Problem 12 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

\( \frac{2}{5}x^{5/2} + \frac{2}{3}x^{3/2} + C \)

Step 1: Distribute the Power of x

Begin by expanding the expression inside the integral. Distribute the term \(x^{1/2}\) across the term \((x + 1)\): \[ x^{1/2} imes (x + 1) = x^{1/2} imes x + x^{1/2} imes 1 = x^{3/2} + x^{1/2}.\]

Step 2: Rewrite the Integral

Rewrite the integral by replacing the original expression with the expanded form from Step 1: \[\int (x^{3/2} + x^{1/2}) \, dx.\]

Step 3: Integrate Each Term Separately

Apply the power rule for integration \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) to each term:- For \(x^{3/2}\), increase the exponent by 1 to get \(x^{5/2}\) and divide by the new exponent:\[\int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2}{5}x^{5/2}\] - For \(x^{1/2}\), increase the exponent by 1 to get \(x^{3/2}\) and divide by the new exponent:\[\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}\]

Step 4: Combine the Results

Add the results of the integrals to get the final expression, including the constant of integration \(C\):\[\int (x^{3/2} + x^{1/2}) \, dx = \frac{2}{5}x^{5/2} + \frac{2}{3}x^{3/2} + C.\]

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental tool in calculus. It allows us to integrate expressions of the form \(x^n\). The general rule states that the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration. This rule is applicable as long as \(n eq -1\). The reason we cannot use it for \(n = -1\) is because it leads to a logarithmic function, specifically, the integral of \(x^{-1}\) or \(\frac{1}{x}\) is \(\ln|x| + C\).
In the given exercise, we apply the power rule to each term in the expanded form of the integral. Using the rule, we transform \(x^{3/2}\) into \(\frac{2}{5}x^{5/2}\) by increasing the exponent by one and then dividing by the new exponent. Similarly, \(x^{1/2}\) becomes \(\frac{2}{3}x^{3/2}\).
By understanding and using the power rule, we can simplify and solve many integration problems efficiently. Keep practicing applying this rule to improve your confidence and speed with integrals.

Integration by Parts

Integration by parts is another powerful method for solving integrals, especially when dealing with products of functions. The technique is based on the product rule for differentiation and is expressed with the formula:
\[ \int u \ dv = uv - \int v \ du\]
Here, \(u\) and \(dv\) are parts of the integrand that you choose to differentiate and integrate, respectively. This technique is helpful when direct application of basic integration rules isn't possible or when dealing with complex functions.
However, in the exercise provided, integration by parts is not needed. The integral simplifies easily with basic algebraic manipulation and the power rule. As you work on different problems, understanding when to use integration by parts will help you tackle more challenging integrals smoothly.

Exponent Rules

Exponent rules are crucial for simplifying expressions before integrating. They make even complex integrals manageable, just like in this exercise. These rules allow us to manipulate powers efficiently, and include principles such as:

• \(x^a \cdot x^b = x^{a+b}\)
• \((x^a)^b = x^{ab}\)
• \(x^{-a} = \frac{1}{x^a}\)

In our exercise, the expression \(x^{1/2}(x+1)\) can be expanded to \(x^{3/2} + x^{1/2}\) using these rules. We multiplied \(x^{1/2}\) by \(x\) and by the constant 1 separately. This transformation simplified the integration process by isolating each term, making it possible to directly apply the power rule.
Mastering these exponent rules not only eases integration but also helps in various algebraic manipulations across different areas of mathematics.

Constant of Integration

When calculating indefinite integrals, we must include the constant of integration, often denoted by \(C\). This constant captures the fact that antiderivatives are determined only up to an additive constant. Essentially, for any function \(F(x)\) that satisfies \(F'(x) = f(x)\), \(F(x) + C\) is also a solution due to the derivative of a constant being zero.
In our specific example, after integrating each term separately, the results: \(\frac{2}{5}x^{5/2}\) and \(\frac{2}{3}x^{3/2}\) are combined into one expression. The sum is completed by adding \(C\), making the solution \(\int (x^{3/2} + x^{1/2}) \, dx = \frac{2}{5}x^{5/2} + \frac{2}{3}x^{3/2} + C\).
Always remember to include \(C\) in indefinite integrals, as it represents an entire family of antiderivative functions and is vital in accurately expressing the solution.