (2x^{\frac{13}{2}}-4x^{\frac{5}{6}})dx$$

Apply the power rule for integration

Use the power rule for integration on both terms: $$\int (2x^{\frac{13}{2}}-4x^{\frac{5}{6}})dx = \frac{2}{\frac{15}{2}}x^{\frac{15}{2}} - \frac{4}{\frac{11}{6}}x^{\frac{11}{6}} + C$$

Simplify the coefficients

Simplify the result by simplifying the coefficients of the integrated terms: $$\frac{2}{\frac{15}{2}}x^{\frac{15}{2}} - \frac{4}{\frac{11}{6}}x^{\frac{11}{6}} + C = \frac{4}{15}x^{\frac{15}{2}} - \frac{24}{11}x^{\frac{11}{6}} + C$$ The indefinite integral is: $$\int \sqrt{x}(2 x^{6}-4 \sqrt[3]{x}) dx = \frac{4}{15}x^{\frac{15}{2}} - \frac{24}{11}x^{\frac{11}{6}} + C$$

Check the work by differentiation

Now, differentiate the result to check our work: $$\frac{d}{dx}\left(\frac{4}{15}x^{\frac{15}{2}} - \frac{24}{11}x^{\frac{11}{6}} + C\right) = \left(\frac{15}{2}\right)\left(\frac{4}{15}\right)x^{\frac{13}{2}} - \left(\frac{11}{6}\right)\left(\frac{24}{11}\right)x^{\frac{5}{6}}$$ Simplify to get the original function: $$\left(\frac{15}{2}\right)\left(\frac{4}{15}\right)x^{\frac{13}{2}} - \left(\frac{11}{6}\right)\left(\frac{24}{11}\right)x^{\frac{5}{6}} = 2x^{6}\sqrt{x} - 4x^{\frac{1}{3}}\sqrt{x} = \sqrt{x}(2 x^{6}-4 \sqrt[3]{x})$$ The differentiation of the indefinite integral gives back the original function, confirming the correctness of the solution.

Key concepts

Power Rule for Integration

Understanding the power rule for integration is an essential stepping stone in calculus, particularly when finding indefinite integrals. The rule, simplistic in form, empowers you to integrate functions that are polynomial-like in nature.

When you encounter a term in the form of \(x^n\), where 'n' is any real number, the power rule instructs you to add 1 to the exponent (n) and then divide by the new exponent. Symbolically, this is represented as the integration of \(x^n\) yielding \(\frac{x^{n+1}}{n+1} + C\), where 'C' denotes the constant of integration. This process, as seen in Step 3 of our exercise solution, is applied term-by-term when dealing with polynomial expressions.

It’s vital not to forget that this rule only applies directly to terms with exponents different from -1, since integrating \(x^{-1}\) or \(\frac{1}{x}\) results in a natural logarithm function rather than a power function. Understanding and applying this rule correctly is key in simplifying the integration process of polynomials and ensuring accurate results.

Radicals and Fractional Exponents

When faced with radicals, like the square root in the exercise above, knowing how to rewrite them as fractional exponents is an invaluable skill. A radical can always be expressed as a number raised to a fraction less than 1; for instance, \(\sqrt[n]{x}\) is equivalent to \(x^{1/n}\).

In our example, \(\sqrt{x}\) becomes \(x^{1/2}\), and \(\sqrt[3]{x}\) is rewritten as \(x^{1/3}\). This manipulation is fundamental because it enables the use of the power rule for integration, as shown in Step 2. Without converting radicals to fractional exponents, you may find yourself stuck, unsure how to proceed with integration. Mastery of this concept is crucial because it not only aids in integrating functions with roots but also in differentiating them, as both operations often require a function to be in an exponentiated form.

Differentiation to Check Integration

In calculus, it’s just as important to verify your results as it is to find them. When it comes to indefinite integrals, one reliable method of verification is by differentiation, as implemented in Step 5 of the solution. After finding an indefinite integral, you should be able to differentiate it and arrive back at the original function you started with.

Differentiation acts as the inverse operation of integration. If the differentiation of an indefinite integral doesn't quite match the original function, this indicates a mistake in the integration process. This technique is not only useful for error-checking but also serves as a strong conceptual reinforcement, bridging the gap between the two fundamental processes of calculus: differentiation and integration.