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Chapter 5: Problem 181

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

Short Answer

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

Step by step solution

01

Identify the Integral

We have the definite integral \( \int_{1}^{16} \frac{d t}{t^{1 / 4}} \). This can be rewritten as \( \int_{1}^{16} t^{-1/4} \, dt \).
02

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: \(F(t) = \frac{t^{3/4}}{3/4} + C = \frac{4}{3}t^{3/4}\).
03

Evaluate the Antiderivative at the Bounds

According to the Fundamental Theorem of Calculus, Part 2, we evaluate \(F(t)\) from 1 to 16: \(F(16) - F(1)\).
04

Calculate \(F(16)\)

Substitute 16 into the antiderivative: \(F(16) = \frac{4}{3}(16)^{3/4} = \frac{4}{3} \times 8 = \frac{32}{3}\).
05

Calculate \(F(1)\)

Substitute 1 into the antiderivative: \(F(1) = \frac{4}{3}(1)^{3/4} = \frac{4}{3}\).
06

Compute the Definite Integral

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

In the following exercises, use a change of variables to show that each definite integral is equal to zero. $$ \int_{0}^{1}(1-2 t) d t $$

Find the antiderivative. $$\int \frac{e^{2 x}}{1+e^{4 x}} d x$$

In the following exercises, use a change of variables to show that each definite integral is equal to zero. $$ \int_{0}^{\sqrt{\pi}} t \cos \left(t^{2}\right) \sin \left(t^{2}\right) d t $$

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In the following exercises, approximate the average value using Riemann sums \(L_{100}\) and \(R_{100} .\) How does your answer compare with the exact given answer? [T] \(y=e^{x / 2}\) over the interval \([0,1] ;\) the exact solution is 2\((\sqrt{e}-1)\)
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