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Chapter 4: Problem 55

Find the average value of the function on the given interval. \(f(x)=2 x-2 x^{2},[0,1]\)

Short Answer

Expert verified
The average value of the function \(f(x)=2 x-2 x^{2}\) over the interval [0,1] is 1/3.

Step by step solution

01

Integrate the Function Over the Given Interval

Firstly, find the definite integral of the function over the interval [0,1]. This integral can be determined using the power rule of integration. So, 鈭玙{0}^{1} (2 x-2 x^{2}) dx = [x^{2} - (2/3)x^{3}]_{0}^{1} = (1 - (2/3)*(1) which equals to 1/3.
02

Compute the Average Value

The average value of a function over an interval [a, b] is calculated as (1/(b-a)) multiplied by the definite integral of the function from 'a' to 'b'. Here, the width of the interval is (1-0)=1. Therefore, the average value is (1/1) * (1/3) = 1/3.

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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 way to calculate the accumulation of quantities, such as area under a curve, with specific limits of integration. When we refer to the accumulation between two points, we are looking at how the function behaves between these limits. For example, with the function given,
  • we determine the area under the curve from point \(a = 0\) to point \(b = 1\).
  • This is done by integrating the function \(f(x) = 2x - 2x^2\) over the interval \([0, 1]\).
Actually evaluating the definite integral gives us a numerical value, which represents the sum of infinitely small areas that make up the complete area under the curve. This value serves as an essential component in finding the average value of the function over the interval.
Power Rule of Integration
The power rule of integration is a fundamental technique for finding integrals, especially when dealing with functions that are polynomials. This rule states:
  • For any function of the form \(x^n\), the integral is \(\frac{x^{n+1}}{n+1} + C\), where \(n eq -1\).
  • \(C\) is the constant of integration, often omitted in definite integrals.
Using this rule helps in solving the given integral:
  • For \(f(x) = 2x - 2x^2\), apply the rule to each term separately as:\[ \int (2x)\,dx = x^2, \text{ and } \int (-2x^2)\,dx = -\frac{2}{3}x^3.\]
  • Add them to get the integral of the function, \( x^2 - \frac{2}{3}x^3 \).
This powerful rule allows us to systematically compute the integral, aiding in finding the average value over the given interval.
Interval [a,b]
The interval \([a,b]\) defines the range for which the definite integral is calculated, effectively setting the boundaries for our area under the curve calculation. In the context of the average value of a function:
  • \([a, b]\) determines the starting and ending points along the x-axis.
  • Here, \(a = 0\) and \(b = 1\) create limits for integration, delineating where accumulation starts and ends.
By clearly defining this interval, we ensure the integration process focuses on the relevant section of the function. This precise focus allows for accurate computation of the average value between the specific points.
Step-by-Step Solution
Following a step-by-step solution is a methodical approach that breaks down the problem into manageable parts. This technique allows students to systematically solve complex calculus problems. When determining the average value of a function:
  • Step 1 involves calculating the definite integral over the interval, which provides the accumulated value under the curve.
  • Step 2 uses this integral value divided by the interval's length \((b-a)\) to find the average.
In the given exercise, these steps are broken down as:
1. Evaluate \(\int_{0}^{1} (2x - 2x^2)\,dx\) to find the accumulated area.
2. Compute the average as \(\frac{1}{1} \times \frac{1}{3} = \frac{1}{3}\).
Using a step-by-step solution helps maintain clarity, allowing for easy verification of each part, thus ensuring an overall understanding of how the average value is calculated.

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Most popular questions from this chapter

The impulse-momentum equation states the relationship between a force \(F(t)\) applied to an object of mass \(m\) and the resulting change in velocity \(\Delta v\) of the object. The equation is \(m \Delta v=\int_{a}^{b} F(t) d t,\) where \(\Delta v=v(b)-v(a) .\) Suppose that the force of a baseball bat on a ball is approximately \(F(t)=9-10^{8}(t-0.0003)^{2}\) thousand pounds, for \(t\) between 0 and 0.0006 second. What is the maximum force on the ball? Using \(m=0.01\) for the mass of a baseball, estimate the change in velocity \(\Delta v\) (in \(\mathrm{f} t / \mathrm{s}\) ).

Suppose that \(R_{L}\) and \(R_{R}\) are the Riemann sum approximations of \(\int_{a}^{b} f(x) d x\) using left- and right-endpoint evaluation rules, respectively, for some \(n > 0 .\) Show that the trapezoidal approximation \(T_{n}\) is equal to \(\left(R_{L}+R_{R}\right) / 2\)

Use a graph to explain why \(\int_{-1}^{1} x^{3} d x=0 .\) Use your knowledge of \(e^{-x}\) to determine whether \(\int_{-1}^{1} x^{3} e^{-x} d x\) is positive or negative.

Suppose you have a 1 -in- 10 chance of winning a prize with some purchase (like a lottery). If you make 10 purchases (i.e., you get 10 tries), the probability of winning at least one prize is \(1-(9 / 10)^{10} .\) If the prize had probability 1 -in-20 and you tried 20 times, would the probability of winning at least once be higher or lower? Compare \(1-(9 / 10)^{10}\) and \(1-(19 / 20)^{20}\) to find out. To see what happens for larger and larger odds, compute \(\lim _{n \rightarrow \infty}\left[1-((n-1) / n)^{n}\right].\)

Find the average value of the function on the given interval. \(f(x)=x^{2}-1,[1,3]\)
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