Problem 80 - 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. \(\int x^{3} e^{x} d x\)

Step-by-Step Solution

Answer

\( \int x^3 e^{x} \, dx = x^3 e^{x} - 3x^2 e^{x} + 6x e^{x} - 6e^{x} + C \).

Step 1: Setup the first integration by parts

For the integration by parts formula \( \int u \, dv = uv - \int v \, du \), choose \( u = x^3 \) and \( dv = e^{x} \, dx \). Then, differentiate \( u \) and integrate \( dv \):\[ du = 3x^2 \, dx, \quad v = e^{x} \].

Step 2: Apply integration by parts formula

Substitute back into the integration by parts formula:\[ \int x^3 e^{x} \, dx = x^3 e^{x} - \int e^{x} (3x^2) \, dx \]. This simplifies to:\[ x^3 e^{x} - 3 \int x^2 e^{x} \, dx \].

Step 3: Setup second integration by parts

For the integral \( \int x^2 e^{x} \, dx \), choose \( u = x^2 \) and \( dv = e^{x} \, dx \). Differentiate \( u \) and integrate \( dv \):\[ du = 2x \, dx, \quad v = e^{x} \].

Step 4: Apply second integration by parts

Substitute these into the integration by parts formula:\[ \int x^2 e^{x} \, dx = x^2 e^{x} - \int e^{x} (2x) \, dx \]. This can be simplified to:\[ x^2 e^{x} - 2 \int x e^{x} \, dx \].

Step 5: Setup third integration by parts

For the integral \( \int x e^{x} \, dx \), choose \( u = x \) and \( dv = e^{x} \, dx \). Differentiate \( u \) and integrate \( dv \):\[ du = dx, \quad v = e^{x} \].

Step 6: Apply third integration by parts

Substitute these into the integration by parts formula:\[ \int x e^{x} \, dx = x e^{x} - \int e^{x} \, dx \]. This simplifies to:\[ x e^{x} - e^{x} \].

Step 7: Consolidate results

Substitute back to obtain the final expression. From Step 6, we have:\[ \int x e^{x} \, dx = x e^{x} - e^{x} \]. Substitute into Step 4 result:\[ \int x^2 e^{x} \, dx = x^2 e^{x} - 2(x e^{x} - e^{x}) = x^2 e^{x} - 2x e^{x} + 2e^{x} \]. Finally, substitute this result into Step 2:\[ \int x^3 e^{x} \, dx = x^3 e^{x} - 3(x^2 e^{x} - 2x e^{x} + 2e^{x}) \].

Step 8: Simplify the final expression

Simplify the last equation to reach the final result.\[ \int x^3 e^{x} \, dx = x^3 e^{x} - 3x^2 e^{x} + 6x e^{x} - 6e^{x} + C \], where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals are an important concept in calculus, allowing us to find the area under a curve over a specific interval. Unlike indefinite integrals, which include a constant of integration because they represent families of functions, definite integrals provide a precise numerical value. This value can be thought of as the net area between the function and the x-axis, within the limits of integration.
To calculate a definite integral, we use the Fundamental Theorem of Calculus. This theorem links differentiation and integration, enabling us to find the exact area under a curve by evaluating the antiderivative of the function at the upper and lower boundaries of the interval. While this exercise primarily uses integration by parts, the concept of definite integrals helps us understand why these techniques are useful for areas and totals.
When dealing with real-world problems, definite integrals can be applied to determine quantities such as:

• Area under a curve.
• Total distance traveled.
• Accumulated quantities, like mass or charge.

Thus, mastering the techniques to solve these integrals, such as integration by parts, becomes essential for precise problem-solving.

Integration Techniques

Integration techniques are tools that help us solve integrals that are not readily solvable by basic methods. The method called "integration by parts" is particularly useful for integrals involving the product of functions, such as polynomials multiplied by exponential or trigonometric functions. This exercise showcases how integration by parts can be applied multiple times to achieve a solution.
The integration by parts formula is derived from the product rule for differentiation and is expressed as:

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

where \( u \) and \( dv \) are parts of the original integral.
In practice, we choose \( u \) and \( dv \) wisely to simplify the problem as much as possible with each step. This often requires some trial and error or insight into function behaviors. Because \( dv \) will become \( v \) through integration, it's advantageous to have a \( dv \) that integrates easily. Similarly, \( u \) should be a function that simplifies when differentiated.
Other integration techniques include substitution, partial fraction decomposition, and trigonometric identities, each useful for different kinds of integrals. Knowing when and how to apply each technique is a major skill in calculus.

Calculus Problem-Solving

Calculus problem-solving involves not only mathematical computation but also strategy and planning. When you encounter a complex integral or problem, like this exercise on repeatedly applying integration by parts, it's important to break down the problem into manageable parts.
To effectively solve calculus problems:

• Start by understanding the problem statement, identifying what's given and what needs to be found.
• Choose appropriate methods or techniques based on the function types involved.
• Consider alternative solutions for verification and a deeper understanding.

In this particular problem, repetition of the integration by parts method was necessary because of the polynomial component \( x^3 \). Each step simplified the function further until reaching a basic integral that was easy to solve.
Practicing this type of strategic thinking can make tackling calculus challenges not only easier but also faster. Furthermore, this problem-solving approach is not limited to calculus but is applicable across mathematical fields and real-life problems where logical and critical thinking is required.