Problem 101 - Calculus and its applications

Question

Evaluate. $$ \int x^{2} \sqrt{x^{3}+1} d x $$

Step-by-Step Solution

Answer

\( \int x^2 \sqrt{x^3+1} \, dx = \frac{2}{9} (x^3+1)^{3/2} + C \)

Step 1: Choose the Substitution

To solve the integral, we will use the substitution method. We choose to let \( u = x^3 + 1 \). The reason for this choice is because the derivative of \( u \) with respect to \( x \) will help simplify the integral involving \( x^2 \).

Step 2: Differentiate the Substitution

Calculate the derivative to find \( du \). Since \( u = x^3 + 1 \), differentiate both sides with respect to \( x \):\[ \frac{du}{dx} = 3x^2 \]This implies that \( du = 3x^2 \, dx \) so \( x^2 \, dx = \frac{1}{3} du \).

Step 3: Substitute in the Integral

Substitute the expressions for \( u \) and \( x^2 \, dx \) into the integral:\[ \int x^2 \sqrt{x^3 + 1} \, dx = \int \, \sqrt{u} \, \frac{1}{3} \, du \]Simplify it:\[ \frac{1}{3} \int \sqrt{u} \, du \]

Step 4: Integrate with Respect to u

Now, integrate \( \frac{1}{3} \int u^{1/2} \, du \):Use the power rule for integration, which states \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \):\[ \frac{1}{3} \times \frac{u^{3/2}}{3/2} = \frac{1}{3} \times \frac{2}{3} u^{3/2} = \frac{2}{9} u^{3/2} + C \]

Step 5: Back-Substitute for x

Substitute \( u = x^3 + 1 \) back into the expression:\[ \frac{2}{9} (x^3 + 1)^{3/2} + C \]

Step 6: Write the Final Answer

The evaluated integral is:\[ \int x^2 \sqrt{x^3 + 1} \, dx = \frac{2}{9} (x^3 + 1)^{3/2} + C \]

Key Concepts

Substitution Method

The substitution method in integral calculus is like a change of variables that makes integration easier. Think of it as simplifying the problem. In our exercise, we're dealing with the integral \( \int x^2 \sqrt{x^3 + 1} \, dx \). By choosing an appropriate substitution, such as \( u = x^3 + 1 \), we make the expression under the square root simpler.

Here’s how substitution works step-by-step:

• First, identify a function inside the integral that, if replaced, will simplify the expression. Often it’s something inside a parenthesis or under a square root.
• Check if the derivative of this function exists elsewhere in the integral. In our case, the derivative of \( u = x^3 + 1 \) is \( 3x^2 \), which, after adjusting factors, matches \( x^2 \, dx \) in the original integral.

By substituting \( x^2 \, dx \) with \( \frac{1}{3} \, du \), we can rewrite the integral in terms of \( u \), which makes it much simpler to evaluate.

Power Rule for Integration

Once the integral is expressed in terms of \( u \), the power rule for integration assists us. The power rule is straightforward: for any real number \( n \), \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), provided \( n eq -1 \).

In our problem, we found the integral form \( \frac{1}{3} \int u^{1/2} \, du \). Applying the power rule here simplifies the integral:

• Identify \( n \). For \( u^{1/2} \), \( n = 1/2 \).
• Apply the formula: \( \frac{1}{3} \times \frac{u^{3/2}}{3/2} = \frac{2}{9} u^{3/2} \).

This step drastically simplifies the process by allowing us to integrate quickly and clearly, moving us towards solving the original problem. Don't forget the constant \( C \), as integration involves indefinite integration, leaving room for any constant addition.

Back-Substitution

After integrating with respect to the new variable \( u \), it's time to revert back to the original variable \( x \) to express the final answer in terms of \( x \). This is the "back-substitution" step.

Here's how to do it:

• Recall your substitution: \( u = x^3 + 1 \).
• Replace \( u \) in your integrated function with \( x^3 + 1 \) to convert back: \( \frac{2}{9} (x^3 + 1)^{3/2} + C \).

Back-substitution ensures the solution reflects the problem's original form, providing a complete answer to the exercise. Remember, integrating wasn't just about playing with numbers — it connects the derived function to the context of the original problem, illustrating the answer within the original framework.