Problem 8 - Thomas Calculus

Question

Evaluate the integrals $$ \int_{1}^{32} x^{-6 / 5} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{5}{2} \).

Step 1: Recognize the Power Rule for Integration

The integral given is \( \int x^n \, dx \), where \( n = -\frac{6}{5} \). We will use the power rule for integration which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), assuming \( n eq -1 \).

Step 2: Apply the Power Rule to the Integral

First, find \( n+1 \) where \( n = -\frac{6}{5} \). You get \( -\frac{6}{5} + 1 = -\frac{1}{5} \). Then, the indefinite integral becomes \( \frac{x^{-1/5}}{-1/5} + C = -5x^{-1/5} + C \).

Step 3: Set Up the Definite Integral

The definite integral from 1 to 32 is given by \[ \int_{1}^{32} x^{-6/5} \, dx = \left[-5x^{-1/5}\right]_{1}^{32}.\]

Step 4: Evaluate the Antiderivative at the Bounds

Substitute the upper limit into the antiderivative: \(-5(32)^{-1/5} = -5 \times \frac{1}{2} = -\frac{5}{2}.\)Substitute the lower limit into the antiderivative: \(-5(1)^{-1/5} = -5.\)

Step 5: Calculate the Definite Integral Result

Subtract the value of the antiderivative at the lower limit from the value at the upper limit: \[-\frac{5}{2} - (-5) = -\frac{5}{2} + 5 = \frac{10}{2} - \frac{5}{2} = \frac{5}{2}.\]

Step 6: Final Result

The value of the integral \( \int_{1}^{32} x^{-6/5} \, dx \) is \( \frac{5}{2} \).

Key Concepts

Power Rule for Integration

The Power Rule for Integration is a fundamental tool that simplifies the process of finding an antiderivative. When you have a function of the form \( x^n \), the power rule allows you to integrate it by increasing the exponent by one and then dividing by the new exponent. It's expressed mathematically as

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

Here, \( C \) is the constant of integration, which appears when dealing with indefinite integrals. However, in our case with definite integrals, this constant cancels out. This simple rule makes it easy to deal with polynomial integrands.
Keep in mind, though, this formula holds true as long as \( n eq -1 \), because division by zero is undefined.

Antiderivatives

An antiderivative is a function whose derivative is the original function. In the context of integration, we seek an antiderivative that, when differentiated, gives back the integrand. For the integral \( \int x^{-6/5} \, dx \), the antiderivative is determined using the power rule. By finding \( n+1 \) where \( n = -\frac{6}{5} \), we get

• \( -\frac{6}{5} + 1 = -\frac{1}{5} \)

Thus, the antiderivative becomes \( -5x^{-1/5} + C \).
This function, \( -5x^{-1/5} \), is what you evaluate at your bounds in a definite integral. Sometimes, even just understanding this step deeply can be the difference between stumbling block and success in calculus.

Bound Evaluation

Evaluating the antiderivative at the bounds of the integral involves plugging the upper and lower limits into the antiderivative and calculating the difference. This step gives the actual area under the curve over the specified interval. Traditionally, this is done by

• Calculating \( F(b) \) for the upper limit \( b \)
• Calculating \( F(a) \) for the lower limit \( a \)

Then, the result is given by \( F(b) - F(a) \).
In our example, inputting the bounds into the antiderivative \( -5x^{-1/5} \):

• At \( x = 32 \), the result is \( -5 \times (32)^{-1/5} = -\frac{5}{2} \)
• At \( x = 1 \), it simplifies to \( -5 \times (1)^{-1/5} = -5 \)

The final evaluation gives us:

• \( \left( -\frac{5}{2} - (-5) \right) = \frac{5}{2} \)

Integration Techniques

Integration techniques are a set of methods applied to solve integrals, and they can range from using the power rule and substitution, to more advanced methods like integration by parts and partial fractions. The technique chosen often depends on the form of the integrand.
For simple power-like functions, as in our example, applying the power rule directly is both the fastest and the simplest method. More complex functions may require additional techniques, such as

• U-substitution: useful for changing the variable of integration when faced with composite functions.
• Integration by parts: helpful for integrands that are products of functions.
• Trigonometric integrals: involve trigonometric identities.

The success of solving an integral often lies in recognizing which technique to apply, starting with straightforward rules like the power rule and advancing to more complex strategies as needed.