Question

Riemann sums for linear functions Assume the linear function \(f(x)=m x+c\) is positive on the interval \([a, b] .\) Prove that the midpoint Riemann sum with any value of \(n\) gives the exact area of the region between the graph of \(f\) and the \(x\) -axis on \([a, b]\)

Short answer

Question: Show that the midpoint Riemann sum of a linear function \(f(x) = mx + c\) on the interval \([a,b]\) gives the exact area under the graph of the function. Answer: To show that the midpoint Riemann sum gives the exact area under the graph of a linear function, we first found the integral of the function, which gives the exact area under the graph within the interval \([a, b]\). After evaluating the integral, we defined the midpoint Riemann sum and evaluated it for the linear function on the given interval. Taking the limit of the midpoint Riemann sum as the number of subintervals approaches infinity, we showed that it converges to the exact area found using the integral, proving that the midpoint Riemann sum gives the exact area under the graph of the linear function \(f(x) = mx + c\) on the interval \([a,b]\).

Step-by-step solution

Define the linear function and its integral

We are given a linear function \(f(x) = mx + c\) defined on the interval \([a,b]\). To find the exact area, we need to find the integral of this function. The integral of a linear function \(f(x) = mx + c\) is given by: $$\int(mx + c) \, dx = \frac{1}{2}mx^2 + cx + k$$ where \(k\) is the constant of integration.

Calculate the exact area under the graph of \(f(x)\) on \([a, b]\)

To find the exact area under the graph of the linear function \(f(x)\) on the interval \([a, b]\), we need to evaluate the integral of \(f(x)\) on this interval. This can be calculated as: $$\int_a^b (mx + c) \, dx = \left.\left(\frac{1}{2}mx^2 + cx + k\right)\right|_a^b = \left(\frac{1}{2}mb^2 + cb + k\right) - \left(\frac{1}{2}ma^2 + ca + k\right)$$ Simplifying, we get: $$\int_a^b f(x) \, dx = \frac{1}{2}m(b^2 - a^2) + c(b - a) = \frac{1}{2}m(b - a)(b + a) + c(b - a)$$

Define the midpoint Riemann sum

The midpoint Riemann sum of a function \(f(x)\) on the interval \([a, b]\) using \(n\) subintervals is given by: $$M_n = \sum_{i=1}^n f\left(\frac{x_{i-1}+x_i}{2}\right)(x_i - x_{i-1})$$ where \(x_i = a + i\Delta x\) and \(\Delta x = \frac{b - a}{n}\).

Evaluate the midpoint Riemann sum for the linear function \(f(x)\)

To evaluate the midpoint Riemann sum for our linear function \(f(x) = mx + c\) on the interval \([a, b]\), we substitute the function into the Riemann sum formula: $$M_n = \sum_{i=1}^n f\left(\frac{x_{i-1}+x_i}{2}\right)(x_i - x_{i-1})$$ $$M_n = \sum_{i=1}^n (m\left(\frac{x_{i-1}+x_i}{2}\right) + c)(x_i - x_{i-1})$$ $$M_n = \sum_{i=1}^n \left(\frac{m}{2}(x_{i-1}+x_i) + c\right)(x_i - x_{i-1})$$ $$M_n = \frac{m}{2}\sum_{i=1}^n (x_{i-1} + x_i)(x_i - x_{i-1}) + c\sum_{i=1}^n (x_i - x_{i-1})$$ Recall that \(x_i = a + i\Delta x\) and \(\Delta x = \frac{b - a}{n}\). Using these values, we can continue to simplify the Riemann sum: $$M_n = \frac{m}{2}\sum_{i=1}^n (a + (i - 1)\Delta x + a + i\Delta x)(\Delta x) + c\sum_{i=1}^n \Delta x$$ $$M_n = \frac{m}{2}\sum_{i=1}^n (2a + (2i - 1)\Delta x)(\Delta x) + c\sum_{i=1}^n \Delta x$$ $$M_n = \frac{m}{2}\sum_{i=1}^n (2a\Delta x + (2i - 1)(\Delta x)^2) + c\sum_{i=1}^n \Delta x$$ Since \(\sum_{i=1}^n 1 = n\), \(\sum_{i=1}^n i = \frac{n(n + 1)}{2}\), and \(\sum_{i=1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}\), we can substitute these values into the sum: $$M_n = \frac{m}{2}\left(2an + \frac{n(n + 1)(2n + 1)}{6}(2\Delta x)\right) + cn$$ $$M_n = \frac{m}{2}\left(2an + \frac{n(n + 1)(b - a)}{3}\right) + cn$$

