Question

Let \(f:[-1,2] \rightarrow[0, \infty)\) be a continuous function such that \(f(x)=f(1-x)\) for all \(x \in[-1,2]\). Let \(R_{1}=\int_{-1}^{2} x f(x) d x\), and \(R_{2}\) be the area of the region bounded by \(y=f(x), x=-1, x=2\), and the \(x\) -axis. Then (A) \(R_{1}=2 R_{2}\) (B) \(R_{1}=3 R_{2}\) (C) \(2 R_{1}=R_{2}\) (D) \(3 R_{1}=R_{2}\)

Short answer

Since \(R_1\) simplifies to 0 due to the symmetry of the integral about x = 1/2, none of the options where R1 is proportional to R2 can be correct since R2 is the area under a positive function. Thus, none of the given options (A), (B), (C), or (D) are correct.

Step-by-step solution

Break down the properties of the function

Identify that the function described is symmetrical with respect to the line x = 1/2. This means that the values of the function for x and 1-x are the same, creating a mirror image around the line x = 1/2.

Expressing the integral for R1

Write the integral for R1 as two separate integrals using the property of the function: \(R_1 = \int_{-1}^{2} xf(x) dx\), splitting it at the line of symmetry, x = 1/2, we get \(R_1= \int_{-1}^{1/2} xf(x) dx + \int_{1/2}^{2} xf(x) dx\).

Use the property of the function on the integrals

By substituting \(1-x\) for \(x\) in the second integral, we can use the symmetry of the function to show that \(R_1 = \int_{-1}^{1/2} xf(x) dx + \int_{-1/2}^{1} (1-x)f(1-x) dx\). Now using the property \(f(x)=f(1-x)\), the second integral becomes \(\int_{-1/2}^{1} (1-x)f(x) dx\).

Combine the integrals using common limits

Now change the limits of the second integral from -1/2 to 1 to 1/2 to 1 by reversing the direction of integration: \(R_1 = \int_{-1}^{1/2} xf(x) dx - \int_{1}^{1/2} xf(x) dx = \int_{-1}^{1/2} xf(x) dx - \int_{1/2}^{-1} xf(x) dx\).

Simplify the expression for R1

Adding the two integrals, we get: \(R_1 = \int_{-1}^{1/2} xf(x) dx - \int_{1/2}^{-1} xf(x) dx = 0\). This simplification is due to the fact that the function is symmetrical and the two integrals cancel each other out as they are over the same interval but with opposite signs.

Calculate the area R2

To find R2, evaluate the integral of the function over the range without weighting by x (since area is just the integral of f(x) over the interval [-1,2]): \(R_2 = \int_{-1}^{2} f(x) dx\).

Compare R1 and R2

Knowing that R1 is zero and R2 is a positive number due to the range of f mapping to \([0, \infty)\), we can compare them to find the correct answer.

Key concepts

Mathematical Symmetry

Symmetry in mathematics refers to a kind of balance or harmony in the arrangement of a function or shape. When a function exhibits symmetry, it means that there is a specific point, line, or plane where the function creates a mirror image of itself. In the context of our exercise, the function is symmetrical with respect to the line x = 1/2. This symmetry has practical implications when calculating integrals and areas under curves, as it can simplify computations or result in certain values being equal or cancelling each other out.

Understanding symmetry is crucial as it can significantly simplify mathematical problems, especially when working with definite integrals. By recognizing symmetry, you can split integrals at the line of symmetry, reducing complex problems into simpler, more manageable parts. In some cases, like with even or odd functions in symmetrical intervals, integrals can be evaluated much more swiftly, sometimes even leading to immediate conclusions about their value without explicit calculation.

Definite Integrals

A definite integral represents the net area under the curve of a function between two specified points. It is one of the central concepts in calculus and serves various purposes such as calculating areas, volumes, central points, among others. The definite integral of a function from a to b is denoted as \(\int_{a}^{b} f(x) dx\), and it is numerically equivalent to the signed area under the graph of the function and above the x-axis, from x=a to x=b.

The computation of a definite integral can be visualized as the summing up of infinitely small rectangles under the curve, and this is the foundation of the Riemann integral. By breaking down an integral into symmetry-based components, as we did with our function, we leverage the properties to solve the integral more efficiently. In cases where we deal with symmetrical functions, we can often use these symmetries to our advantage to either simplify or directly evaluate the integrals.

Area Under Curve

The concept of 'area under the curve' stems directly from definite integrals in calculus. Essentially, when we talk about the area under the curve, we refer to the size of the region bounded by the graph of a given function, the x-axis, and the vertical lines corresponding to the boundaries of the integration interval. For non-negative functions, this area is exactly the value of the definite integral of the function over that interval.

In the problem we are considering, the integral \(R_2 = \int_{-1}^{2} f(x) dx\) directly represents the area under the curve of the function f(x) from x = -1 to x = 2. A primary use of integrals in real-world applications is to calculate such areas, which in turn can represent physical quantities like distance traveled, the work done by a force, or any other quantity where the integral adds up a distribution along an interval. Recognizing the geometric meaning of these calculations often aids in their interpretation and appreciation of their practical relevance.