Problem 23 - Applied Calculus

Question

Find the integrals .Check your answers by differentiation. $$\int y^{2}(1+y)^{2} d y$$

Step-by-Step Solution

Answer

The integral is \( \frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5} + C \).

Step 1: Expand the Expression

First, we need to expand the expression \( y^2 (1+y)^2 \). This requires us to expand \( (1+y)^2 \). The expanded form of \( (1+y)^2 \) is \( 1 + 2y + y^2 \).Then, multiply the expanded expression by \( y^2 \):\[ y^2 (1 + 2y + y^2) = y^2 + 2y^3 + y^4 \].

Step 2: Integrate Each Term Separately

The integral we need to solve is now \( \int (y^2 + 2y^3 + y^4) \, dy \).We integrate each term:\( \int y^2 \, dy = \frac{y^3}{3} + C_1 \)\( \int 2y^3 \, dy = \frac{2y^4}{4} = \frac{y^4}{2} + C_2 \)\( \int y^4 \, dy = \frac{y^5}{5} + C_3 \).

Step 3: Combine the Integrals

Combine all the integrated terms:\[ \frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5} + C \] where \( C = C_1 + C_2 + C_3 \) is the constant of integration.

Step 4: Differentiate to Verify the Solution

Differentiate the obtained solution to ensure it matches the original integrand:Differentiating \( \frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5} \), we get:\( \frac{d}{dy} \left( \frac{y^3}{3} \right) = y^2 \)\( \frac{d}{dy} \left( \frac{y^4}{2} \right) = 2y^3 \)\( \frac{d}{dy} \left( \frac{y^5}{5} \right) = y^4 \)Combining these derivatives results in the original expression: \[ y^2 + 2y^3 + y^4 \], confirming the solution is correct.

Key Concepts

Polynomial Expansion

When faced with an expression like \( y^2 (1+y)^2 \), one efficient way to handle it is through polynomial expansion. Polynomial expansion allows us to convert a complex polynomial into a simpler form. This makes it easier to integrate each term separately.

Consider the expression \((1+y)^2\), which is a binomial. By applying the binomial expansion formula, we get:

• \((1+y)^2 = 1 + 2y + y^2\).

Once expanded, we then multiply each term by \(y^2\):

• \(y^2(1+2y+y^2) = y^2 + 2y^3 + y^4\).

Polynomial expansion simplifies the expression to a form that is ready for easy integration. This foundational concept is crucial for doing many calculus-related problems, turning potentially difficult integrands into manageable parts.

Differentiation

Differentiation is an essential concept in calculus, allowing us to find the derivative of a function, which represents the rate of change of a function with respect to a variable. In this exercise, after integrating the expression, differentiation is used to verify our solution.

Once we have integrated the polynomial and obtained the expression:

• \(\frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5}\),

we then differentiate each term to check our work:

• The derivative of \(\frac{y^3}{3}\) is \(y^2\),
• The derivative of \(\frac{y^4}{2}\) is \(2y^3\),
• The derivative of \(\frac{y^5}{5}\) is \(y^4\).

Adding these derivatives gives us back the original polynomial, \(y^2 + 2y^3 + y^4\), confirming the process was correctly followed. Differentiation serves as a useful tool for verifying integrations by checking if the calculated integral is consistent with the original function.

Integral Calculus

Integral calculus is concerned with the accumulation of quantities and the areas under and between curves. A vital concept is the indefinite integral, which involves finding a function whose derivative is the given function.

In our exercise, the task was to integrate \(y^2 + 2y^3 + y^4\). Integrating each term separately, we find:

• \(\int y^2 \, dy = \frac{y^3}{3}\),
• \(\int 2y^3 \, dy = \frac{y^4}{2}\),
• \(\int y^4 \, dy = \frac{y^5}{5}\).

After calculating the indefinite integrals, we accumulate them along with a constant of integration, \( C \), resulting in:

• \(\frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5} + C\).

This final expression encompasses all possible antiderivatives of the given polynomial, demonstrating the fundamental importance of integral calculus in determining such functions.