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Chapter 5: Problem 39

Evaluate the integral. $$ \int_{1}^{8} \frac{2+t}{\sqrt[3]{t^{2}}} d t $$

Short Answer

Expert verified
The integral evaluates to 17.25.

Step by step solution

01

Simplify the Integrand

First, simplify the integrand \( \frac{2+t}{\sqrt[3]{t^{2}}} \). Recognize that \( \sqrt[3]{t^{2}} = t^{2/3} \). So the integrand becomes \( (2+t) t^{-2/3} \), which can be separated into \( 2t^{-2/3} + t^{1 - 2/3} = 2t^{-2/3} + t^{1/3} \).
02

Integrate Each Term

Now, integrate each term separately:1. For \( \int 2t^{-2/3} \, dt \), use the power rule: the integral is \( 2 \cdot \frac{t^{1/3}}{1/3} = 6t^{1/3} \).2. For \( \int t^{1/3} \, dt \), apply the power rule again to get \( \frac{t^{4/3}}{4/3} = \frac{3}{4}t^{4/3} \).
03

Write the Result of Integration

Combine the results from Step 2:\[ \int (2t^{-2/3} + t^{1/3}) \, dt = 6t^{1/3} + \frac{3}{4}t^{4/3} + C \]
04

Evaluate the Definite Integral

Evaluate the definite integral from 1 to 8.- Plug in the upper limit 8: \[ 6(8^{1/3}) + \frac{3}{4}(8^{4/3}) = 6 \times 2 + \frac{3}{4} \times 16 = 12 + 12 = 24 \]- Plug in the lower limit 1: \[ 6(1^{1/3}) + \frac{3}{4}(1^{4/3}) = 6 \times 1 + \frac{3}{4} \times 1 = 6 + \frac{3}{4} = 6.75 \]
05

Subtract the Results

Subtract the value of the integral at the lower limit from the value at the upper limit:\[ 24 - 6.75 = 17.25 \]

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Key Concepts

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

Power Rule for Integration
When we talk about the `Power Rule for Integration`, we refer to a key method used to integrate functions that are powers of a variable. It's similar to the power rule in differentiation, but we're doing the reverse operation. Here鈥檚 a simple way to apply it:
  • If you have a term like \( t^n \), where \( n \) is not equal to -1, you can integrate it by adding 1 to the exponent and then dividing by the new exponent.
  • Mathematically, the integral of \( t^n \) is \( \frac{t^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.
  • For example, integrating \( t^{1/3} \) gives \( \frac{t^{4/3}}{4/3} \), which simplifies to \( \frac{3}{4}t^{4/3} \).
This rule is incredibly powerful for breaking down complex expressions into manageable pieces that we can integrate one term at a time.
Simplifying Integrands
Before integrating, it's often necessary to simplify the integrand to make the integration process easier. Simplifying involves re-writing expressions in a way that is easier to integrate. Let's walk through how we simplify an integrand:
  • First, express any radicals or roots in terms of exponents. For example, \( \sqrt[3]{t^2} = t^{2/3} \).
  • Then, distribute and simplify the expression, such as rewriting \( \frac{2+t}{t^{2/3}} \) as \( 2t^{-2/3} + t^{1/3} \).
  • This simplification allows us to clearly identify the power of each term, making the use of the power rule straightforward.
By simplifying the integrand, we transform complex expressions into a sum of simpler terms that follow straightforward integration rules.
Evaluating Integrals
`Evaluating Integrals` means finding the area under the curve for a given function, especially between two points, which is often called a definite integral. Here鈥檚 how we go about evaluating definite integrals:
  • First compute the integral of the function, known as an antiderivative.
  • Apply the limits of integration by plugging in the upper and lower bounds into the antiderivative.
    In our example, we evaluate \( 6(8^{1/3}) + \frac{3}{4}(8^{4/3}) \) versus \( 6(1^{1/3}) + \frac{3}{4}(1^{4/3}) \).
  • Subtract the lower bound evaluation from the upper bound evaluation to get the final result.
  • In the exercise example, this process gives us \( 24 - 6.75 = 17.25 \).
Evaluating definite integrals not only gives us a numerical value representing the area under a curve but also helps us understand the behavior of functions within certain bounds.

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