Problem 8 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-28\). $$\int_{1}^{32} x^{-6 / 5} d x$$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{5}{2} \).

Step 1: Understanding the Integral

We need to evaluate the definite integral \( \int_{1}^{32} x^{-6/5} \, dx \). This is an integral where we need to find the antiderivative of \( x^{-6/5} \) and then evaluate it from 1 to 32.

Step 2: Finding the Antiderivative

To find the antiderivative of \( x^{-6/5} \), we use the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \). Here, \( n = -\frac{6}{5} \), so the antiderivative is \[ \frac{x^{-6/5 + 1}}{-6/5 + 1} = \frac{x^{-1/5}}{-1/5}. \]

Step 3: Simplifying the Antiderivative

Simplify the expression \( \frac{x^{-1/5}}{-1/5} \). This results in \[ -5x^{-1/5}. \] This is the antiderivative of \( x^{-6/5} \).

Step 4: Evaluate the Definite Integral

Now, evaluate the definite integral from 1 to 32. This involves calculating \[ \left[ -5x^{-1/5} \right]_{1}^{32} = -5(32)^{-1/5} + 5(1)^{-1/5}. \]

Step 5: Calculating the Values

Calculate \( 32^{-1/5} \). Since \( 32 = 2^5 \), \( 32^{-1/5} = (2^5)^{-1/5} = 2^{-1} = \frac{1}{2} \). Similarly, calculate \( 1^{-1/5} = 1 \).

Step 6: Substitute and Simplify

Substitute these values back into the expression: \[ -5 \times \frac{1}{2} + 5 \times 1 = -\frac{5}{2} + 5 = \frac{5}{2}. \] Thus, the value of the definite integral is \( \frac{5}{2} \).

Key Concepts

Power Rule for Integration

In calculus, the power rule for integration is an essential tool that simplifies finding integrals of power functions. This rule is especially useful and forms the foundation for integrating polynomial expressions.
To apply the power rule, recall that for any function of the form \( x^n \), where \( n eq -1 \), its integral is given by:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

Here, \( C \) represents the constant of integration, as indefinite integrals admit a family of functions differing by a constant.
In the original problem, we encountered the function \( x^{-6/5} \). Using the power rule, we added 1 to the exponent, resulting in \( x^{-1/5} \), and divided by the new exponent, \( -1/5 \), to find the antiderivative. This step transformed \( x^{-6/5} \) into \( -5x^{-1/5} \), preparing it for further evaluation.

Antiderivative

An antiderivative is another name for the indefinite integral of a function. It represents a function \( F(x) \) whose derivative is equal to a given function \( f(x) \).
Essentially, finding an antiderivative reverses the process of differentiation, allowing us to move from \( f(x) \) back to its original function before it was differentiated.

• \( F(x) = \int f(x) \, dx \)

In our exercise, we determined the antiderivative of \( x^{-6/5} \) to be \( -5x^{-1/5} \). Once you have this antiderivative, you can evaluate definite integrals by substituting the upper and lower limits of integration to compute the net area under the curve.
This method captures how antiderivatives play a pivotal role in connecting calculus concepts of differentiation and integration.

Calculus

Calculus is the branch of mathematics that studies how things change. It consists of two major parts: differentiation and integration, both of which are processes that represent and deal with change and accumulation. Differentiation:

• Concerns deriving change rates using derivatives.
• Examines slopes and tangents of functions at any point.

Calculus revolutionized our ability to compute limits and instantaneous rates of change, developed through the concept of derivatives. Integration:

• Focuses on summation and area calculation under curves.
• Uses antiderivatives to solve definite integrals, which can represent total accumulation.

In the original exercise, we engaged with the integral's part of calculus, using antiderivatives and the power rule to simplify and solve the given integral.
Integration often answers questions about the total quantity from rates of change, just as differentiation provides rates of change from total quantities.