Problem 14 - Calculus

Question

Use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{1}\left(x^{4 / 3}-2 x^{1 / 3}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{15}{14}\).

Step 1: Identify the Function to Integrate

The function we need to integrate is \(f(x) = x^{4/3} - 2x^{1/3}\). We will apply the Fundamental Theorem of Calculus to evaluate its definite integral over \([0, 1]\).

Step 2: Find the Antiderivative

Calculate the antiderivative of \(f(x)\). This involves integrating each term separately. The integral of \(x^{4/3}\) is \(\frac{x^{7/3}}{7/3} = \frac{3}{7}x^{7/3}\), and the integral of \(-2x^{1/3}\) is \(-2\cdot\frac{x^{4/3}}{4/3} = -\frac{3}{2}x^{4/3}\). Thus, the antiderivative is \(F(x) = \frac{3}{7}x^{7/3} - \frac{3}{2}x^{4/3}\).

Step 3: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that \(\int_{a}^{b} f(x)\, dx = F(b) - F(a)\), where \(F\) is an antiderivative of \(f\). Substitute \(x = 1\) and \(x = 0\) into \(F(x)\).

Step 4: Evaluate \(F(1)\) and \(F(0)\)

Calculate \(F(1) = \frac{3}{7}(1)^{7/3} - \frac{3}{2}(1)^{4/3} = \frac{3}{7} - \frac{3}{2}\). Calculate \(F(0) = \frac{3}{7}(0)^{7/3} - \frac{3}{2}(0)^{4/3} = 0\).

Step 5: Calculate the Definite Integral

Substitute the values from Step 4 into the theorem: \[F(1) - F(0) = \left(\frac{3}{7} - \frac{3}{2}\right) - 0\]. Simplify: \[\frac{3}{7} - \frac{3}{2} = \frac{3}{7} - \frac{21}{14} = \frac{3}{7} - \frac{21}{14} = -\frac{15}{14}.\]

Step 6: Final Result

The value of the definite integral \(\int_{0}^{1} (x^{4/3} - 2x^{1/3}) \, dx\) is \(-\frac{15}{14}\).

Key Concepts

Definite Integrals

Definite integrals are a key part of calculus that help us find the exact area under a curve over a certain interval. In the exercise, we calculated the definite integral of the function \( f(x) = x^{4/3} - 2x^{1/3} \) over the interval \([0, 1]\). This integral gives us the total net area between the curve and the x-axis from 0 to 1. When the curve is below the x-axis, the area is considered negative.

To tackle a definite integral, you can break down the process into these simple steps:

• Identify the function that needs integrating.
• Determine the limits of integration, which are the starting and ending points on the x-axis.
• Calculate the antiderivative of the function.
• Apply the Second Fundamental Theorem of Calculus to find the definite integral.

After evaluating the antiderivative at the upper and lower limits, we find the definite integral by subtracting these values. This provides us with the net area under the curve, which in this case was calculated as \(-\frac{15}{14}\).

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse process of taking derivatives. Finding an antiderivative is crucial for solving definite integrals.

In this exercise, we needed the antiderivative of the function \( f(x) = x^{4/3} - 2x^{1/3} \), which involves integrating each term separately. For a term like \( x^n \), its antiderivative is \( \frac{x^{n+1}}{n+1} \).

• The antiderivative for \( x^{4/3} \) is \( \frac{3}{7}x^{7/3} \).
• The antiderivative for \(-2x^{1/3} \) is \(-\frac{3}{2}x^{4/3} \).

Combining these, the overall antiderivative is \( F(x) = \frac{3}{7}x^{7/3} - \frac{3}{2}x^{4/3} \).

This antiderivative function allows you to evaluate the definite integral by using the limits \([0, 1]\).

Integration Techniques

While working through calculus problems, having a solid understanding of various integration techniques is essential. These techniques incorporate recognizing patterns, employing substitution, or using the integration of standard forms.

In the given exercise, direct integration was used because the terms \(x^{4/3}\) and \(x^{1/3}\) are easily integrable based on their exponents. Here’s a quick refresher on basic integration rules:

• Power Rule: The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\) if \(n eq -1\).
• Constant Factor Rule: Constants can be factored out of the integral, simplifying integration.

By applying these rules correctly, the antiderivatives were found without the need for more complex techniques such as integration by parts or substitution. Understanding when and how to use these basic rules helps to simplify and efficiently solve calculus problems.