Question

Integration by parts to find the indefinite integral. $$ \int x^{2} e^{-x} d x $$

Short answer

The indefinite integral \(\int x^2 e^{-x} dx = -x^2e^{-x} - 2x e^{-x} -2e^{-x} + C\).

Step-by-step solution

Identifying u and dv

Firstly, we identify \(u\) and \(dv\). According to the \(ILIATE\) rule, Algebraic function comes before the Exponential. Thus, we let \(u = x^2\) and \(dv = e^{-x} dx\). Next, we find \(du\) and \(v\), where \(du = 2x dx\) and \(v = -e^{-x}\). By applying the formula, \(\int udv = uv - \int vdu\), we can substitute these values in.

First application of Integration by Parts

Substituting the above values into the equation gives us: \(-x^2e^{-x} - \int(-e^{-x})(2x dx) = -x^2e^{-x} + 2 \int xe^{-x} dx\). Now, observe that the integration on the right necessitates another iteration of integration by parts.

Second application of Integration by Parts

Repeat step 1 for our new integral \(\int xe^{-x} dx\). We let \(u = x\) and \(dv = e^{-x} dx\). Hence \(du = dx\) and \(v = -e^{-x}\). Substitute these values into the formula: \(-x e^{-x} -\int(-e^{-x}) dx = -x e^{-x} - e^{-x}\). Then substitute this result for the integral in our equation from Step 2.

Simplification

Substitute the result from step 3 back into equation from step 2: \( -x^2e^{-x} + 2(-x e^{-x} - e^{-x}) = -x^2e^{-x} - 2x e^{-x} -2e^{-x}\). This is the indefinite integral of \(x^2 e^{-x}\), we add \(C\) for the constant of integration.

Key concepts

Indefinite Integral

In calculus, an indefinite integral is a fundamental concept that represents the set of all antiderivatives of a function. Unlike definite integrals, which provide a numerical value, indefinite integrals include an arbitrary constant of integration, typically denoted by \( C \). This constant accounts for all possible antiderivatives of the function. When dealing with an indefinite integral, such as \( \int x^{2} e^{-x} \, dx \), the process involves finding a function whose derivative matches the integrand quite exactly.

To solve an indefinite integral, one often applies various techniques depending on the form of the function to be integrated. Here, the strategy of integration by parts becomes handy. The result, once fully integrated, represents a family of functions differing only by their \(C\), giving us a broad spectrum of solutions for differential problems.

Integration Techniques

Integration techniques involve various methods that aid in solving integrals that cannot be directly computed using elementary antiderivatives. One such technique is Integration by Parts, which follows the formula \( \int u \, dv = uv - \int v \, du \).

This method is particularly useful when the integrand is a product of functions that appear complicated or do not have straightforward antiderivatives. For example, to handle the indefinite integral \( \int x^{2} e^{-x} \, dx \), we use integration by parts twice due to its complex form. Initially, we choose our "\( u \)" and "\( dv \)" using the \( ILIATE \) rule, which helps decide the order: Inverse, Logarithmic, Algebraic, Trigonometric, Exponential functions. In this case, assigning \( u = x^2 \) and \( dv = e^{-x} \, dx \) follows this precedence.

Through careful selection and iterative application of the method, integration by parts decomposes the integral into more tractable pieces, enabling the meticulous calculation required to find the indefinite integral.

Calculus Integration

Calculus integration is the reverse process of differentiation, concerned with finding the original function given its derivative. This aspect of calculus is versatile, covering techniques like substitution, partial fractions, and notably, integration by parts. Each of these tackles different types of functions and combinations, expanding the toolkit for solving integrals.

Furthermore, calculus integration provides solutions not just in mere calculations but in understanding the area under curves, total accumulation of quantities, and solutions to differential equations. In the specific example of \( \int x^{2} e^{-x} \, dx \), integration by parts demonstrates how calculus integration exceeds beyond simple reversals of derivatives. It requires strategic thinking and repeated applications, highlighting the intricate dance between different functions within the realms of mathematical analysis.

Mastery of integration, including recognizing when and how to apply techniques like integration by parts, uncovers deeper insights into mathematical problems and offers precise methods to tackle real-world applications proficiently.