Problem 21 - Thomas Calculus

Question

Use the Substitution Formula in Theorem 6 to evaluate the integrals in Exercises \(1-24 .\) $$ \int_{0}^{1}\left(4 y-y^{2}+4 y^{3}+1\right)^{-2 / 3}\left(12 y^{2}-2 y+4\right) d y $$

Step-by-Step Solution

Answer

The value of the integral is 3.

Step 1: Identify and Choose Substitution

The integral given is \( \int_{0}^{1} (4y - y^2 + 4y^3 + 1)^{-2/3} (12y^2 - 2y + 4) \, dy \). To apply the substitution formula effectively, identify a substitution that simplifies the integrand. Here, let \( u = 4y - y^2 + 4y^3 + 1 \).

Step 2: Derive the Differential and Substitute

Differentiate \( u = 4y - y^2 + 4y^3 + 1 \) with respect to \( y \): \( \frac{du}{dy} = 12y^2 - 2y + 4 \). This implies \( du = (12y^2 - 2y + 4) \, dy \). Substitute \( u \) and \( du \) into the integral to replace \( (12y^2 - 2y + 4) \, dy \), making the integral simpler.

Step 3: Change Limits of Integration

Since the original variable is \( y \) and we substituted \( u \), convert the limits of integration from \( y \) values to \( u \) values. When \( y = 0 \), \( u = 4(0) - 0^2 + 4(0)^3 + 1 = 1 \). When \( y = 1 \), \( u = 4(1) - 1^2 + 4(1)^3 + 1 = 8 \). So, the limits change from \( 0 \) to \( 1 \) to \( 1 \) to \( 8 \).

Step 4: Integrate with Respect to \( u \)

With the substitution, the integral becomes \( \int_{1}^{8} u^{-2/3} \, du \). To integrate, use the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \).

Step 5: Calculate the Indefinite Integral

Apply the power rule to integrate \( u^{-2/3} \): \( \int u^{-2/3} \, du = \frac{u^{(-2/3) + 1}}{(-2/3) + 1} = \frac{u^{1/3}}{1/3} = 3u^{1/3} \).

Step 6: Evaluate the Definite Integral

Evaluate the antiderivative from \( u = 1 \) to \( u = 8 \): \( 3[u^{1/3}]_{1}^{8} = 3(8^{1/3}) - 3(1^{1/3}) \). Since \( 8^{1/3} = 2 \), calculate \( 3 \times 2 - 3 \times 1 = 6 - 3 = 3 \).

Key Concepts

Substitution Method

The substitution method is a crucial technique in integral calculus that simplifies the process of integration by substituting part of the integrand with a new variable. This technique is particularly helpful when dealing with complicated integrals where the integral can be rewritten in a simpler form.

• First, identify a part of the integrand that, when substituted, will make the integral easier to evaluate. This part is often a complicated expression inside a function or multiplied by a derivative.
• Choose a substitution, typically denoted as \( u \), and express it in terms of the original variable. For example, in the given problem, we choose \( u = 4y - y^2 + 4y^3 + 1 \).
• Differentiate \( u \) with respect to the original variable \( y \) to find \( \frac{du}{dy} \), then solve for \( dy \) in terms of \( du \).

The objective is to replace every appearance of the original variable and its differentials in the integral with the new variable \( u \) and \( du \). Once substituted, the integral is generally simpler to solve.

Definite Integral

A definite integral computes the net area under the curve of a function along an interval. It is represented by the integral sign with upper and lower limits of integration.

• After substitution, it's crucial to also modify the limits of integration to correspond with the new variable \( u \). In our example, the original limits were from \( y = 0 \) to \( y = 1 \). After substitution, these limits changed to \( u = 1 \) to \( u = 8 \).
• The definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \) and represents the accumulation of quantities or, in geometric terms, the net area between the curve \( y = f(x) \), the x-axis, and the vertical lines \( x = a \) and \( x = b \).

After integrating with respect to \( u \), evaluate the definite integral using the new limits. The result will give you the total change or the net area in the context of the problem. This becomes especially practical in evaluating complex integrals after simplification.

Power Rule for Integration

The power rule is one of the fundamental rules in calculus used to find antiderivatives of power functions. It states that for any real number \( n ot= -1 \), the integral of \( x^n \) is given by \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).

• In our example, once substitution is done, the integral \( \int u^{-2/3} \, du \) becomes straightforward with the power rule. Integrate it as \( \frac{u^{1/3}}{1/3} = 3u^{1/3} \).
• Using this rule requires careful handling of negative exponents and fractional powers, which is common in integration challenges. Hence, rewriting with positive exponents post-integration is often done for clarity.
• The power rule simplifies evaluating antiderivatives significantly, which are then used in applying the Fundamental Theorem of Calculus to compute definite integrals.

This rule allows us to easily find the antiderivative, which upon substitution of limits, gives the measurement of the given integral problem, simplifying previously daunting tasks efficiently.