Question

Mean Value Theorem for Integrals Find or approximate all points at which the given function equals its average value an the given interval. \(f(x)=\frac{\pi}{4} \sin x\) on \([0, \pi] \quad\) 43. \(f(x)=1-|x|\) on [-1,1]

Short answer

Question: For the given functions \(f(x)=\frac{\pi}{4} \sin x\) on \([0, \pi]\) and \(f(x)=1-|x|\) on [-1,1], find all points at which the function equals its average value on the given interval. Answer: For the function \(f(x)=\frac{\pi}{4} \sin x\) on \([0, \pi]\), there are no points at which the function equals its average value. For the function \(f(x)=1-|x|\) on [-1,1], the points at which the function equals its average value are -\(\frac{1}{2}\) and \(\frac{1}{2}\).

Step-by-step solution

Function 1: Calculate the Average Value

For the function \(f(x)=\frac{\pi}{4}\sin x\) on the interval \([0, \pi]\), the average value can be calculated as: \(Avg(f)=\frac{1}{\pi-0} \int_{0}^{\pi} \frac{\pi}{4}\sin x dx\) Calculate the integral: \(\int_{0}^{\pi} \frac{\pi}{4}\sin x dx = \frac{\pi}{4}[-\cos x]_{0}^{\pi} = \frac{\pi}{4}(2) = \frac{\pi}{2}\) Now, the average value of \(f(x)\) on \([0, \pi]\) is \(\frac{\pi}{2}\).

Function 1: Find Points Equal to Average Value

To find points where \(f(x)\) is equal to its average value \(\frac{\pi}{2}\), we will set \(f(x) = \frac{\pi}{2}\): \(\frac{\pi}{4} \sin x = \frac{\pi}{2}\) Now, divide both sides by \(\pi\): \(\frac{1}{4} \sin x = \frac{1}{2}\) Solve for x: \(\sin x = 2\) But since the sine function has a range of [-1,1], there are no points on the interval \([0, \pi]\) where \(f(x)\) equals its average value of \(\frac{\pi}{2}\).

Function 2: Calculate the Average Value

For the function \(f(x)=1-|x|\) on the interval \([-1,1]\), the average value can be calculated as: \(Avg(f)=\frac{1}{1-(-1)} \int_{-1}^{1} (1-|x|) dx\) We will evaluate this integral by breaking it into two parts, one for \([-1,0]\) and one for \([0,1]\): \(\int_{-1}^{1} (1-|x|) dx = \int_{-1}^{0} (1+x) dx + \int_{0}^{1} (1-x) dx\) Calculate the integrals: \(\int_{-1}^{0} (1+x) dx = [x+\frac{1}{2}x^2]_{-1}^{0}=0-(-\frac{1}{2}) = \frac{1}{2}\) \(\int_{0}^{1} (1-x) dx = [x-\frac{1}{2}x^2]_{0}^{1}=1-\frac{1}{2} = \frac{1}{2}\) Now, add the results from the two integrals: \(Avg(f) = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{2} \right) = \frac{1}{2}\) Now, the average value of \(f(x)\) on [-1,1] is \(\frac{1}{2}\).

Function 2: Find Points Equal to Average Value

To find points where \(f(x)\) is equal to its average value \(\frac{1}{2}\), we will set \(f(x) = \frac{1}{2}\): \(1- |x| = \frac{1}{2}\) Solve for x when \(x<0\): \(1-(-x)=\frac{1}{2} \Rightarrow x=-\frac{1}{2}\) Solve for x when \(x\geq0\): \(1-x=\frac{1}{2} \Rightarrow x=\frac{1}{2}\) So, there are two points on the interval \([-1,1]\) where \(f(x)\) equals its average value of \(\frac{1}{2}\): \(-\frac{1}{2}\) and \(\frac{1}{2}\).

Key concepts

average value of a function

The average value of a function provides a single number that represents the overall behavior of the function across an interval. It kind of tells us, on average, how high or low the function is rising or dropping. To find this value between any two points, say from point \( a \) to point \( b \), you'll perform a definite integral of the function over this interval. This operation sums up all the continuous changes that the function experiences over this path.

Its expression is given by dividing the result of the integral by the length of the interval, which is \( b-a \). Here’s the formula:

\[\text{Average value of } f(x) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]

When you see this in action, like with the function \(f(x)=\frac{\pi}{4}\sin x\) on the interval \([0,\pi]\), it provides the average value that this sine function takes across the given path. This approach is fundamental in understanding what the "mean value" means in the context of integrals.

definite integral

A definite integral is a key concept in calculus that aids in finding the total area under the curve of a function over a specified interval from point \(a\) to point \(b\).

Think of it as a powerful tool to measure how a function behaves across a segment. Unlike an indefinite integral, which provides a general form of antiderivatives, a definite integral gives a specific numerical result.

If you consider a function \(f(x)\) resembling a rollercoaster across a track, then the definite integral gives you the complete experience of the ride across your chosen boundaries.

The calculation involves finding the antiderivative and computing the difference between its values at \(b\) and \(a\). Mathematically, it is expressed as:

\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]

For example, integrating \(\frac{\pi}{4}\sin x\) from \(0\) to \(\pi\) as shown in the solution helps evaluate the total change experienced via this sin curve. The result leads to knowing specific values like the average, providing insight into the aggregate behavior from start to finish.

integration techniques

Integration techniques are the strategies employed to solve integrals effectively. Not all functions are straightforward to integrate, and sometimes we utilize particular methods depending on the function's complexity.

Methods such as substitution, integration by parts, and partial fraction decomposition are frequently used strategies to simplify and calculate integrals accurately. Each technique calls for understanding the nature of the function you're working with.

• Substitution: This method is analogous to reversing the chain rule where you pick part of the integral to substitute with a new variable, simplifying the expression.
• Integration by Parts: Useful when dealing with products of functions, it stems from the product rule of differentiation and often involves choosing which function part to differentiate and integrate.
• Partial Fraction Decomposition: Mainly used for rationals with complex fractions, where the fraction is broken down into simpler fractions to integrate separately.

In our solutions, simple substitutions might have been used to handle terms like \(\sin x\), allowing the integral calculation to be more direct and easier to solve, demonstrating how these techniques are pivotal in finding precise integral outcomes.