Problem 11 - Thomas Calculus

Question

Evaluate the integrals. \(\int x^{3} e^{x} d x\)

Step-by-Step Solution

Answer

The integral is \( x^3 e^x - 3x^2 e^x + 6xe^x - 6e^x + C \).

Step 1: Identify Technique

To evaluate the integral \( \int x^{3} e^{x} dx \), we need to use integration by parts. Since this is a product of a polynomial and an exponential function, integration by parts is appropriate. Integration by parts is given by the formula: \( \int u \, dv = uv - \int v \, du \).

Step 2: Choose Functions

For integration by parts, choose \( u = x^3 \) because differentiating reduces the power, and \( dv = e^{x} dx \) because the integral of \( e^{x} \) is straightforward. Thus, \( du = 3x^2 dx \) and \( v = e^x \).

Step 3: Apply Integration by Parts

Substitute into the integration by parts formula: \( \int x^{3} e^{x} dx = x^{3} e^{x} - \int 3x^2 e^{x} dx \). This reduces the integral by one degree of \( x \).

Step 4: Repeat Process

Now, apply integration by parts again to \( \int 3x^2 e^{x} dx \). Choose \( u = x^2 \) and \( dv = 3e^{x} dx \). This gives \( du = 2x dx \) and \( v = 3e^{x} \). Substitute into the formula: \( \int 3x^2 e^{x} dx = 3x^2 e^{x} - \int 6xe^{x} dx \).

Step 5: Apply Integration by Parts Again

For \( \int 6xe^{x} dx \), choose \( u = x \) and \( dv = 6e^{x} dx \). Then \( du = dx \) and \( v = 6e^{x} \). Substitute into the formula: \( \int 6xe^{x} dx = 6xe^{x} - \int 6e^{x} dx \).

Step 6: Integrate Final Term

The integral \( \int 6e^{x} dx = 6e^{x} + C \), where \( C \) is the constant of integration.

Step 7: Combine Results

Combine all parts together: \( \int x^3 e^x dx = x^3 e^x - [3(x^2 e^x - (6xe^x - 6e^x))] \). Simplify to get \( x^3 e^x - 3x^2 e^x + 6xe^x - 6e^x + C \).

Key Concepts

Integral Calculus

Integral calculus provides the tools to calculate areas under curves and find antiderivatives, which are essential for understanding accumulated quantities over a continuous interval. Its main idea is to take a derivative and find the original function.

When dealing with complex functions, such as products of polynomial and exponential functions, integration by parts becomes a key technique. Integration by parts is essentially the reverse of the product rule for differentiation.

• The formula is: \( \int u\, dv = uv - \int v\, du \), where you need to identify parts of the integrand as \( u \) and \( dv \).
• Choosing \( u \) and \( dv \) wisely simplifies the integration process. The goal is to make \( du \) simpler and the resulting integral easier to solve.

Integration by parts often needs to be applied multiple times, as seen in integrating \( \int x^3 e^x dx \), gradually reducing the polynomial until the integral is resolved.

Polynomial Functions

Polynomial functions are mathematical expressions involving terms with variables raised to whole number exponents. They are foundational in algebra and calculus due to their wide range of applications and properties.

In the integration of \( \int x^3 e^x dx \), the polynomial \( x^3 \) plays a significant role:

• Choosing a function like \( u = x^3 \) helps simplify the integral with each application of integration by parts, as differentiating reduces the power of the variable.
• Since polynomials are easy to differentiate, they are typically chosen as \( u \) in the integration by parts process.

Remembering basic polynomial differentiation rules is crucial, as it helps in finding \( du \). For instance, the derivative of \( x^3 \) is straightforwardly \( 3x^2 \). Handling polynomials effectively in calculus problems often involves breaking down into smaller, more manageable pieces.

Exponential Functions

Exponential functions always have the form \( e^x \) or similar bases, where the variable is in the exponent. They are unique because their rate of growth is proportional to their current value, making them crucial in modeling real-world phenomena like population growth or radioactive decay.

When evaluating \( \int x^3 e^x dx \), the exponential component is \( e^x \):

• The integration of \( e^x \) remains \( e^x \), which simplifies the process. This property makes \( e^x \) an ideal choice for \( dv \) in integration by parts.
• Exponential functions retain their form upon differentiation and integration, lending ease to calculations and steps in integration-by-parts, reducing possible errors.

Understanding how exponential functions integrate and differentiate is key to tackling calculus problems, as they frequently appear in both theoretical and applied mathematics.