Problem 42 - Calculus for Biology and Medicine

Question

First make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals. $$ \int x^{5} e^{x^{2}} d x $$

Step-by-Step Solution

Answer

\( \int x^{5} e^{x^{2}} \, dx = \frac{x^4}{2} e^{x^2} - x^2 e^{x^2} + e^{x^2} + C \)

Step 1: Choose the Substitution

Observe the integral \( \int x^{5} e^{x^{2}} \, dx \). To begin with a substitution, let's set \( u = x^2 \), which implies \( du = 2x \, dx \). Thus, \( x^4 e^{x^2} \) needs to be adjusted to this substitution by breaking \( x^5 \) as \( x^4 \cdot x \). This gives us \( \int x^4 \cdot x e^{x^2} \, dx = \frac{1}{2} \int x^4 e^u \, du \).

Step 2: Simplify with Substitution

Substitute \( x^4 \) as \( (u^2)^{2} \) to continue converting the entire integral in terms of \( u \). Taking the substitution from Step 1 fully transforms the integral to \( \int \frac{(u)^2}{2} e^u \, du \).

Step 3: Integration by Parts Setup

Use integration by parts, \( \int u \, dv = uv - \int v \, du \). Assign \( \frac{u^2}{2} \) as \( u \) and \( e^u \, du \) as \( dv \). This makes \( du = u \, du \) and \( v = e^u \). Apply the formula accordingly.

Step 4: Perform Integration by Parts

Evaluating using the parts setup: \( uv = \frac{u^2}{2} e^u \). The integral becomes: \( \int \frac{u^2}{2} e^u \, du = \frac{u^2}{2} e^u - \int u e^u \, du \).

Step 5: Integrate Remaining Part

For the remaining integral \( \int u e^u \, du \), apply integration by parts again, letting \( u = u \) and \( dv = e^u \, du \), which results in \( v = e^u \) and \( du = du \). Thus, \( \int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u + C \).

Step 6: Compile the Complete Solution

Substitute back the \( u \) terms through the integration by parts steps gives: \[ \int x^{5} e^{x^{2}} \, dx = \frac{x^4}{2} e^{x^2} - \left( x^2 e^{x^2} - e^{x^2} \right) + C \] which simplifies to \( \frac{x^4}{2} e^{x^2} - x^2 e^{x^2} + e^{x^2} + C \).

Key Concepts

Substitution Method

The substitution method is a powerful technique for simplifying integrals. The aim is to change the integrand into a form that is easier to evaluate. In this exercise, we began by observing the integral \( \int x^{5} e^{x^{2}} \, dx \). We noticed that the expression \((x^2)\) appears in the exponent. This hints at a natural substitution, \( u = x^2 \). This simplifies our equation because the differential \( du \) becomes \( 2x \, dx \), which is closely related to the \( x^5 \) term we originally had.
To make it fit the substitution perfectly, we decompose \( x^5 \) as \( x^4 \cdot x \). This lets us express the integral as \( \int x^4 \cdot x e^{x^2} \, dx = \int \frac{1}{2} u^2 e^u \, du \).
Employing substitution like this transforms the integral into a simpler problem, allowing us to solve through other methods such as integration by parts.

Indefinite Integrals

Indefinite integrals represent the family of all antiderivatives of a function. They're expressed without boundaries, thus including a constant \( C \) to represent any constant difference between particular solutions.
In the exercise, the final result \( \int x^{5} e^{x^{2}} \, dx = \frac{x^4}{2} e^{x^2} - x^2 e^{x^2} + e^{x^2} + C \) showcases how we find an antiderivative of the original function. Integration by parts, in combination with substitution, revealed the function structure, which wasn't initially evident.

• To understand indefinite integrals, always remember: the derivative of the integral should bring you back to the original function.
• Your results will contain a constant \( C \), reflecting the indefinite nature of the solution.

Integrals like these build the foundation for deeper concepts in calculus, especially in solving differential equations.

Exponential Functions

Exponential functions are defined by their bases raised to a power, typically seen in expressions like \( e^x \). In the context of calculus, these functions have unique properties, such as their derivatives being proportional to the function itself.
In the integral \( \int x^{5} e^{x^{2}} \, dx \), the exponential part \( e^{x^2} \) added complexity to the integration process. When transformed through substitution, \( e^{x^2} \) becomes \( e^u \), which is much easier to handle.

• Exponential growth or decay models in real-life phenomena often rely on \( e \), the base of natural logarithms, due to its mathematical properties.
• The derivative and integral of \( e^u \) are quite straightforward, remaining \( e^u \) up to a multiplicative constant.

In relation to the problem, understanding exponential functions helped us apply methods like integration by parts effectively, allowing us to unravel the integral through logical steps.