/*! 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 14 Find \(\int\left(x^6+x^3\right) ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 14

Find \(\int\left(x^6+x^3\right) \sqrt[3]{x^3+2} d x\)

Short Answer

Expert verified
\(\frac{1}{3} \left( \frac{3}{10} (x^3 + 2)^{10/3} - \frac{9}{7} (x^3 + 2)^{7/3} + \frac{6}{4}(x^3 + 2)^{4/3} \right) + C\).

Step by step solution

01

- Identify and Simplify the Integrand

Rewrite the integrand \(\int\left(x^6+x^3\right) \sqrt[3]{x^3+2} d x\) for easier manipulation. Recognize that \(\sqrt[3]{x^3 + 2} = (x^3 + 2)^{1/3}\). Therefore, the integrand becomes \(\int (x^6 + x^3)(x^3 + 2)^{1/3} dx\).
02

- Use Substitution Method

Let \( u = x^3 + 2 \). Then, \( du = 3x^2 dx \). This means \( x^2 dx = \frac{1}{3} du \). Notice that \( x^6 = (u - 2)^2\) and \( x^3 = u - 2 \).
03

- Rewrite the Integrand in Terms of the Substitution

Substitute the expressions involving \( u \) to reframe the integral. The integrand \( (x^6 + x^3)(x^3 + 2)^{1/3} dx \) becomes \( ((u - 2)^2 + (u - 2))(u)^{1/3} \frac{1}{3} du \).
04

- Simplify the New Integral

Distribute and combine like terms: \( ((u - 2)^2 + (u - 2))(u)^{1/3} \frac{1}{3} du \) becomes \( (u^2 - 4u + 4 + u - 2)u^{1/3} \frac{1}{3} du \ = (u^2 -3u +2)u^{1/3} \frac{1}{3} du \).
05

- Integrate Term-by-Term

Separate the terms and integrate: \( \frac{1}{3} \int (u^2 u^{1/3} - 3u u^{1/3} + 2u^{1/3}) du \). Integrate each term: \(\frac{1}{3} \left( \int u^{7/3} du - 3 \int u^{4/3} du + 2 \int u^{1/3} du \right)\).
06

- Perform the Integration

Find the antiderivatives: \( \frac{1}{3} \left( \frac{3}{10} u^{10/3} - 3 \cdot \frac{3}{7} u^{7/3} + 2 \cdot \frac{3}{4} u^{4/3} \right) + C \= \frac{1}{3} \left( \frac{3}{10} u^{10/3} - \frac{9}{7} u^{7/3} + \frac{6}{4}u^{4/3} \right) + C \).
07

- Substitute Back in Terms of x

Replace \( u \) with \( x^3 + 2 \) to convert the expression back in terms of x: \( \frac{1}{3} \left( \frac{3}{10} (x^3 + 2)^{10/3} - \frac{9}{7} (x^3 + 2)^{7/3} + \frac{6}{4}(x^3 + 2)^{4/3} \right) + C \).

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Key Concepts

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

integration techniques
Integration is a cornerstone of calculus, employed to find the area under curves. Different methods exist, each suited for particular types of functions.
One such technique involves recognizing patterns or breaking integrals into simpler parts. Here, we recognize that the integrand \(\int (x^6 + x^3)(x^3 + 2)^{1/3} dx\) can be simplified. Substitution, integration by parts, and partial fractions are just a few methods we use.
The first step usually involves simplifying the integrand, making it easier to integrate.
substitution method
The substitution method helps when direct integration is tough. It involves substituting a part of the integrand with a new variable. This reduces complexity.
For example, in \(\int (x^6 + x^3)(x^3 + 2)^{1/3}dx\), we let \(\u = x^3 + 2\). This shifts the variable of integration from \(x\) to \(u\).
Calculating the differential gives \(\du = 3x^2 dx\). Thus, \(\dx = \frac{1}{3} \frac{du}{x^2} \). Now anything in terms of \(\x\) is substituted.
antiderivatives
Antiderivatives reverse differentiation and are vital in integration.
The process entails finding functions whose derivative equals the given function. For instance, the antiderivative of \( u^{n} \) is \[ \frac{u^{n+1}}{n+1} + C \].
In our solved problem, after rewriting, we integrate term by term:
  • \int u^{7/3} du = \[ \frac{3}{10} u^{10/3} \]
  • -3 \int u^{4/3} du = \[ -\frac{9}{7} u^{7/3} \]
  • 2 \int u^{1/3} du = \[ \frac{6}{4} u^{4/3} \]
This step-by-step process helps reach the solution.

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Most popular questions from this chapter

Fast Eddie needs to double his money; he can only do so by playing a certain win-lose game, in which the probability of winning is \(p\). However, he can play this game as many or as few times as he wishes, and in a particular game he can bet any desired fraction of his bankroll. The game pays even money (the odds are one-to-one). Assuming he follows an optimal strategy if one is available, what is the probability, as a function of \(p\), that Fast Eddie will succeed in doubling his money?

Note that the three positive integers \(1,24,120\) have the property that the sum of any two of them is a different perfect square. Do there exist four positive integers such that the sum of any two of them is a perfect square and such that the six squares found in this way are all different? If so, exhibit four such positive integers; if not, show why this cannot be done.

Consider the equation \(x^2+\cos ^2 x=\alpha \cos x\), where \(\alpha\) is some positive real number. a. For what value or values of \(\alpha\) does the equation have a unique solution? b. For how many values of \(\alpha\) does the equation have precisely four solutions?

Find all real polynomials \(p(x)\), whose roots are real, for which $$ p\left(x^2-1\right)=p(x) p(-x). $$

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