Problem 27 - The Calculus with Analytic Geometry

Question

Evaluate \(\int(2 x+1)^{3} d x\) by two methods: (a) Expand \((2 x+1)^{3}\) by the binomial theorem, and apply Formulas 1,4, and \(5 ;(b)\) make the substitution \(u=2 x+1\). Explain the difference in appearance of the answers obtained in (a) and (b).

Step-by-Step Solution

Answer

Using the binomial theorem yields: \(2x^4 + 4x^3 + 3x^2 + x + C\); Using substitution gives: \frac{(2x+1)^4}{8} + C\.

Step 1: Understanding the Goal

The objective is to evaluate the integral \(\int (2x+1)^3 \, dx\) using two different methods: expanding using the binomial theorem and using the substitution method.

Step 2: Method (a): Expand Using the Binomial Theorem

First, expand \( (2x+1)^3 \) using the binomial theorem. The binomial theorem states \( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \). For \((2x+1)^3\), let \(a = 2x\) and \(b = 1 \), and \(n = 3\): \[(2x+1)^3 = \binom{3}{0}(2x)^3(1)^0 + \binom{3}{1}(2x)^2(1)^1 + \binom{3}{2}(2x)^1(1)^2 + \binom{3}{3}(2x)^0(1)^3 = 8x^3 + 12x^2 + 6x + 1 \]

Step 3: Integrate Each Term

Now integrate each term individually: \[\begin{aligned} \int 8x^3 \, dx &= 2x^4 + C_1 \ \int 12x^2 \, dx &= 4x^3 + C_2 \ \int 6x \, dx &= 3x^2 + C_3 \ \int 1 \, dx &= x + C_4 \ \end{aligned}\] Combining results: \(2x^4 + 4x^3 + 3x^2 + x + C\)

Step 4: Method (b): Use Substitution

Let \(u = 2x+1\). Then \(du = 2 \, dx\) or \(dx = \frac{du}{2}\), adjusting the integral: \[\begin{aligned} \int (2x+1)^3 \, dx &= \int u^3 \, \frac{du}{2} \ &= \frac{1}{2} \int u^3 \, du \ &= \frac{1}{2} \left( \frac{u^4}{4} \right) + C \ &= \frac{u^4}{8} + C \ &= \frac{(2x+1)^4}{8} + C \end{aligned}\]

Step 5: Comparison of the Two Methods

Both methods lead to equivalent expressions for the integral, but the expanded polynomial is simpler to integrate term-by-term, while the substitution method directly simplifies the integral form.

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool to expand expressions raised to a power. It states that for any numbers \(a\) and \(b\), and a non-negative integer \(n\): \[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \].
In this exercise, we used the binomial theorem to expand \((2x+1)^3\). By identifying \(a = 2x\), \(b = 1\), and \(n = 3\), we can expand:
\[(2x+1)^3 = \binom{3}{0}(2x)^3(1)^0 + \binom{3}{1}(2x)^2(1)^1 + \binom{3}{2}(2x)^1(1)^2 + \binom{3}{3}(2x)^0(1)^3 \].
The binomial coefficients \(\binom{3}{k}\) are derived from Pascal's Triangle. This expands to: \[ (2x+1)^3 = 8x^3 + 12x^2 + 6x + 1 \]. By expanding, complex expressions are simplified into a summation of polynomial terms, making integration easier.

Polynomial Expansion

Polynomial expansion involves expressing a polynomial, like \((2x + 1)^3\), as a sum of simpler polynomial terms. In our exercise, we used the binomial theorem to rewrite \((2x + 1)^3\) into individual polynomial terms: \[ (2x + 1)^3 = 8x^3 + 12x^2 + 6x + 1 \].
Each term of the expanded polynomial can then be integrated separately:

• \int 8x^3 \, dx = 2x^4 + C_1 \
• \int 12x^2 \, dx = 4x^3 + C_2 \
• \int 6x \, dx = 3x^2 + C_3 \
• \int 1 \, dx = x + C_4 \

The constants of integration combine into one, giving us the result:
\[ 2x^4 + 4x^3 + 3x^2 + x + C \].
This method helps in visualizing and solving integrals efficiently.

Substitution Method

The substitution method simplifies integration by substituting a part of the integral with a new variable. For \((2x + 1)^3\), letting \(u = 2x + 1\) transforms the integral.
By differentiating, we get \(du = 2 \, dx\) or \(dx = \frac{du}{2}\). This changes the original integral into a simpler form: \[\int (2x+1)^3 \, dx = \int u^3 \, \frac{du}{2} = \frac{1}{2} \int u^3 \, du \].
We perform the integration:
\[ \frac{1}{2} \left( \frac{u^4}{4} \right) + C = \frac{u^4}{8} + C \].
Finally, substituting \(u\) back, we obtain: \[ \frac{(2x+1)^4}{8} + C \].
This method is useful for integrals with compositions of functions, allowing for easier computation by changing variables.

Definite Integrals

Definite integrals calculate the exact area under a curve between two bounds. Unlike indefinite integrals, definite integrals evaluate to a specific number.
Suppose we want to evaluate \int_{a}^{b} (2x+1)^3 \, dx\. Using substitution, let \(u = 2x + 1\), hence the bounds change accordingly.
For \int_{a}^{b} (2x+1)^3 \, dx\ with \( u = 2x + 1 \):

• \int_{a}^{b} f(x) \, dx = \int_{u(a)}^{u(b)} f(g(u)) \, du \

This ensures the values respect the transformed variable.
Exact bounds must be calculated from the original limits.
Definite integrals can often be simplified using properties of antiderivatives, making it straightforward to compute areas and physical quantities under a curve.