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Finding an Indefinite Integral In Exercises \(5-26\) , find the indefinite integral and check the result by differentiation. $$ \int x^{2}\left(x^{3}-1\right)^{4} d x $$

Short Answer

Expert verified
The indefinite integral of \(x^{2}\left(x^{3}-1\right)^{4} dx\) is \((1/15)(x^{3} -1)^{5} + C\), and the subsequent differentiation of this result confirms the correctness of the solution.

Step by step solution

01

Express in Proper Form

Given the expression \(\int x^{2}\left(x^{3}-1\right)^{4} dx\), one of the best approaches is to start by letting \(u = x^{3}-1\) for simplification and then calculate the derivative \(du\).
02

Find du

Differentiating \(u = x^{3} - 1\) gives \(du = 3x^{2} dx\). We can then rearrange to get \(dx = du / 3x^{2}\).
03

Substitute dx

We can substitute the \(dx\) from step 2 into the original integral, giving us \(\int x^{2} u^{4} du / 3x^{2}\). The \(x^{2}\) terms cancel out, simplifying the expression to \(\int u^{4} du / 3\). This can be integrated easily.
04

Integrate

The integral of \(u^{4}\) with respect to \(u\) is \((1/5)u^{5}\). Therefore, \(\int u^{4} du = (1/5)u^{5}\). Adding the denominator from the previous step, we get \((1/15)u^{5} + C\), where \(C\) is the constant of integration.
05

Substitute u back

Substitute back for \(u\) from our original substitution (\(u = x^{3} - 1\)). This gives the final answer, \((1/15)(x^{3} -1)^{5} + C\).
06

Verification by Differentiation

To check our result, differentiate the resulting expression using the chain rule. Specifically, the derivative of \((1/15)(x^{3} -1)^{5} + C\) is \((1/3)x^{2}(x^{3}-1)^{4}\), which is the integrand of the initial expression, verifying our solution.

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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 fundamental concept in calculus, essential for solving problems involving the area under a curve, the accumulation of quantities, and in many other applications. One of the key skills in mastering integration is to become familiar with various integration techniques. These techniques include, but are not limited to, the substitution method, integration by parts, partial fraction decomposition, and trigonometric integration.

Each technique has its ideal scenarios. For straightforward functions, direct antiderivatives may be possible. For complex expressions involving products or compositions of functions, more powerful techniques like substitution or integration by parts might be necessary. Choosing the right technique not only simplifies the calculation but can also make the difference between finding a solution or not. It's crucial to familiarize oneself with different scenarios where each method is most effective.
Substitution Method
The substitution method is often the go-to technique for simplifying integrals that involve a composite function—think of a function within another function. The method involves identifying a part of the integrand as a new variable, say 'u,' which simplifies the integral.

The power of substitution comes from its connection to the chain rule in differentiation. Essentially, you're working backward through the chain rule: after identifying 'u', you find 'du', and then replace parts of the original integral with 'u' and 'du'. This transforms the integral into one in terms of 'u', which is often simpler to solve. Once you find the antiderivative in terms of 'u', you substitute the original expressions back in to return to your original variable.
Chain Rule
The chain rule is a rule in calculus used to find the derivative of the composition of two or more functions. In the context of integration, it allows us to understand the mechanics behind the substitution method. When you reverse the process of the chain rule, you can break down a complicated function into simpler parts that are easier to integrate.

For example, when you have a composite function like \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \). The chain rule is what we utilize in reverse when we work out the antiderivative of a composite function. By choosing \( u = g(x) \), we effectively reverse engineer this process to find the indefinite integral of complex expressions.
Antidifferentiation
In calculus, antidifferentiation, or finding the indefinite integral, is the reverse process of differentiation. If differentiation gives you the rate at which a quantity changes, antidifferentiation gives you the original quantity. The goal of antidifferentiation is to find a function \( F(x) \) whose derivative is the given function \( f(x) \).

The constant of integration, represented by \( C \), is a critical component, as there are infinitely many antiderivatives for any given function, each differing by a constant. When approaching antidifferentiation, one can often find patterns or use techniques like substitution to convert the integral into a more familiar form, facilitating the finding of an antiderivative. As the last step in any integration problem, we must always remember to add \( C \) to account for all potential solutions.

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