/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Determine the following indefini... [FREE SOLUTION] | 91影视

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

Step by step solution

01

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: $$6\int x^{\frac{1}{3}} dx = 6\left(\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right) + C$$
02

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 $$\left(\frac{9}{2}\right)x^{\frac{4}{3}} + C.$$
03

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.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. \(F(x)=x^{3}-4 x+100\) and \(G(x)=x^{3}-4 x-100\) are antiderivatives of the same function. b. If \(F^{\prime}(x)=f(x),\) then \(f\) is an antiderivative of \(F\) c. If \(F^{\prime}(x)=f(x),\) then \(\int f(x) d x=F(x)+C\) d. \(f(x)=x^{3}+3\) and \(g(x)=x^{3}-4\) are derivatives of the same function. e. If \(F^{\prime}(x)=G^{\prime}(x),\) then \(F(x)=G(x)\)

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

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First Derivative Test is not exhaustive Sketch the graph of a (simple) nonconstant function \(f\) that has a local maximum at \(x=1,\) with \(f^{\prime}(1)=0,\) where \(f^{\prime}\) does not change sign from positive to negative as \(x\) increases through \(1 .\) Why can't the First Derivative Test be used to classify the critical point at \(x=1\) as a local maximum? How could the test be rephrased to account for such a critical point?
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