Question

Use the approaches discussed in this section to evaluate the following integrals. $$\int_{0}^{2} \frac{2}{x^{3}+3 x^{2}+3 x+1} d x$$

Short answer

Question: Evaluate the integral $$\int_{0}^{2} \frac{2}{x^{3}+3 x^{2}+3 x+1} d x$$. Answer: $$\frac{8}{9}$$

Step-by-step solution

Analyze the denominator

Let's take a close look at the denominator: $$x^{3}+3 x^{2}+3 x+1$$ This denominator can be factored as $$(x+1)^3$$ This observation does not lead to any obvious integration techniques, so we will use a substitution method.

Substitution

Let's make the substitution $$u = x + 1$$ Then, $$du = dx$$ The limits of integration changes as well: When \(x=0\), \(u=1\), and when \(x=2\), \(u=3\). This changes our integral to: $$\int_{1}^{3} \frac{2}{(u)^3} du$$

Integrate with respect to u

Now, we integrate the new integral with respect to \(u\): $$\int_{1}^{3} \frac{2}{u^3} du = 2 \int_{1}^{3} u^{-3} du$$ To integrate it, apply the power rule: $$ 2 \cdot \frac{u^{-2}}{-2} \Big|_{1}^{3}$$

Evaluate the integral

Now we can evaluate our integral by substituting the limits: $$\frac{-1}{u^2} \Big|_{1}^{3}= -\frac{1}{3^2} + \frac{1}{1^2} = -\frac{1}{9} + 1 = \frac{8}{9}$$ The result is: $$\int_{0}^{2} \frac{2}{x^{3}+3 x^{2}+3 x+1} d x = \frac{8}{9}$$

Key concepts

Definite Integration

Definite integration refers to finding the value of an integral with specific upper and lower limits. It computes the net area between a function and the x-axis over a specified interval.
When we perform definite integration, we are essentially calculating the accumulated value of a function from the lower limit to the upper limit of the interval.

• For the integral \( \int_{a}^{b} f(x) \, dx \), \( a \) is the lower limit and \( b \) is the upper limit.
• The outcome is a real number that represents a precise area or accumulated change.
• At times, the result may be positive, negative, or even zero, depending on the behavior of the function over the interval.

In the example given, the integral is evaluated from \( 0 \) to \( 2 \), and after finding the antiderivative, we substitute these limits to find a specific numerical value of \( \frac{8}{9} \). This means the net area under the curve from 0 to 2 is \( \frac{8}{9} \).

Substitution Method in Integration

The substitution method is a technique used to simplify integrals, particularly when faced with complex algebraic expressions. It's akin to the "chain rule" in reverse, where you substitute part of the expression with a single variable.
The steps in the substitution method often look like this:

• Identify a part of the integral as a new variable \( u \). In the example above, \( u = x + 1 \).
• Derive \( du \) in terms of \( dx \) or the original variable. Here, \( du = dx \).
• Adjust the integration limits according to this substitution. For this problem, the limits change from \( x=0 \) to \( u=1 \) and from \( x=2 \) to \( u=3 \).

Using substitution can significantly simplify the original integral, making it easier to evaluate. Once we make the substitution, the integral turns into \( \int_{1}^{3} \frac{2}{u^3} \, du \), which is a simpler form that we can integrate using basic rules.

Factorization in Calculus

Factorization in calculus involves rewriting expressions as products to simplify their integration or differentiation. It can reveal hidden patterns or facilitate easier manipulation. The process may sometimes enable us to utilize specific integration techniques.
In practice, factorization:

• Reduces complex polynomial expressions into simpler, recognizable forms.
• Allows easier application of techniques, such as substitution or partial fraction decomposition.

For our exercise, the polynomial in the denominator, \( x^3+3x^2+3x+1 \), is factored into \( (x+1)^3 \). Although this observation didn’t immediately solve the integral, recognizing this factorization was essential for applying the substitution method effectively. Understanding these factoring techniques can also aid in identifying simplifiable structures in other calculus problems. Factorization transforms challenging integrals into manageable forms, offering a clearer path to obtaining solutions.