Problem 14 - Calculus: Early Transcendentals

Question

Evaluate the indefinite integral. \( \displaystyle \int y^2 (4 - y^3)^{2/3} \, dy \)

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{5} (4 - y^3)^{5/3} + C\).

Step 1: Identify the Integration Technique

To solve the integral \( \int y^2 (4 - y^3)^{2/3} \, dy \), we can use substitution. This method simplifies the expression inside the integral by substituting a part of it with a single variable.

Step 2: Choose a Suitable Substitution

Let's choose \( u = 4 - y^3 \). Then, find the derivative of \( u \) in terms of \( y \), so \( du = -3y^2 \, dy \). From this, we can express \( y^2 \, dy \) as \( -\frac{1}{3} du \).

Step 3: Substitute and Simplify the Integral

Replace \( y^2 (4 - y^3)^{2/3} \, dy \) with the substitution terms: \( -\frac{1}{3} \int u^{2/3} \, du \). This simplifies the integral to \( -\frac{1}{3} \int u^{2/3} \, du \).

Step 4: Perform the Integration

Integrate \( u^{2/3} \) to get \( \int u^{2/3} \, du = \frac{3}{5} u^{5/3} + C \), where \( C \) is the constant of integration. Multiplying by \(-\frac{1}{3}\), the integral becomes \( -\frac{1}{3} \cdot \frac{3}{5} u^{5/3} = -\frac{1}{5} u^{5/3} + C \).

Step 5: Substitute Back the Original Variable

Replace \( u \) with \( 4 - y^3 \) to get the integral in terms of \( y \): \(-\frac{1}{5} (4 - y^3)^{5/3} + C\).

Step 6: Simplified Solution

Thus, the indefinite integral \( \int y^2 (4 - y^3)^{2/3} \, dy \) evaluates to \(-\frac{1}{5} (4 - y^3)^{5/3} + C\).

Key Concepts

Integration Technique

In calculus, choosing the right integration technique is crucial for evaluating an indefinite integral efficiently. This particular integral, \( \int y^2 (4 - y^3)^{2/3} \, dy \), can be tackled using the substitution method, a common technique. When faced with complex integrands, this method helps by simplifying the problem into a more manageable form. Other techniques include integration by parts and partial fractions, but substitution is often the go-to strategy when dealing with expressions that can be rewritten in terms of a single new variable.

Substitution Method

The substitution method is like a helper in disguise during integration tasks, simplifying parts of the integral using a new variable. For this exercise, we set \( u = 4 - y^3 \). This choice is based on finding a part of the integrand whose derivative appears elsewhere in the expression. The derivative, \( du = -3y^2 \, dy \), allows us to rewrite the differential \( y^2 \, dy \) as \( -\frac{1}{3} du \). This transformation is the crux of the method, reducing the complexity and turning the integral into a simpler form involving \( u \) instead of \( y \). Successfully choosing a substitution is key to a smooth integration process.

Integral Calculus

Integral calculus is a fundamental area of calculus focusing on the concept of integration. It deals with finding a function when its derivative is known, aiming to calculate areas under curves, among other applications. In this step-by-step solution, after substituting and simplifying the integral to \( -\frac{1}{3} \int u^{2/3} \, du \), we used basic integral calculus to find the antiderivative. The power rule for integration, which says \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), was applied here, giving us \( \int u^{2/3} \, du = \frac{3}{5} u^{5/3} + C \). This illustrates the foundational process of finding antiderivatives to solve integrals.

Constant of Integration

In indefinite integrals, such as \( \int y^2 (4 - y^3)^{2/3} \, dy \), a constant of integration, denoted usually as \( C \), is added. This constant represents all possible vertical shifts of the antiderivative function, as the derivative of any constant is zero. Therefore, when integrating, we include \( C \) to account for this family of antiderivatives. In our solution, \( -\frac{1}{5} (4 - y^3)^{5/3} + C \), \( C \) ensures that the antiderivative is as general as possible, covering all potential functions whose derivative returns to the original integrand. This inclusion makes indefinite integrals distinct from definite integrals, which calculate one specific numerical value.