Problem 47 - Applied Calculus

Question

Find in two different ways and check that your answers agree. \(\int x(x-2)^{5} d x\) a. Use integration by parts. b. Use the substitution \(u=x-2\) (so \(x\) is replaced by \(u+2\) ) and then multiply out the integrand.

Step-by-Step Solution

Answer

The integral is \( \frac{x(x-2)^6}{6} - \frac{(x-2)^7}{42} + C \) using both methods.

Step 1: Integration by Parts - Choose Functions

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int x(x-2)^5 \, dx \), let \( u = x \) and \( dv = (x-2)^5 \, dx \).

Step 2: Differentiate and Integrate Components

Differentiate \( u \) to get \( du = dx \), and integrate \( dv \) to get \( v = \frac{(x-2)^6}{6} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u, du, v, dv \) into the integration by parts formula: \[ \int x(x-2)^5 \, dx = x \cdot \frac{(x-2)^6}{6} - \int \frac{(x-2)^6}{6} \, dx \].

Step 4: Simplify and Integrate the Remaining Integral

The integral becomes \[ \int x(x-2)^5 \, dx = \frac{x(x-2)^6}{6} - \frac{1}{6} \int (x-2)^6 \, dx \]. Now, solve \( \int (x-2)^6 \, dx \) as \( \frac{(x-2)^7}{7} \).

Step 5: Write Complete Solution for Integration by Parts

Combine everything: \[ \int x(x-2)^5 \, dx = \frac{x(x-2)^6}{6} - \frac{(x-2)^7}{42} + C \].

Step 6: Substitution - Define Variables

Let \( u = x - 2 \), then \( x = u + 2 \), and \( du = dx \). This changes the integral to \( \int (u+2)u^5 \, du \).

Step 7: Expand and Integrate the Substitution Expression

Expand the integrand: \( (u+2)u^5 = u^6 + 2u^5 \). Now integrate: \( \int (u^6 + 2u^5) \, du = \frac{u^7}{7} + \frac{2u^6}{6} + C \).

Step 8: Simplify and Equate Substitution Integration

This becomes \( \frac{u^7}{7} + \frac{u^6}{3} + C \), which simplifies to \( \frac{(x-2)^7}{7} + \frac{(x-2)^6}{3} + C \).

Step 9: Compare Results from Both Methods

Both methods lead to expressions for \( \int x(x-2)^5 \, dx \) that agree upon simplification and reordering terms, confirming the solution is consistent.

Key Concepts

Integration by Parts

When faced with an integral that is a product of functions, the integration by parts technique becomes quite handy. It can be a great alternative when standard integration techniques do not suffice. The core idea is derived from the product rule for differentiation, and it is mathematically expressed as:\[\int u \, dv = uv - \int v \, du\]Here's how you can apply it effectively:

• Identify the parts: Choose 'u' as a function that simplifies upon differentiation, and 'dv' that easily integrates.
• Differentiate 'u' to find 'du', and integrate 'dv' to find 'v'.
• Substitute into the integration by parts formula.
• Simplify and integrate any remaining terms.

For example, for the integral \( \int x(x-2)^5 \, dx \), setting \( u = x \) and \( dv = (x-2)^5 \, dx \) makes the process straightforward. This choice reduces the difficulty of the integrand after applying the integration by parts formula. After the substitution and integrating the remaining integral, the solution simplifies to \[ \frac{x(x-2)^6}{6} - \frac{(x-2)^7}{42} + C \].
Ultimately, this technique is structured and helps decompose complex integrals into workable parts.

Substitution Method

The substitution method is another popular technique used to simplify integrals, especially when dealing with composite functions. The idea here is to change variables to make the integral easier to solve.Here's a simple plan:

• Identify a substitution that simplifies the integrand. Typically, this might be an expression within a power or a denominator.
• Change variables: Set \( u \) equal to the chosen substitution expression, replace \( x \) appropriately, and find \( du \) in relation to \( dx \).
• Rewrite the integral in terms of \( u \) and integrate.
• Substitute back the original variables, if necessary.

In our exercise, choosing \( u = x-2 \) transforms the integral \( \int x(x-2)^5 \, dx \) into \( \int (u+2)u^5 \, du \). The substitution step simplifies the polynomial part into a manageable form that can be expanded and integrated term by term. The result is \[ \frac{u^7}{7} + \frac{2u^6}{6} + C \], which simplifies to agree with the integration by parts method after returning to the original variable.

Polynomial Integration

Polynomial integration involves finding the antiderivative of polynomial expressions. Among the integration topics, it is one of the most straightforward, due to power rules.Here's how you tackle polynomial integrals:

• Identify the polynomial expression within the integral.
• Apply the power rule for integration: increase the power by one and divide by the new power for each term.
• Simplify and add the constant of integration, \( C \).

For the integral of a function like \((x-2)^6\), consider it as individual terms raised to a power. The integral of \((x-2)^6\) is computed as \( \frac{(x-2)^7}{7} \).
When combined with other terms, like in our exercise, expanding and integrating each term separately allows us to ultimately piece back together a complete polynomial solution after integration by parts or substitution. This method underscores the simplicity and elegance of handling polynomial expressions in calculus.