Problem 20 - Calculus Single Variable

Question

Calculate the indefinite integral. $$ \int x^{2}(\sqrt{x}+1)^{2} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{x^{4}}{4} + \frac{4}{7} x^{3.5} + \frac{x^{3}}{3} + C \).

Step 1: Expand the Inside of the Integral

We have the integral \( \int x^{2} (\sqrt{x} + 1)^{2} \, dx \). Start by expanding the squared term: \((\sqrt{x} + 1)^{2} = (\sqrt{x})^{2} + 2 \cdot \sqrt{x} \cdot 1 + 1^{2} = x + 2\sqrt{x} + 1\). This simplifies the integral to \( \int x^{2}(x + 2\sqrt{x} + 1) \, dx \).

Step 2: Distribute to Form a Single Polynomial

Next, distribute the \(x^{2}\) inside the integral: \(x^{2}(x + 2\sqrt{x} + 1) = x^{3} + 2x^{2.5} + x^{2}\). The integral becomes \( \int (x^{3} + 2x^{2.5} + x^{2}) \, dx \).

Step 3: Integrate Each Term Individually

Now, integrate each term separately: 1. \( \int x^{3} \, dx = \frac{x^{4}}{4} + C_{1} \) 2. \( \int 2x^{2.5} \, dx = 2 \cdot \frac{x^{3.5}}{3.5} + C_{2} \) 3. \( \int x^{2} \, dx = \frac{x^{3}}{3} + C_{3} \) Combine these using a single constant of integration \(C\): \( \frac{x^{4}}{4} + \frac{2}{3.5}x^{3.5} + \frac{x^{3}}{3} + C \).

Step 4: Simplify the Result

Simplify the expression: - \( \frac{2}{3.5} = \frac{4}{7} \), so the second term becomes \( \frac{4}{7} x^{3.5} \).Thus, the indefinite integral simplifies to \[ \frac{x^{4}}{4} + \frac{4}{7} x^{3.5} + \frac{x^{3}}{3} + C \].

Key Concepts

Polynomial Expansion

Polynomial expansion is a critical first step to tackle when solving integrals like this one. Expanding a polynomial means expressing a squared or cubed expression in its expanded form by multiplying the terms as needed. It simplifies complex terms into a sum of simpler terms, making integration easier.

In this exercise,

• you start with \[ ( \sqrt{x} + 1)^2 \] which needs expansion.
• By applying the formula \[ (a + b)^2 = a^2 + 2ab + b^2, \] we find \[ ( \sqrt{x})^2 + 2 \sqrt{x} \cdot 1 + 1^2 \] simplifying into \[ x + 2 \sqrt{x} + 1. \]

This expansion allows easier handling of each term when you start integrating.

Integration

Integration is the process of finding the integral of a function. It can be considered "reversing" differentiation. The indefinite integral of a function is found without specific limits, focusing on finding a general form of the antiderivative, which includes all possible constants.

In the exercise, after expanding the polynomial, you write the integral in terms of separate, simplified terms like \[\int (x^3 + 2x^{2.5} + x^2) \, dx.\]Then, integrate each term individually.

• Start with simple power rules,\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C_n,\) for solving each term in the integral.

This process results in each term being integrated separately before summing them up towards the end.

Exponents

Exponents are a way to represent repeated multiplication. In integration, handling varying exponents correctly ensures accurate functions after integration. The power rule \(x^n \)derivatives yield integral functions \[\frac{x^{n+1}}{n+1} + C.\]

For instance, take \(x^{2.5},\)an exponent that isn't a whole number. Integrating such terms requires careful handling. You raise the base \(\sqrt{x}\) to new powers during polynomial expansion, understanding each term's exponent is key to integrating correctly.

• The term in \(2\times\int x^{2.5}\,dx\)elevates correctly to achieve the functional form \(\frac{x^{3.5}}{3.5}.\)

Understanding exponents helps execute precise integration.

Constant of Integration

The constant of integration, represented by \(C,\)is crucial in indefinite integrals. Because integrals translate into antiderivatives, the differentiation loses any constant terms that should be in the original function. By applying a constant \(C,\)you ensure all potential functions are accounted for.

When solving the integral,

• you integrate all terms separately and each gets its own constant, \(C_1, C_2,\)
• but ultimately only one \(C\) is used in the final solution.

The constant of integration is essential for expressing the family of all antiderivatives a function might have, reflecting that an infinite number of functions can have the same derivative.