Problem 75 - Applied Calculus

Question

Sometimes an integral requires two or more integrations by parts. As an example, we apply integration by parts to the integral \(\int x^{2} e^{x} d x\). \(\int \underbrace{x^{2} e^{x} d x}=x^{2} e^{x}-\int_{u} \underbrace{e^{x} 2 x d x}=x^{2} e^{x}-2 \int x e^{x} d x\) \(\left[\begin{array}{cc}u=x^{2} & d v=e^{x} d x \\ d u=2 x d x & v=\int e^{x} d x=e^{x}\end{array}\right]\) The new integral \(\int x e^{x} d x\) is solved by a second integration by parts. Continuing with the previous solution, we choose new \(u\) and \(d u\) : \(=x^{2} e^{x}-2\left(\int x e^{x} d x\right) \quad\left[\begin{array}{c}u=x \quad d v=e^{x} d x \\ d u=d x \quad v=e^{x}\end{array}\right]\) \(=x^{2} e^{x}-2\left(x e^{x}-\int e^{x} d x\right)\) \(=x^{2} e^{x}-2\left(x e^{x}-e^{x}\right)+C\) \(=x^{2} e^{x}-2 x e^{x}+2 e^{x}+C\) After reading the preceding explanation, find each integral by repeated integration by parts. \(\int x^{2} e^{-x} d x\)

Step-by-Step Solution

Answer

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

Step 1: Choose First u and dv

Select functions for integration by parts, let \( u = x^2 \) and \( dv = e^{-x} dx \). Then, differentiate \( u \) to obtain \( du = 2x \, dx \) and integrate \( dv \) to get \( v = \int e^{-x} dx = -e^{-x} \). We will use these in the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \].

Step 2: Apply Integration by Parts

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula:\[ \int x^2 e^{-x} dx = -x^2 e^{-x} - \int -e^{-x} (2x) dx \]. This simplifies to:\[ -x^2 e^{-x} + 2 \int x e^{-x} dx \] .We now focus on the remaining integral \( \int x e^{-x} dx \).

Step 3: Choose Second u and dv

For the integral \( \int x e^{-x} dx \), choose \( u = x \) and \( dv = e^{-x} dx \). Then, differentiate \( u \) to get \( du = dx \) and integrate \( dv \) to obtain \( v = -e^{-x} \).

Step 4: Apply Second Integration by Parts

Substitute the newly chosen \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula again:\[ \int x e^{-x} dx = -x e^{-x} - \int -e^{-x} dx \]. This simplifies to:\[ -x e^{-x} + \int e^{-x}dx = -x e^{-x} - e^{-x} \].

Step 5: Combine Results

Substitute back the result from Step 4 into the expression obtained in Step 2:\[ -x^2 e^{-x} + 2(-x e^{-x} - e^{-x}) = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} \].Add the constant of integration, \( C \).

Step 6: Write Final Expression

The completed integral, which incorporates all steps of integration by parts, is given by:\[ \int x^2 e^{-x} dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + C \].

Key Concepts

Definite Integration

Definite integration involves calculating the integral of a function between two specified limits, thereby providing a numerical value. Unlike its indefinite counterpart, which represents a family of functions, definite integration calculates the exact area under a curve, between given points on the x-axis.
To solve a definite integral, follow these steps:

• Evaluate the indefinite integral of the function to find the antiderivative.
• Calculate the value of the antiderivative at the upper limit of the integral.
• Calculate the value of the antiderivative at the lower limit of the integral.
• Subtract the value at the lower limit from the value at the upper limit.

This process corresponds to the Fundamental Theorem of Calculus, which bridges the concept of differentiation and integration. If you're working with integration by parts, as seen in the example, you might first need to find the indefinite integral before applying the limits.

Indefinite Integration

Indefinite integration is the process of finding a function whose derivative is the given function. This leads to the concept of antiderivatives, which introduce a constant of integration, denoted by `C`, due to the indefinite nature of the calculation.

The notation for indefinite integration is simply to omit the limits, as seen with:

• \( \int f(x) \, dx \)
• The integration produces a function plus a constant: \( F(x) + C \)

This indefinite integral denotes a collection of functions differing by only a constant. When integrating by parts, the indefinite integration is crucially applied to break down more complex expressions into simpler ones.

Calculus Techniques

Calculus has various techniques to solve different kinds of integrals, with integration by parts being a key method. This technique is particularly useful for integrals where the product of two functions is present. The integration by parts formula is derived from the product rule of differentiation and is given by:

• \( \int u \, dv = uv - \int v \, du \)

In practice, you:

• Select appropriate functions for \( u \) and \( dv \) that make \( \int v \, du \) easier to evaluate.
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \).
• Substitute these into the formula to simplify the integral.
• If necessary, perform integration by parts more than once to solve complex integrals, as shown in the given example.

By mastering this and other calculus techniques, such as substitution, students can tackle a wide range of integration problems effectively.