Problem 9 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int x^{2} e^{x} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \((x^2 - 2x + 2)e^x + C\).

Step 1: Identify Parts for Integration by Parts Formula

The integration by parts formula is given by \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) such that integration becomes simpler. Choose \( u = x^2 \) and \( dv = e^x \, dx \).

Step 2: Differentiate and Integrate

Compute \( du \) by differentiating \( u \): \( du = 2x \, dx \). Compute \( v \) by integrating \( dv \): \( v = e^x \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \( \int x^2 e^x \, dx = x^2 e^x - \int e^x 2x \, dx \).

Step 4: Simplify and Apply Integration by Parts Again

Simplify the expression: \( x^2 e^x - 2 \int x e^x \, dx \). For \( \int x e^x \, dx \), apply integration by parts again choosing \( u = x \) and \( dv = e^x \, dx \).

Step 5: Repeat Differentiation and Integration

Differentiate \( u = x \) to get \( du = dx \), and integrate \( dv = e^x \, dx \) to get \( v = e^x \).

Step 6: Apply Integration by Parts Again

Substitute into the formula: \( \int x e^x \, dx = x e^x - \int e^x \, dx \).

Step 7: Complete the Simplification

Integrate \( \int e^x \, dx \) to get \( e^x \). Substitute back into the previous expression: \( \int x e^x \, dx = x e^x - e^x \).

Step 8: Substitute Back for Full Solution

Use \( \int x e^x \, dx = x e^x - e^x \) to substitute back: \( \int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) = x^2 e^x - 2x e^x + 2e^x \).

Step 9: Final Answer with Simplification

After simplifying, the integral becomes: \( (x^2 - 2x + 2)e^x + C \), where \( C \) is the constant of integration.

Key Concepts

Integration Techniques

Integration by parts is one of several methods used to solve integrals that can't be easily integrated at a glance. The core idea is to transform the integral of a product of functions into a form that is simpler to solve. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Here:

• \( u \) is the part of the integrand that, when differentiated, simplifies the integral.
• \( dv \) is the part of the integrand that, when integrated, complements this simplification.

Choosing \( u \) and \( dv \) carefully affects the ease of solving the integral. The simplification often requires repeating the method more than once, especially for expressions involving higher powers or complex functions like exponentials. Integrating by parts is often combined with other strategies when solving various types of integrals. Methods like substitution, partial fraction decomposition, or trigonometric substitution also play significant roles in the comprehensive toolkit for solving integrals.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is key when dealing with calculus problems. An **indefinite integral** is essentially the antiderivative of a function. It represents a family of functions and is expressed with a constant of integration \( C \), because there isn't a particular endpoint involved. For instance, When evaluating \( \int x^2 e^x \, dx \), we obtain the indefinite integral \((x^2 - 2x + 2)e^x + C\), indicating there could be infinitely many functions depending on the constant \( C \). In contrast, a **definite integral** evaluates the area under a curve between two specific bounds. This concept gives a precise numerical value rather than an expression. Definite integrals do not include a constant of integration because you're calculating a specifics area, not all possible areas. The process to solve each starts similarly by finding the antiderivative, but differs in that definite integrals require evaluating the result at the bounds of integration.

Differentiation

Differentiation is the mathematical process of finding the derivative, or rate of change, of a function. In the integration by parts method, differentiation helps simplify the integral. It involves:

• Finding \( du \) from the function \( u \) by taking its derivative.
• The choice of \( u \) is critical, as it affects \( du \) and can drastically change the complexity of the solution.

For example, when integrating \( x^2 e^x \), \( u = x^2 \) because its derivative, \( du = 2x \, dx \), is simpler than the original expression, enabling us to reduce the integral's complexity. Differentiation not only simplifies parts of an integral but also lays the foundation for integral calculus. It is the reverse process of integration, serving as a reminder that calculus is a study of how functions change and accumulate values over intervals.

Exponential Functions

Exponential functions are prevalent in calculus and have the distinct characteristic of a constant rate of growth or decay. This unique property, expressed as \( e^x \), means that the function remains fundamentally unchanged by differentiation or integration. In the context of integration by parts, the exponential function often appears in the expression for \( dv \) because:

• Its derivative and integral are both \( e^x \), simplifying the steps involved in solving the integral.
• This quality helps minimize the complexity of problems involving products of polynomials and exponentials, like \( x^2 e^x \).

Calculating the integral of products that include exponentials often strategically uses integration by parts to exploit the unchanged nature of \( e^x \) as it cycles through differentiation and integration. Such functions underpin many real-world phenomena like population growth, radioactive decay, and continuously compounded interest, emphasizing the importance of understanding them thoroughly in calculus.