Question

Find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) $$ \int(x-1) e^{x} d x $$

Short answer

The indefinite integral of \(∫(x-1) e^{x} dx = xe^{x} - 2e^{x} + C\)

Step-by-step solution

Write down the formula for the integral of a product of two functions

The integral of a product of two functions \(u\) and \(v\) is given by \(u ∫v dx - ∫u' ∫v dx dx \). In this case, \(u=x-1\) and \(v=e^{x}\)

Identify u and v and their derivatives and integrals

Here, \(u=(x-1)\), so \(du=dx\). \(v=e^{x}\), so \(\int v dx = e^{x}\). Now we are equipped with all the parts to apply the formula from Step 1

Apply the formula

The formula for the integral of a product of two functions is substituted with identified functions and their integrals and derivatives. Therefore, we get \(∫(x-1) e^{x} dx = (x-1)\int e^{x} dx - ∫dx \int e^{x} dx dx = (x-1) e^{x} - ∫e^{x} dx = (x-1) e^{x} - e^{x} + C \) where C is the constant of integration

Simplify the result

The result from Step 3 can be rewritten in a simplified elegant manner by taking out a common factor. The final answer will be \(xe^{x} - 2e^{x} + C \)

Key concepts

Integration by Parts

Integration by parts is a powerful technique used to compute the integral of the product of two functions when the integral cannot be easily found by an elementary formula. It is based on the rule for differentiation of a product. The formula for integration by parts is \[ \text{If } u = f(x) \text{ and } v = g(x), \text{then } \text{Integrate}[u dv] = uv - \text{Integrate}[v du]. \]
This approach requires us to choose functions 'u' and 'dv' such that 'dv' is easy to integrate and the derivative 'du' simplifies the integral. In our exercise, the student correctly noted that integration by parts was not necessary for the given function \((x-1)e^x\), indicating a proficient understanding of when to apply this method.

It is essential for students to practice discerning when integration by parts is the optimal path versus when simpler methods are available. This insight can save considerable time and lead to fewer errors in calculations.

Integral of Exponential Functions

The integral of exponential functions is straightforward due to the unique property of the exponential function. Exponential functions have the form \(e^{kx}\), where 'e' is the base of the natural logarithm, and 'k' is a constant. When we find the integral \( \text{Integrate}[e^{kx} dx], \)
the antiderivative is simply \(\frac{1}{k}e^{kx}\), when 'k' is not zero. For our specific case where \(k = 1\), the antiderivative of \(e^x\) is itself \(e^x\), making the computation of the integral much more manageable.

Understanding the behavior of exponential functions and their antiderivatives is crucial for students tackling integrals involving exponential terms. Comprehending this concept allows a quick and error-free application of the integral to a variety of related problems.

U-Substitution

U-substitution, often referred to as the reverse chain rule, is a method used to simplify certain integrals. The technique involves substituting a part of the integral with a new variable 'u', which turns a complex expression into a simpler form. The substitution is chosen such that \(du\) corresponds to another part of the integral, making the integral easily solvable.

In the case of our exercise, u-substitution is not required, but understanding when and how to apply u-substitution is a valuable skill. It gives students the ability to break down complex integrals into simpler parts, which ultimately makes integration a less daunting task. With practice, recognizing patterns that lend themselves well to u-substitution becomes second nature.

Constant of Integration

The constant of integration, represented by 'C', is an important part of integrating functions. It represents the indefinite nature of antiderivatives — for any constant value of 'C', the derivative of the function would be the same. Hence, when we integrate a function, we must add 'C' to account for all potential antiderivatives.
In the provided exercise solution, when simplifying the integral of \((x-1)e^x\), the student correctly included the constant of integration after performing the integration operation. Remembering to include 'C' is crucial because it demonstrates comprehension that integration can produce a family of functions, all differing by a constant. It's a small but essential detail that signifies the completeness of an indefinite integral.