Problem 60 - Calculus and its applications

Question

Evaluate. $$ \int_{0}^{1} 12 x \sqrt[5]{1-x^{2}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to 5.

Step 1: Identify the Integral and its Components

The given integral is \( \int_{0}^{1} 12x \sqrt[5]{1-x^2} \; dx \). It is a definite integral from 0 to 1, where the integrand is \( 12x \sqrt[5]{1-x^2} \). To solve it, we might need a substitution method.

Step 2: Choose a Suitable Substitution

Select a substitution to simplify the integrand. Let \( u = 1 - x^2 \), which implies \( du = -2x \, dx \). Solve for \( x \, dx \), we get \( x \, dx = -\frac{1}{2} du \). Substitute these in the integral.

Step 3: Change the Limits of Integration

With the substitution \( u = 1 - x^2 \), the limits will change from \( x = 0 \) to \( x = 1 \). When \( x = 0, \) \( u = 1 - 0^2 = 1 \), and when \( x = 1, \) \( u = 1 - 1^2 = 0 \). Thus, the limits in terms of \( u \) are from 1 to 0.

Step 4: Rewrite the Integral with Substitution

After substitution, the integral becomes \( \int_{1}^{0} -6 u^{1/5} \, du \). Notice the negative sign due to the substitution \( x \, dx = -\frac{1}{2} du \). Simplify by reversing the limits and removing the negative sign.

Step 5: Simplify and Integrate

Reverse the limits to get \( \int_{0}^{1} 6 u^{1/5} \, du \). Integrate using the formula \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C \). Here, \( n = \frac{1}{5} \), so the integral is \( \frac{6}{6/5} u^{6/5} = 5u^{6/5} \).

Step 6: Evaluate the Integral at the New Limits

Evaluate \( 5u^{6/5} \) from 0 to 1. Substituting the upper limit, \( u = 1 \), gives \( 5(1)^{6/5} = 5 \). Substituting the lower limit, \( u = 0 \), gives \( 5(0)^{6/5} = 0 \). Thus, the integral evaluates to \( 5 - 0 = 5 \).

Key Concepts

Substitution Method

The substitution method is a powerful technique often used for solving integrals. It's particularly useful when the function inside the integral, known as the integrand, is complex and not easily integrable in its current form. By making a substitution, you transform the original variable into a new variable, simplifying the integral. In our example, we have the definite integral \( \int_{0}^{1} 12x \sqrt[5]{1-x^2} \; dx \). The expression \( \sqrt[5]{1-x^2} \) complicates direct integration.

• **Choosing the Right Substitution:** The key to using the substitution method effectively is selecting a substitution that simplifies the integrand. Here, we choose \( u = 1 - x^2 \) because the derivative \( du = -2x \, dx \) matches a part of the content of the original integral.
• **Simplifying the Integral:** By substituting, we express \( x \, dx \) as \( -\frac{1}{2} du \). This step simplifies the intricate parts of the integral that involve \( x \).

By transforming the variables, not only does the integration process become easier, but it also allows you to apply more straightforward integration rules. Remember that practice makes perfect when it comes to choosing effective substitutions.

Integration Limits

Integration limits change along with the substitution. This is because the limits define the boundary within which we evaluate the integral for the original variable. When we switch to a new variable, we must adjust these boundaries accordingly. In the example, the limits were initially from \( x = 0 \) to \( x = 1 \).

• **Transforming the Limits:** With the substitution \( u = 1 - x^2 \), the corresponding new limits are calculated by evaluating \( u \) at the original limit values. For \( x = 0 \), \( u = 1 \), and for \( x = 1 \), \( u = 0 \).
• **Order of Limits:** Notice the role reversal of limits, resulting in limits going from \( u = 1 \) to \( u = 0 \). This sequence usually indicates that you will have to account for a negative within the integral. In this scenario, we can reverse these limits to maintain the integral assessment from smaller to larger numbers, removing the negative sign by doing so.

Understanding the impact of substitutions on limits is essential. It ensures accuracy in both the transformation of the integral and the final evaluation result.

Integral Evaluation

Once substitution and limits are sorted, the final step is to evaluate the transformed integral. The integral has been simplified to \( \int_{0}^{1} 6 u^{1/5} \, du \). It's now ready for integration.

• **Applying Integration Rules:** Use the formula for integrating powers, \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n = \frac{1}{5} \). Here, it calculates to \( \frac{6}{6/5} u^{6/5} \), simplifying further to \( 5u^{6/5} \).
• **Evaluating the Integral:** With integration complete, apply the new limits, substituting these into your antiderivative \( 5u^{6/5} \). For \( u = 1 \), \( 5(1)^{6/5} = 5 \), and for \( u = 0 \), \( 5(0)^{6/5} = 0 \). The evaluated integral is \( 5 - 0 = 5 \), providing the final result.

Definite integrals, when evaluated correctly, give the net "accumulation" of the given function's area between the specified limits. With the steps of substitution and careful limit handling, integration becomes manageable and leads to accurate results.