Problem 181 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{1}^{16} \frac{d t}{t^{1 / 4}} $$

Step-by-Step Solution

Answer

The value of the definite integral is \( \frac{28}{3} \).

Step 1: Identify the Integral

We have the definite integral \( \int_{1}^{16} \frac{d t}{t^{1 / 4}} \). This can be rewritten as \( \int_{1}^{16} t^{-1/4} \, dt \).

Step 2: Find the Antiderivative

To apply the Fundamental Theorem of Calculus, we first need to find the antiderivative. Add 1 to the exponent \(-1/4\) to get \(3/4\) and divide by this new exponent: \(F(t) = \frac{t^{3/4}}{3/4} + C = \frac{4}{3}t^{3/4}\).

Step 3: Evaluate the Antiderivative at the Bounds

According to the Fundamental Theorem of Calculus, Part 2, we evaluate \(F(t)\) from 1 to 16: \(F(16) - F(1)\).

Step 4: Calculate \(F(16)\)

Substitute 16 into the antiderivative: \(F(16) = \frac{4}{3}(16)^{3/4} = \frac{4}{3} \times 8 = \frac{32}{3}\).

Step 5: Calculate \(F(1)\)

Substitute 1 into the antiderivative: \(F(1) = \frac{4}{3}(1)^{3/4} = \frac{4}{3}\).

Step 6: Compute the Definite Integral

Subtract the lower bound result from the upper bound result: \(F(16) - F(1) = \frac{32}{3} - \frac{4}{3} = \frac{28}{3}\).

Key Concepts

Definite Integral Evaluation

Evaluating a definite integral is a process that helps us find the net area under a curve between two points on the x-axis. In this exercise, we used the Fundamental Theorem of Calculus to evaluate the definite integral \( \int_{1}^{16} \frac{d t}{t^{1 / 4}} \). This theorem bridges the concept of differentiation and integration, enabling us to tackle problems involving the area.To evaluate this definite integral:

• First, we rewrite the integrand \( \frac{d t}{t^{1 / 4}} \) as \( t^{-1/4} \).
• Then, we identify and calculate the antiderivative, which is essential for using the theorem.
• Finally, we calculate the antiderivative at the two bounds of the integral and find their difference, giving us the value of the definite integral.

Each step requires a good grasp of integration techniques and the rules for exponentiation, which are vital for success in calculus.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative results in the original function we intend to integrate. In this exercise, we find the antiderivative of \( t^{-1/4} \), which is a crucial step before evaluating the definite integral.To calculate the antiderivative, follow these steps:

• Add 1 to the exponent of the term, changing \( -1/4 \) to \( 3/4 \).
• Divide the function by the new exponent, leading to \( \frac{t^{3/4}}{3/4} \) or expressed differently, \( \frac{4}{3}t^{3/4} \).
• Remember to add the constant of integration, \( C \), even though it cancels out during definite integration.

Through this process, we derive a function that describes the accumulation of the original function's value, which sets the stage for the next phase of integration.

Power Rule Integration

The power rule integration is a straightforward technique to find antiderivatives of functions in the form \( t^n \). This rule aids in solving problems like this exercise, where the goal is to integrate \( t^{-1/4} \) using its antiderivative.Here's how to apply the power rule for integration:

• Identify the power of the variable "\( t \)," in this instance \( -1/4 \).
• Increase this exponent by 1 resulting in \( 3/4 \).
• Divide the term by this new exponent, yielding our antiderivative: \( \frac{t^{3/4}}{3/4} \).

This power rule makes it easier to tackle integrals of simple power functions, with the principle being a fundamental tool for solving a wide range of more complex calculus problems.