Question

Find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) $$ \int x e^{-2 x} d x $$

Short answer

-\frac{1}{2}x e^{-2x} -\frac{1}{4}e^{-2x} + C

Step-by-step solution

Select u and dv

According to rule 'LIATE', Algebra (here \( x \)) should be chosen as \( u \), because it is of higher priority than Exponent (here \( e^{-2x} \)). Thus, \( u = x \), and the remaining part \( e^{-2x} dx \) becomes \( dv \).

Find du and v

Differentiate \( u \) to get \( du \), and integrate \( dv \) to get \( v \). That is, \( du = dx \) and \( v = \int e^{-2x} dx = -\frac{1}{2}e^{-2x} \). Note that the integration of \( e^{-2x} dx \) requires using the formula for integration of \( e^{kx} dx \) which gives \( \frac{1}{k} e^{kx} \). Here, \( k = -2 \).

Apply the integration by parts formula

The formula for integration by parts is given by: \( \int udv = uv - \int vdu \). On substitution, we obtain: \( \int x e^{-2 x} dx = uv - \int vdu = x(-\frac{1}{2}e^{-2x}) - \int (-\frac{1}{2}e^{-2x}) dx = -\frac{1}{2}x e^{-2x} - \int (-\frac{1}{2}e^{-2x}) dx\).

Evaluate the integral remaining

The integral left is now smaller and simpler than the original one, and we can solve it by basic integral properties. \(\int (-\frac{1}{2}e^{-2x}) dx = -\frac{1}{4}e^{-2x} + C\)where \(C\) is the constant of integration.

Combine all parts

Combine all values from the steps above into a final answer: \( -\frac{1}{2}x e^{-2x} -\frac{1}{4}e^{-2x} + C \).

Key concepts

Integration by Parts

Integration by parts is a technique used to solve integrals that typically cannot be computed through direct methods. It's based on the product rule for differentiation and is applied when an integral contains a product of two functions, each of which is a part of a different class of functions. There's a mnemonic, 'LIATE', which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions, used to decide which function to choose as 'u' (the one that will be differentiated) and which as 'dv' (the one that will be integrated).

The formula for integration by parts is given by:
\[ \int u\, dv = uv - \int v\, du \]
where 'u' is the function to be differentiated, 'dv' is the function to be integrated, 'v' is the integral of 'dv', and 'du' is the derivative of 'u'. In the given exercise, the integration by parts is applied to the integral of \( x e^{-2 x} \), with 'x' chosen for differentiation as it simplifies upon taking the derivative.

Exponential Function Integration

Integrating exponential functions is a common task in calculus. An exponential function has the form \( e^{kx} \), where 'k' is a constant. When integrating such functions, one typically utilizes the integral formula:\[ \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C \]
where 'C' represents the constant of integration. In the context of our exercise, we have \( e^{-2x} \), which, upon integration, yields \( -\frac{1}{2}e^{-2x} \), following the aforementioned formula with \( k = -2 \). This principle simplifies the integration process when dealing with exponential terms by providing a straightforward method to find the antiderivative.

Constant of Integration

When calculating an indefinite integral, we include a 'constant of integration', denoted as 'C'. This is fundamental because the process of differentiation erases any constant term since the derivative of a constant is zero. Thus, when we integrate, we must account for all possible constants that could have been present.

In essence, \( C \) encapsulates all possible constant values that the original function might have had before differentiation, leading to an entire family of functions that are the antiderivatives of the given function. For example, in the step-by-step solution, the final integration step includes this constant of integration, resulting in \( -\frac{1}{4}e^{-2x} + C \), signifying all possible antiderivatives of the exponential function.