Riemann sum is used to estimate the area under a curve in an interval [a, b]. Its formula is
`A ~~ sum_(i=1)^n f(x_i ) Delta x`.
To apply this formula, the interval [a, b] is subdivided into subintervals/rectangles of equal width (refer to figure 1). The width of each rectangle is
`Delta x = (b - a)/n`,
where n is the number of subintervals/rectangles.
To determine the height of each rectangle, plug in the value of xi to the function of the curve f(x).
`height = f(x_i)`
In midpoint Riemman sum, the xi is the middle x value of each subinterval.
`x_i= (x_i + x_(i+1))/2`
When the height of each rectangle is known, compute the area of each rectangle by multiplying the height and width.
`A_i = f(x_i) Delta x`
To get the estimate of the area under the curve, add the area of all the rectangles present from A_1 to A_n.
`A ~~ A_1+A_2+ ... + A_n`
`A~~ f(x_1)Delta x + f(x_2)Delta x+ ... + f(x_n) Delta x`
Factoring out the width, this becomes
`A~~ Delta x ( f(x_1) + f(x_2)+...+f(x_n))`
Thus, `A~~ sum _(i=1)^n f(x_i) Delta x = Delta x(f(x_1) + f(x_2) + ... + f(x_n)).`
To show its application, consider this example.
Example: Use midpoint Riemann sum to estimate the area between the function `f(x) = x^2+2` and the x-axis on the interval [0, 6]. Use 3 subintervals (n=3).
To solve, graph the given function. Then, subdivide the region between the curve and the x-axis to three rectangles. (Refer to Figure 2.)
The width of each rectangle is
`Delta x = (b - a)/n = (6-0)/3 = 2`.
Then, determine the midpoint of each subinterval.
The first subinterval is [0, 2]. So the midpoint of the first subinterval is
`x_1 = (0+2)/2 = 1`.
The second subinterval is [2, 4]. So the midpoint of the second subinterval is
`x_2 = (2+4)/2 = 3`.
The third subinterval is [4, 6]. So the midpoint of the third subinterval is
`x_3 = (4+6)/2 = 5`.
Use these midpoints of each subinterval to get the height of each rectangle.
Height of first rectangle: `f(x_1) = 1^2 + 2 = 3`
Height of the second rectangle: `f(x_2) = 3^2+2 = 11`
Height of the third rectangle: `f(x_3) = 5^2+2 = 27`
Applying the formula above, the estimated area is
`A~~ sum_(i=1)^3 f(x_i) Delta x = Delta x[ f(x_1+ f(x_2) + f(x_3)]`
`A ~~ 2*(3 + 11 + 27)`
`A~~ = 2 * 41`
`A~~ = 82`
Therefore, the area of the region between f(x) and the x-axis on the interval [0, 6] is approximately 82 square units.