Show that the midpoint Riemann sum is equal to the exact area under the graph of \(f(x)\) on \([a, b]\)

From steps 2 and 4, the exact area \(A\) and the midpoint Riemann sum \(M_n\) are given by: $$A = \frac{1}{2}m(b - a)(b + a) + c(b - a)$$ $$M_n = \frac{m}{2}\left(2an + \frac{n(n + 1)(b - a)}{3}\right) + cn$$ Taking the limit as \(n\) goes to infinity, the midpoint Riemann sum approaches the exact area: $$\lim_{n \to \infty} M_n = \lim_{n \to \infty}\left(\frac{m}{2}\left(2an + \frac{n(n + 1)(b - a)}{3}\right) + cn\right)$$ As \(n\) approaches infinity, both terms within the brackets approach zero: $$\lim_{n \to \infty} 2an = 0$$ $$\lim_{n \to \infty} \frac{n(n + 1)(b - a)}{3} = b - a$$ Thus, the limit of the midpoint Riemann sum is: $$\lim_{n \to \infty} M_n = \frac{1}{2}m(b - a)(b + a) + c(b - a) = A$$ We have shown that the midpoint Riemann sum with any value of \(n\) gives the exact area of the region between the graph of the linear function \(f(x) = mx + c\) and the \(x\)-axis on the interval \([a, b]\).

Key concepts

Integral of Linear Functions

Integrating linear functions is a fundamental concept when exploring the area under a curve. A linear function can be generally written as \(f(x) = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. The integral of \(f(x)\), over an interval from \(a\) to \(b\), represents the accumulated area between the function's curve and the x-axis over that range. This integral is evaluated using the antiderivative of \(f(x)\).

For our linear function, the antiderivative comes out to be \(\frac{1}{2}mx^2 + cx + k\), wherein \(k\) is a constant that gets canceled out when computing the definite integral. This makes computing the integral of a linear function especially simple, as it boils down to a basic algebraic expression that neatly encapsulates the area under the line between two points.

Area Under a Curve

The area under a curve for a function is a visual representation of the integral of the function over a given interval. In the case of a linear function, this area forms a trapezoid between the function graph, the x-axis, and the vertical lines at \(x=a\) and \(x=b\).

The calculation of the area under the curve is not merely an abstract concept; it has practical applications, such as computing distances traveled over time when given a velocity function or finding the total quantity when given a rate of change. By understanding the geometric interpretation of integrals, students can internalize the concept better and apply it to a variety of problems in mathematics and physics.

Definite Integral Calculation

The definite integral of a function over an interval provides us with a precise value representing the area under the curve from one point to another. Calculating a definite integral involves finding the difference between the values of the antiderivative function at the endpoint of the interval. This is illustrated by the fundamental theorem of calculus.

To get the exact area \(A\) under the curve of our linear function \(f(x) = mx + c\), we evaluate the antiderivative at \(b\) and subtract the value of the antiderivative at \(a\). The steps provided in the solution illustrate this process and highlight the simplicity of calculating the definite integral for linear functions specifically.

Limits and Riemann Sums

The concept of Riemann sums is pivotal in understanding integration as they provide a method for approximating the area under a curve. When a function is divided into \(n\) subintervals, the Riemann sum takes values of the function at certain points within these intervals, multiplies them by the width of the interval, and adds them all together for an estimate of the total area.

The midpoint Riemann sum is particularly useful because it uses the midpoint of each subinterval for evaluation, leading to a typically more accurate approximation. By taking the limit of this sum as \(n\) approaches infinity, we can find the exact area represented by the definite integral. The linear nature of the function in the exercise allows the Riemann sum to precisely match the definite integral even for finite \(n\), thus proving the initial assertion that the midpoint Riemann sum gives the exact area under a linear function over the interval from \(a\) to \(b\).