Problem 15 - Thomas Calculus

Question

Evaluate the integrals. \(\int x^{5} e^{x} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \[e^x(x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120) + C.\]

Step 1: Choose Integration Technique

The integral \(\int x^{5} e^{x} d x\) involves a polynomial \(x^5\) and an exponential function \(e^x\). For integrals involving products of polynomial and exponential functions, the integration technique of choice is integration by parts.

Step 2: Apply Integration by Parts Formula

The integration by parts formula states \[\int u \, dv = uv - \int v \, du.\]Let's set \(u = x^5\), which makes and \(dv = e^x \, dx\). We then differentiate and integrate to find:\(du = 5x^4 \, dx\) and \(v = e^x\).

Step 3: Substitute Values into Integration by Parts Formula

Substitute \(u\), \(v\), \(du\), and \(dv\) into the formula:\[\int x^5 e^x \, dx = x^5 e^x - \int e^x \cdot 5x^4 \, dx.\]This simplifies to:\[\int x^5 e^x \, dx = x^5 e^x - 5\int x^4 e^x \, dx.\]

Step 4: Repeat Integration by Parts

Now apply integration by parts to \(\int x^4 e^x \, dx\), setting \(u = x^4\) and \(dv = e^x \, dx\). Differentiate and integrate again:\(du = 4x^3 \, dx\) and \(v = e^x\). Substitute into the formula:\[\int x^4 e^x \, dx = x^4 e^x - \int e^x \cdot 4x^3 \, dx.\]Combining gives:\[\int x^5 e^x \, dx = x^5 e^x - 5(x^4 e^x - \int 4x^3 e^x \, dx).\]

Step 5: Continue Repeated Integration by Parts

Continue this process for \(\int x^3 e^x \, dx\), \(\int x^2 e^x \, dx\), \(\int x e^x \, dx\), and \(\int e^x \, dx\). Apply integration by parts repeatedly, reducing the polynomial degree each time, until terms become constants. The repetitive pattern reduces to a formula similar to above: \[\cdots - 4(3\cdots - 3(2\cdots - 2(x e^x - e^x))).\]

Step 6: Combine Results and Simplify

Combine and simplify all obtained terms and their integrals to yield the solution. The final integrated function should reintegrate to:\[ e^x(x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120) + C,\] where \(C\) is the constant of integration.

Key Concepts

Integration by Parts

Integration by Parts is a powerful method used to evaluate integrals where a product of functions is involved, such as a polynomial and an exponential function. This technique is particularly helpful when the integral seems too complex to solve by standard methods.

The formula for Integration by Parts is given as: \[\int u \, dv = uv - \int v \, du.\]This formula essentially transforms the original integral into a possibly simpler one by dissecting one part of the product (function) through differentiation and the other through integration.

Here's how it generally works:

• You select two components from the given integral: \(u\) (which you differentiate) and \(dv\) (which you integrate).
• The choices for \(u\) and \(dv\) are often guided by the "LIATE" rule: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential functions.
• Then, find \(du\) and \(v\) as needed, and substitute these into the Integration by Parts formula.

For example, in the integral \(\int x^5 e^x \, dx\), setting \(u = x^5\) allows you to reduce the polynomial degree with each iteration.

Polynomial and Exponential Integration

Polynomial and Exponential Integration often go hand in hand when dealing with integrals of the form \(\int x^n e^x \, dx\). The polynomial \(x^n\) represents the algebraic part, and \(e^x\) represents the exponential.

To integrate such functions, Integration by Parts plays a crucial role because:

• It helps reduce the degree of the polynomial step by step, making the integral simpler with each successive iteration.
• Repeated application can turn what seems to be a daunting task into a structured process, steadily working towards the solution.

For example, starting with the integral \(\int x^5 e^x \, dx\) and performing integration by parts, will break down into smaller integrals like \(\int x^4 e^x \, dx\), then \(\int x^3 e^x \, dx\), and so forth, until reaching \(\int e^x \, dx\), which is straightforward.

Definite and Indefinite Integrals

When working with integrals, it's crucial to distinguish between definite and indefinite integrals.

**Indefinite integrals** do not have specific bounds, and thus include a constant of integration denoted as \(C\). These are essential for finding general antiderivatives of functions (e.g., \(\int f(x) \, dx = F(x) + C\)). The exercise we're discussing is an indefinite integral due to the lack of specified limits.

**Definite integrals**, on the other hand, evaluate the net area under the curve of a function between two specified points \(a\) and \(b\). They provide a numeric value (e.g., \(\int_{a}^{b} f(x) \, dx\)).

When performing a solution involving indefinite integrals, always remember to add \(+ C\) at the end of your solution to account for the family of possible solutions derived from different constants of integration.

• This is vital for expressing all potential antiderivatives of a function.
• Always re-check bounds (if any) and remember that when moving from indefinite to definite, the constant \(C\) cancels itself out.

Understanding the difference is crucial for correctly solving problems and interpreting integral results. This is especially true for integrations which follow complex steps like the one outlined, ensuring it results in a valid family of solutions.