Question

Integration by parts to find the indefinite integral. $$ \int x e^{-x} d x $$

Short answer

The indefinite integral of \(x e^{-x} dx\) is \(-x e^{-x} - e^{-x} + C\).

Step-by-step solution

Identify \(u\) and \(dv\)

Assign \(u\) to \(x\) and \(dv\) to \(e^{-x} dx\). Now differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\). This results in \(du = dx\) and \(v = -e^{-x}\).

Apply the integration by parts formula

Now substitute \(u\), \(v\), and \(du\) into the integration by parts formula: \[\int udv = uv - \int v du\] This gives: \[ \int x e^{-x} dx = -x e^{-x} - \int -e^{-x} dx\] After simplifying, we get \[ \int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx\]

Final steps

The remaining integral is a simple integral with a known solution. So we can write: \[ \int x e^{-x} dx = -x e^{-x} - e^{-x} + C\] where \(C\) is the constant of integration. So the indefinite integral of \(x e^{-x}\) with respect to \(x\) is \(-x e^{-x} - e^{-x} + C\).

Key concepts

Indefinite Integral

When we talk about indefinite integrals, what we mean is the process of finding the antiderivative or original function from its derivative. It's like unwinding the clock to see what the face looked like before time passed. An antiderivative is always accompanied by a '+ C' at the end, which represents the constant of integration. This constant reflects the fact that there could be an infinite number of possible functions, each differing only by a constant value.

Now, engaging with indefinite integrals means engaging with a core aspect of calculus. Every function that has a derivative has an indefinite integral. For our example, \(\int xe^{-x}dx\), we're searching for a function, which, when derived, will give us the product of \(x\) and \(e^{-x}\). Focusing on the landscape of calculus, indefinite integrals are crucial for understanding area under curves, accumulation functions, and even physics applications like motion.

Exponential Functions

Exponential functions are fascinating creatures in the realm of mathematics. They represent processes that change rapidly — growth or decay — representing everything from population dynamics to radioactive decay in physics. An exponential function generally has the form \( f(x) = a e^{kx} \), where \( e\) is the base of the natural logarithm, an irrational and transcendental number approximately equal to 2.71828. The power of \( e \) determines the function's growth or decay, so \( e^{-x} \) in our example represents a decreasing pattern.

Understanding such functions is vital, not just in mathematics but also in real-world applications — think economics or biology. Their unique property of being proportional to their own derivative makes them extremely important in solving differential equations that model these constantly changing phenomena.

Integration Techniques

Diving into integration techniques is like adding a set of special power tools to your math toolbox. These techniques are designed to tackle the vast variety of integrals that can't be solved by simple, straightforward methods. Integration by parts is one such powerful method based on the product rule for derivatives. It provides a strategy to integrate the product of two functions, as we've seen in the above exercise with \( x\) and \( e^{-x}\).

The formula, \( \int u dv = uv - \int v du \), is an algebraic expression reflecting the intertwining of two functions. This technique is particularly useful when you face an integral that includes a polynomial multiplied by an exponential function, a logarithm, or a trigonometric function — as is the case in our practice problem. Mastery of this method opens doors to solving complex integrals that appear not only in homework problems but also in advanced fields such as engineering, physics, and statistics.