Question

Find the weighted average \(\bar{u}\) of \(u(x)\) over \([a, b]\) with respect to \(v,\) and find a point \(c\) in \([a, b]\) such that \(u(c)=\bar{u}\). \(\begin{array}{ll}\text { (a) } u(x)=x, & v(x)=x, & {[a, b]=[0,1]}\end{array}\) (b) \(u(x)=\sin x, \quad v(x)=x^{2}, \quad[a, b]=[-1,1]\) (c) \(u(x)=x^{2}, \quad v(x)=e^{x}, \quad[a, b]=[0,1]\)

Short answer

Question: Calculate the weighted average 𝜇 of the function u(x) for each given condition and find a point c in the interval [a, b] such that u(c) = 𝜇. a) u(x) = x, v(x) = x, [a, b] = [0, 1] b) u(x) = sin x, v(x) = x^2, [a, b] = [-1, 1] c) u(x) = x^2, v(x) = e^x, [a, b] = [0, 1] Answer: a) The weighted average 𝜇 for u(x) = x, v(x) = x, [a, b] = [0, 1] is 2/3, and the point c in the interval [0, 1] such that u(c) = 𝜇 is c = 2/3. b) The weighted average 𝜇 for u(x) = sin x, v(x) = x^2, [a, b] = [-1, 1] is 0, and the points c in the interval [-1, 1] such that u(c) = 𝜇 are c = 0, and c = 𝜋. c) The weighted average 𝜇 for u(x) = x^2, v(x) = e^x, [a, b] = [0, 1] is (2 - e - e^2 + 3e^1) / (e^1 - 1), and the point c in the interval [0, 1] such that u(c) = 𝜇 is approximately c ≈ 0.7823.

Step-by-step solution

Find Integral u(x)v(x)

Integrate u(x)v(x) with respect to x over [0, 1]: $$\int_0^1 x \cdot x dx = \int_0^1 x^2 dx$$.

Find Integral v(x)

Integrate v(x) with respect to x over [0, 1]: $$\int_0^1 x dx$$.

Calculate Weighted Average 𝜇

Compute the weighted average with: $$\bar{u} = \frac{\int_0^1 x^2 dx}{\int_0^1 x dx} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$.

Find a Point c in [a, b] Such That u(c) = 𝜇

We want to find a point c such that u(c) = 𝜇, which means we need to find a c such that $$x = \frac{2}{3}$$. This means our point c is: $$c = \frac{2}{3}$$. ## (b) ## For the second part, we have \(u(x) = \sin x, \space v(x) = x^2, [a, b] = [-1, 1]\).

Find Integral u(x)v(x)

Integrate u(x)v(x) with respect to x over [-1, 1]: $$\int_{-1}^1 \sin x \cdot x^2 dx$$.

Find Integral v(x)

Integrate v(x) with respect to x over [-1, 1]: $$\int_{-1}^1 x^2 dx$$.

Calculate Weighted Average 𝜇

Compute the weighted average with: $$\bar{u} = \frac{\int_{-1}^1 \sin x \cdot x^2 dx}{\int_{-1}^1 x^2 dx} = \frac{0}{\frac{2}{3}} = 0$$.

Find a Point c in [a, b] Such That u(c) = 𝜇

We want to find a point c such that u(c) = 𝜇, which means we need to find a c such that $$\sin x = 0$$. There are two points in the given interval that satisfies this condition: $$c = 0, c = \pi$$. ## (c) ## For the third part, we have \(u(x) = x^2, \space v(x) = e^x, \space [a, b] = [0, 1]\).

Find Integral u(x)v(x)

Integrate u(x)v(x) with respect to x over [0, 1]: $$\int_0^1 x^2 \cdot e^x dx$$.

Find Integral v(x)

Integrate v(x) with respect to x over [0, 1]: $$\int_0^1 e^x dx$$.

Calculate Weighted Average 𝜇

Compute the weighted average with: $$\bar{u} = \frac{\int_0^1 x^2 \cdot e^x dx}{\int_0^1 e^x dx} = \frac{2 - e - e^2 + 3e^1}{e^1 - 1}$$.

Find a Point c in [a, b] Such That u(c) = 𝜇

To find a point c such that u(c) = 𝜇, we need to find a c such that $$x^2 = \frac{2 - e - e^2 + 3e^1}{e^1 - 1}$$. This value of c can be found numerically or using a computer. Approximately, $$c ≈ 0.7823$$.

Key concepts

Real Analysis

Real Analysis focuses on the study of real numbers and real-valued functions. It involves concepts such as sequences, series, limits, and functions. Real Analysis provides a rigorous foundation for understanding properties and behaviors of functions. When considering isotonic functions, for example, we look at functions that either consistently increase or decrease on specified intervals.

1. Real numbers: Denoted as \( \mathbb{R} \), these include all numbers on the number line, from negatives to positives, including irrational numbers.
2. Basic properties: Concepts such as closure, associativity, and distributivity are fundamental in real analysis.
3. Limits and continuity: These concepts are crucial in understanding the changes and behaviors of functions over intervals. Limits involve approaching a value, while continuity ensures no breaks in a function's graph.
4. Derivative and Integration: Both concepts are core to calculus, helping in understanding function rates and areas under curves.

This foundation is pivotal when solving problems involving weighted averages, as seen in exercises involving real numbers and their mappings through functions.

Integration

Integration is a powerful mathematical tool used to calculate the area under a curve or the accumulation of quantities. It is often seen as the inverse operation of differentiation. The process of integration can be used for calculating things such as

• Area under a function,
• Work done by a force over distance,
• Center of mass in physics, among other applications.

When integrating a function, you are essentially adding up infinitely many infinitesimal slices to find a total. The definite integral, \[\int_a^b f(x) \; dx,\]represents the total accumulation from \( x = a \) to \( x = b \). The integration process requires selecting a method of integration like substitution or integration by parts, depending on the integrands.For example, in evaluating the weighted average, we integrate the function \( u(x)\times v(x)\) over the interval \([a, b]\), as seen in the original problem. Integration enables us to consider the entire influence of every portion of the curve.

Weighted Integrals

Weighted integrals incorporate an extra function, known as the weight function, during integration. This allows different parts of the integrand to contribute differently to the total integral value.For example, given the functions \( u(x) \) and \( v(x) \), the weighted integral is expressed as\[\int_a^b u(x) v(x) \; dx.\]The function \( v(x) \) plays the role of the weight; regions where \( v(x)\) is larger have a greater contribution to the integral than regions where it is smaller.A weighted average uses these concepts to modify typical averaging to account for varying importance of different parts of the data set.The equation to find a weighted average \(\bar{u}\) would look like\[\bar{u} = \frac{\int_a^b u(x)v(x) \; dx}{\int_a^b v(x) \; dx}.\]This formulation highlights how each segment of the interval \([a, b]\) has varying contributions based on corresponding weights, providing a more tailored average depending on the context and nature of \( v(x) \). Through understanding weighted integrals, students can address various problems involving non-uniform data weightage distributions.