Question

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

Short answer

The indefinite integral of the given function is \( \int \frac{x}{e^{x}} dx = -x e^{-x} - e^{-x} + C \).

Step-by-step solution

Apply the Integration by Parts Formula

In order to solve the integral \( \int \frac{x}{e^{x}} dx \), it is suitable to use the formula for integration by parts: \( \int u dv = uv - \int v du \). The choice of u and dv is crucial in integration by parts. Here, we will set \( u = x \) and \( dv = e^{-x} dx \). Now, let's find du and v by differentiating and integrating \( u \) and \( dv \) respectively. We will have \( du = dx \) and \( v = -e^{-x} \).

Substitute into Integration by Parts Formula

Substitute \( u, dv, v, du \) into the formula: \( \int u dv = uv - \int v du \). After substituting, we get \( \int x e^{-x} dx = -x e^{-x} - \int (-e^{-x}) dx \).

Simplify the Expression

The right hand side simplifies to \( -x e^{-x} + \int e^{-x} dx \). This simplifies further to \( -x e^{-x} -e^{-x} \). Adding the constant of integration C, the final answer becomes \( -x e^{-x} -e^{-x} + C \)

Key concepts

Integration by Parts

When approaching indefinite integrals, one powerful tool at our disposal is the technique known as integration by parts. It's a method derived from the product rule for differentiation and is generally used when the integral involves a product of two functions that are not easily integrated independently. The idea is to transform the integral of a product into a (hopefully) simpler integral. The basic formula for integration by parts is \[ \text{If }\text{u} = f(x)\text{ and dv} = g(x) dx, \text{ then }\int u dv = uv - \int v du, \] where \( u \) is a function of \( x \), and \( dv \) is another function times \( dx \). After selecting \( u \) and \( dv \), we differentiate \( u \) to get \( du \) and integrate \( dv \) to get \( v \).

To enhance your understanding, it's important to choose \( u \) and \( dv \) wisely, considering the function that will simplify after the differentiation. Remember, the goal is to make the resulting integral easier to evaluate. In our exercise, we chose \( u = x \) and \( dv = e^{-x} dx \), which through the integration by parts formula, gave us a more manageable integral to work with.

Integrating Exponential Functions

Working with exponential functions can often prove to be a bit tricky in calculus. When integrating exponential functions, it is crucial to recognize the general form of an exponential function, which can be described as \( a^x \), where \( a \) is a constant. More formally, integrating an exponential function involves finding the indefinite integral \( \int e^{kx} dx \), where \( k \) is a constant. The result of such an integral is \( \frac{1}{k}e^{kx} + C \), with \( C \) being the constant of integration.

In the given exercise, we are dealing with the function \( e^{-x} \), which is an exponential function. Here, the constant \( k \) is \( -1 \). The integration of \( e^{-x} \) by itself is straightforward - you would simply apply the general rule and obtain \( -e^{-x} + C \). This step makes the solution easier to reach and illustrates the importance of recognizing and efficiently integrating exponential terms independently.

Calculus Methods

To succeed in calculus, one needs an arsenal of calculus methods for solving different types of problems, such as derivatives, limits, and integrals. Beyond the standard substitution and integration by parts, methods such as partial fraction decomposition or trigonometric substitution are also commonly used. It's vital for students to understand when to apply each method and how to execute them effectively.

In contexts like our exercise, judging whether integration by parts is necessary is part of the skill set needed. For example, had the function been simply an exponential function without the \( x \) multiplier, a simple integration would suffice. However, calculus methods are not one-size-fits-all and require careful consideration of the function form. Enhancing problem-solving skills comes from practicing various methods and understanding the underlying principles of calculus, such as continuity, the behavior of functions, and the graphical interpretation of derivatives and integrals. Over time, identifying the appropriate method becomes more intuitive and less of a hurdle, paving the way for successful problem solving in calculus.