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Magazine Chapter 13: Problem 10
Evaluate the given integral using the substitution (or method) indicated. $$ \int(x-1)^{2} e^{(x-1)^{3}} d x ; u=(x-1)^{3} $$
Short Answer
Expert verified
The short version of the answer is:
$$
\int (x-1)^2 e^{(x-1)^3}\, dx = \frac{1}{3} e^{(x-1)^3} + C
$$
Step by step solution
01
Identify the given integral and substitution
We are given the integral to evaluate:
$$
\int (x-1)^2 e^{(x-1)^3} dx
$$
The substitution function is given by:
$$
u = (x-1)^3
$$
02
Find the derivative of the substitution function
Calculate \(du/dx\) of the substitution \(u = (x-1)^3\):
$$
\frac{du}{dx} = 3(x-1)^2
$$
Now, calculate \(dx\) by rearranging this equation:
$$
dx = \frac{du}{3(x-1)^2}
$$
03
Rewrite the integral in terms of u
Using the substitution function, we can see that:
$$
(x-1)^2 = \sqrt[3]{u}
$$
Replace \((x-1)^2\) with \(\sqrt[3]{u}\) and \(dx\) with \(\frac{du}{3\sqrt[3]{u^2}}\) in the integral:
$$
\int (x-1)^2 e^{(x-1)^3}\, dx = \int \sqrt[3]{u} e^{u} \cdot \frac{du}{3\sqrt[3]{u^2}}
$$
04
Simplify the integral
Notice that we can simplify the expression in the integral:
$$
\sqrt[3]{u} \cdot \frac{1}{3\sqrt[3]{u^2}} = \frac{1}{3\sqrt[3]{u}}
$$
Now, we can rewrite the integral as:
$$
\int \frac{1}{3\sqrt[3]{u}} e^u\, du
$$
05
Evaluate the new integral
We can pull the constant 1/3 out of the integral and simply integrate \(e^u\) with respect to \(u\):
$$
\frac{1}{3} \int \frac{1}{\sqrt[3]{u}} e^u\, du = \frac{1}{3} \int e^u\, du
$$
Now, integrate:
$$
\frac{1}{3} \int e^u\, du = \frac{1}{3} e^u + C
$$
Where \(C\) is the constant of integration.
06
Substitute back the original variable x
Now, we substitute back our original substitution \(u = (x-1)^3\) to find the resulting function of \(x\):
$$
\frac{1}{3} e^u + C = \frac{1}{3} e^{(x-1)^3} + C
$$
So, the final result is:
$$
\int (x-1)^2 e^{(x-1)^3}\, dx = \frac{1}{3} e^{(x-1)^3} + C
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
U-Substitution Method
The u-substitution method is a technique used to evaluate integrals, and it works similarly to finding the inverse of a function. The key idea is to take a complex integral and simplify it by substituting a part of the integral with a new variable, typically denoted as 'u'. This makes the integral easier to work with, especially when dealing with composite functions.
For the integral \(\int(x-1)^{2} e^{(x-1)^{3}} dx\), the substitution \(u = (x-1)^{3}\) is chosen to simplify the integral. To implement u-substitution, you calculate the derivative of u with respect to x, denoted \(du/dx\), and solve for \(dx\). This allows you to rewrite the entire integral in terms of u. The ultimate goal is to convert the given integral into a simpler form, where known integration techniques can be readily applied.
For the integral \(\int(x-1)^{2} e^{(x-1)^{3}} dx\), the substitution \(u = (x-1)^{3}\) is chosen to simplify the integral. To implement u-substitution, you calculate the derivative of u with respect to x, denoted \(du/dx\), and solve for \(dx\). This allows you to rewrite the entire integral in terms of u. The ultimate goal is to convert the given integral into a simpler form, where known integration techniques can be readily applied.
Definite Integrals
Definite integrals are used to calculate the exact area under a curve bounded by two points on the x-axis. The integration process for definite integrals involves finding the antiderivative of a function and evaluating it at the upper and lower limits. It鈥檚 important to know that even though our original exercise doesn't showcase a definite integral, the method of u-substitution remains an integral component in their evaluation.
If we were to turn our integral into a definite one by adding lower and upper limits, say \(a\) and \(b\), after finding the antiderivative using the u-substitution method, we would then need to evaluate \(e^{u}\) at points \(u(a)\) and \(u(b)\). The u-substitution technique simplifies the function to make finding the antiderivative and subsequently, the final value of the definite integral, more straightforward.
If we were to turn our integral into a definite one by adding lower and upper limits, say \(a\) and \(b\), after finding the antiderivative using the u-substitution method, we would then need to evaluate \(e^{u}\) at points \(u(a)\) and \(u(b)\). The u-substitution technique simplifies the function to make finding the antiderivative and subsequently, the final value of the definite integral, more straightforward.
Exponential Functions
Exponential functions, such as \(e^x\), are functions where the variable x is an exponent. These functions have unique properties, including constant rate of growth or decay, which makes them appear frequently in real-world applications, such as in calculations of compound interest or radioactive decay. Integral calculus involving exponential functions often requires a detailed understanding of their behavior since their integration can be less intuitive than polynomial functions.
In our worked example, the expression \(e^{(x-1)^3}\) is an exponential function. Integrating exponential functions is relatively straightforward鈥攖he integral of \(e^x\) is \(e^x+C\), where \(C\) is the constant of integration. The u-substitution method is particularly useful in this context because it converts complex exponential functions into a simpler form that can be integrated directly to reveal the antiderivative.
In our worked example, the expression \(e^{(x-1)^3}\) is an exponential function. Integrating exponential functions is relatively straightforward鈥攖he integral of \(e^x\) is \(e^x+C\), where \(C\) is the constant of integration. The u-substitution method is particularly useful in this context because it converts complex exponential functions into a simpler form that can be integrated directly to reveal the antiderivative.
