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Chapter 5: Problem 3

Find the average value of the function over the given interval. $$g(t)=1+t \text { over }[0,2]$$

Short Answer

Expert verified
The average value of the function is 2.

Step by step solution

01

Write the Formula for Average Value

The average value of a function \( g(t) \) over an interval \([a, b]\) is given by the formula: \[ \frac{1}{b-a} \int_{a}^{b} g(t) \, dt \]. In this problem, we have \( g(t) = 1 + t \) over the interval \([0, 2]\).
02

Identify a and b

For the interval \([0, 2]\), we identify \( a = 0 \) and \( b = 2 \). These values will be used in the formula to calculate the average value.
03

Set Up the Integral

Plug \( a = 0 \) and \( b = 2 \) into the formula: \[ \frac{1}{2-0} \int_{0}^{2} (1 + t) \, dt \]. This simplifies to: \[ \frac{1}{2} \int_{0}^{2} (1 + t) \, dt \].
04

Evaluate the Integral

We need to find \( \int_{0}^{2} (1 + t) \, dt \). Start by integrating the function: \[ \int (1 + t) \, dt = t + \frac{t^2}{2} + C \]. Now, evaluate this from 0 to 2: \[ \left[ t + \frac{t^2}{2} \right]_0^2 = \left(2 + \frac{2^2}{2}\right) - \left(0 + \frac{0^2}{2}\right) = 2 + 2 = 4 \].
05

Calculate the Average Value

Substitute the result of the integral back into the formula for the average value: \[ \frac{1}{2} \times 4 = 2 \].

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Key Concepts

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

Definite Integral
The definite integral is a fundamental concept in calculus used to find the accumulation of quantities and areas under curves. It is represented by the integral symbol with limits of integration, such as \( \int_{a}^{b} g(t) \, dt \). This notation signifies that we're integrating the function \( g(t) \) from a starting point \( a \) to an endpoint \( b \).
  • The result of a definite integral is a number that can be interpreted as the net area under the curve described by the function between the specified limits.
  • Unlike indefinite integrals, definite integrals produce an exact numerical output because of these specified limits.
  • In the context of the average value of a function, the definite integral helps us calculate the total accumulation over a certain interval, which can then be averaged out.
Understanding definite integrals is key to solving many real-world problems, like finding distances, areas, volumes, and other quantities that accumulate over time or space.
Interval of Integration
The interval of integration in calculus refers to the range over which we perform the integration of a function. In the exercise discussed, the interval is given as \([0, 2]\).
  • The symbols \( a \) and \( b \) signify the lower and upper bounds of this interval, respectively.
  • These boundaries set the limits within which we want to calculate the area under the curve of the function \( g(t) = 1 + t \).
  • Choosing different values for \( a \) and \( b \) changes the scope and result of the integration process.
The interval of integration is vital for exact calculations as it tells us where to start and stop our integration process, ensuring that we get the right segment of the curve for our calculations.
Function Evaluation
Function evaluation involves substituting specific values into a function to obtain results. It is an important step in calculus and helps in interpreting the integral.
  • In this context, evaluating the function \( t + \frac{t^2}{2} \) at the bounds \( 0 \) and \( 2 \) allows us to calculate the area under the curve.
  • This step involves substitution, where we replace the variable \( t \) with the values of the upper and lower limits of the integration interval.
  • The result shows us the total contribution of the function over that interval, essential for finding averages or differences.
Function evaluation bridges the gap between an abstract formula and its concrete numerical outcome, helping us to solve practical problems in calculus more effectively.
Calculus Problem-Solving
Calculus problem-solving often involves the systematic application of calculus concepts and techniques to find solutions to mathematical queries. It requires a good understanding of definitions, formulas, and procedures.
  • Solving calculus problems typically starts by identifying the relevant information from the problem statement.
  • Next, one must select the appropriate mathematical formula or technique, such as integration, to apply.
  • The process usually ends with actual calculations and evaluations, as seen in the step-by-step solution for the average value of a function.
Effective problem-solving in calculus hinges on clarity and precision in understanding the problem requirements and accurately performing mathematical operations to gain a useful result.
Integration Techniques
Integration techniques encompass various methods used to solve integration problems in calculus. Understanding these techniques is crucial for effectively solving integrals, including definite integrals.
  • The basic technique involves direct antiderivatives, where integrals are solved using known formulas like \( \int (1 + t) \, dt \).
  • For more complex functions, other techniques such as substitution, integration by parts, or partial fraction decomposition might be used.
  • The choice of technique often depends on the form of the function to be integrated and the limits of integration.
By mastering integration techniques, you'll be prepared to solve a wide variety of calculus challenges, including finding the average value of functions over given intervals.

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Most popular questions from this chapter

World annual natural gas \(^{8}\) consumption, \(N,\) in millions of metric tons of oil equivalent, is approximated by \(N=\) \(1770+53 t,\) where \(t\) is in years since 1990 (a) How much natural gas was consumed in \(1990 ?\) In \(2010 ?\) (b) Estimate the total amount of natural gas consumed during the 20 -year period from 1990 to 2010 .

Explain in words what the integral represents and give units. \(\int_{0}^{6} a(t) d t,\) where \(a(t)\) is acceleration in \(\mathrm{km} / \mathrm{hr}^{2}\) and \(t\) is time in hours.

The net worth, \(f(t)\), of a company is growing at a rate of \(f^{\prime}(t)=2000-12 t^{2}\) dollars per year, where \(t\) is in years since \(2005 .\) How is the net worth of the company expected to change between 2005 and \(2015 ?\) If the company is worth \(\$ 40.000\) in \(2005,\) what is it worth in \(2015 ?\)

Use an integral to find the specified area. Under \(y=6 x^{3}-2\) for \(5 \leq x \leq 10\)

Use a calculator or computer to evaluate the integral. $$\int_{1}^{4} \frac{1}{\sqrt{1+x^{2}}} d x$$
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