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Find the indefinite integral. $$ \int(1-3 x)^{1.4} d x $$

Short Answer

Expert verified
The short answer for the indefinite integral is: \(\int(1-3 x)^{1.4} d x = -\dfrac{(1-3x)^{2.4}}{7.2} + C\).

Step by step solution

01

Choose a substitution variable

Choose the expression inside the integral to substitute as a new variable: Let \(u = 1-3x\).
02

Find the derivative of the substitution variable

Now we need to find the derivative of this new variable, \(u\), with respect to \(x\): \(\frac{du}{dx} = -3\).
03

Express \(dx\) in terms of \(du\)

Now, we need to find the value of \(dx\) in terms of \(du\): \(du = -3 dx\), then we have: \(dx = \dfrac{du}{-3}\).
04

Substitute in the integral and solve

Now, we will substitute \(u\) and \(dx\) in the integral: \[\int(1-3 x)^{1.4} d x = \int u^{1.4} \cdot \frac{du}{-3}\] Now we can solve this new integral by using the basic formula for the power rule: \(\int u^{n} du = \frac{u^{n+1}}{n+1} + C\) In our case \(n = 1.4\), so: \[\int u^{1.4} du = \frac{u^{2.4}}{2.4} + C\] Now we substitute back the original expression for \(u\): \[\frac{(1-3x)^{2.4}}{2.4} + C\] And then, we need to remember that we have \(\frac{du}{-3}\) in our integral: \[\frac{1}{-3} \cdot \frac{(1-3x)^{2.4}}{2.4} + C\]
05

Final answer

Finally, we simplify the expression to find the indefinite integral: \[\int(1-3 x)^{1.4} d x = -\dfrac{(1-3x)^{2.4}}{7.2} + C\]

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Key Concepts

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

Substitution Method in Integration
The substitution method, also known as u-substitution, is a technique used to simplify the process of finding the indefinite integral of more complex functions. This method is similar in spirit to using a change of variables to simplify an algebraic expression.

How does it work? Imagine an integral that is difficult to evaluate. By identifying a portion of the integrand (the function being integrated) that could be replaced with a single variable, you transform the integral into a simpler form. In our exercise, we let \(u = 1-3x\) to simplify the function before integration.

This choice is strategic; the derivative of \(u\) with respect to \(x\) appears elsewhere in the integrand. After finding the derivative \(\frac{du}{dx}\) and solving for \(dx\), you substitute all \(x\)-terms in the integral with terms in \(u\) and \(du\). This yields a simpler integral that is easier to evaluate using basic integration rules. After integrating with respect to \(u\), we substitute back to the original variable to get the solution in terms of \(x\), completing the method of substitution.
Power Rule of Integration
The power rule for integration is one of the most fundamental rules in calculus. It provides a direct formula to integrate functions of the form \(x^n\), where \(n\) is any real number except for \(-1\). The rule states that \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

In the context of our exercise, after employing the substitution method, we end up with an integral in the form of \(u^{1.4}\). The power rule is perfectly suited for this scenario. Applying the power rule, we integrate to get \(\frac{u^{2.4}}{2.4} + C\). Remembering that \(u\) is a stand-in for \(1-3x\), we substitute back to express the antiderivative in the original terms. This rule is crucial as it simplifies the integration process for polynomial expressions.
Derivative Calculation
The calculation of derivatives is an essential operation in calculus, typically denoted as \(\frac{du}{dx}\) or \(\frac{df}{dx}\) for functions \(u\) or \(f\) with respect to the variable \(x\). The derivative represents the rate at which a function changes at any given point and is foundational for both differential and integral calculus.

In our exercise, the derivative calculation comes into play during the substitution method. After setting \(u = 1-3x\), we calculate its derivative to be \(-3\), which we then use to re-express \(dx\) as \(\frac{du}{-3}\). By acknowledging this relationship between \(u\) and \(x\), we are able to transform the original integral into one in terms of \(u\), making it possible to apply the power rule of integration. The knowledge of how to calculate derivatives is thus indispensable in performing successful variable substitution within an integral.

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