/*! 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 10 Integrate each of the given expr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 10

Integrate each of the given expressions. $$\int 6 \sqrt[3]{x} d x$$

Short Answer

Expert verified
The integral is \(\frac{9}{2} x^{4/3} + C\).

Step by step solution

01

Rewrite the Expression

The given integral is \(\int 6 \sqrt[3]{x} \, dx\). To integrate, let's first rewrite the cube root using an exponent. We know that \(\sqrt[3]{x} = x^{1/3}\), so the integral becomes \(\int 6x^{1/3} \, dx\).
02

Apply the Power Rule for Integration

The power rule for integration states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(n eq -1\). In our case, \(n = \frac{1}{3}\). So, we integrate: \[\int 6x^{1/3} \, dx = 6 \cdot \frac{x^{4/3}}{4/3} + C.\]
03

Simplify the Result

Now simplify \(6 \cdot \frac{x^{4/3}}{4/3}\). This is equivalent to multiplying by the reciprocal, so we have: \[6 \cdot \frac{3}{4} x^{4/3} = \frac{18}{4} x^{4/3} = \frac{9}{2} x^{4/3}.\] Add the integration constant: \(\frac{9}{2} x^{4/3} + C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule for integration is an essential technique in calculus. It simplifies the process of integrating functions that are in the format of a power of x. If you encounter an expression like \( x^n \), you can easily integrate it using the power rule.
  • The basic form of the power rule is: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
  • Here, \( n \) can be any real number, except \(-1\).
  • Once you add one to the exponent and divide by the new exponent, don't forget to include the constant of integration, \( C \).
This constant \( C \) represents the infinite number of possible antiderivatives that differ by a constant. So, in our example, \( \int 6x^{1/3} \, dx \), we applied the power rule with \( n = \frac{1}{3} \) to find the solution. Calculating \( \frac{x^{4/3}}{4/3} \), and then multiplying by 6, led us to our answer.
Cube Root Integration
In calculus, cube root integration might seem daunting at first, but it becomes straightforward when you convert roots into exponents. A cube root, \( \sqrt[3]{x} \), is the same as \( x^{1/3} \).
  • Always convert roots into exponents for simpler integration. For instance, \( \sqrt[n]{x} = x^{1/n} \).
  • Once the expression is in the exponential form, you can apply the power rule directly.
  • This transformation makes it clear how to manipulate the function algebraically.
In our example, rewriting \( 6\sqrt[3]{x} \) as \( 6x^{1/3} \) was the key first step. This transformation set the stage for straightforward application of the power rule.
Indefinite Integrals
Indefinite integrals are the most fundamental type of integral in calculus. When you integrate a function indefinitely, you find a general formula for all antiderivatives of that function.
  • The result of an indefinite integral is a function plus a constant, denoted as \( C \).
  • This constant is crucial because it accounts for all possible vertical shifts of the antiderivative.
  • Symbolically, an indefinite integral is represented as \( \int f(x) \, dx \).
For the exercise \( \int 6\sqrt[3]{x} \, dx \), we calculated the integral to be \( \frac{9}{2}x^{4/3} + C \). Here, our answer is an indefinite integral, showing all functions whose derivative gives us the original expression \( 6x^{1/3} \). Therefore, the integral isn't just one function, but several, differing by a constant.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the given problems. Explain your answers. $$\text { Is } \int 3 x^{2} d x=x^{3} ?$$

Determine the value of a that makes \(F(x)\) an antiderivative of \(f(x)\) $$f(x)=3 x^{2}, F(x)=a x^{3}$$

Find the exact area under the given curves between the indicated values of \(x\). The functions are the same as those for which approximate areas were found in Exercises \(5-14\). \(y=3 x,\) between \(x=0\) and \(x=3\)

Solve the given problems. Find the general form of the function whose second derivative is \(\sqrt{x}\).

Determine the value of a that makes \(F(x)\) an antiderivative of \(f(x)\) $$f(x)=10 x^{1 / 4}, F(x)=a x^{5 / 4}$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.