Calculus > Riemann Sum

## What is a Riemann Sum?

A Riemann sum is used in calculus as one way to approximate the area under a curve — which is the same as calculating an integral. There are five main ways to calculate Riemann sums:

The **left-hand Reimann sum **and the **right-hand Reimann sum** use the left and the right endpoints of each subinterval to approximate the area. With the right-hand sum, each rectangle is drawn so that the upper-right corner touches the curve; with the left-hand sum, the upper-left corner touches the curve.

The trapezoid rule uses an average of the left- and right-hand values. The trapezoids hug the curve better than left- or right- hand rule rectangles and so give you a better estimate of the area.

The midpoint rule uses the midpoint of the rectangles for the estimate. A midpoint rule is a much better estimate of the area under the curve than either a left- or right- sum. As a rule of thumb, midpoint sums are twice as good than trapezoid estimates.

**Simpson’s rule** uses parabolas and is an extremely accurate approximation method. It will give the exact area for any polynomial of third degree or less.

Simpson’s rule uses a combination of the midpoint rules and trapezoid rules, so if you have already calculated the midpoint and trapezoid areas, it’s a simple way to get a more accurate approximation. The subscript 2n in the equation means that if you use M_{1} and T_{1}, you get S_{2}, if you use M_{2} and T_{2}, you get S_{4}.

S_{2n}=(^{Mn+Mn+Tn})⁄_{3}

## Sample Problem (Right Hand Riemann)

**Sample problem:** Find the area under the curve from x=0 to x=2 for the function x^{3} using the right endpoint rule.

Step 2: **Draw a series of rectangles** under the curve, from the x-axis to the curve. We’ll use four rectangles for this example. The question asks for the right endpoint rule, so draw your rectangles using points furthest to the right. Place your pen on the endpoint (the first endpoint to the right is 0.5), draw up to the curve and then draw left to the y-axis to form a rectangle.

Step 3: **Calculate the area of each rectangle** by multiplying the height by the width.

Interval |
0 to 0.5 | 0.5 to 1 | 1 to 1.5 | 1.5 to 2 |

Height |
0.125 | 1 | 3.375 | 8 |

Width |
.5 | .5 | .5 | .5 |

W * H |
.0625 | .5 | 1.6875 | 4 |

Step 4: **Add all of the rectangle’s areas **together to find the area under the curve: .0625 + .5 + 1.6875 + 4 = 6.25

*That’s it!*

**Tip:**The number of rectangles is arbitrary — you can use as many, or as few, as you want. However, the more rectangles you use, the better the approximation will be to the actual area.

**Tip:**The right-hand curve is an overestimate of the actual area and the left-hand rule is an underestimate.

**Tip:**While the left-hand rule, the right-hand rule and the midpoint rule use rectangles, The trapezoid rule uses trapezoids.

## Example 2: Midpoint Riemann Sum

**Sample question**: Calculate a Riemann sum for f(x)=x^{2}+2 on the interval [2,4] using n=8 rectangles and midpoint evaluation.

Step 1: **Divide the interval into segments.** For this sample problem, divide the x-axis into 8 intervals.

Step 2: **Find the midpoints of those segments.** The midpoints for the segments (in red on the picture below) are: 2.125, 2.375, 2.625, 2.875, 3.125, 3.375, 3.625, and 3.875. These will be your inputs (x-values) for the Riemann sum.

Step 3: **Plug the midpoints into the function**, and then **multiply by the interval length**, which is 0.25:

f(2.125)0.25 + f(2.375)0.25 + f(2.625)0.25 + f(2.875)0.25 + f(3.125)0.25 + f(3.375)0.25 + f(3.625)0.25 + f(3.875)0.25

*Using algebra to rewrite:*

= [f(2.125) + f(2.375) + f(2.625) + f(2.875) + f(3.125) + f(3.375) + f(3.625) + f(3.875)]0.25

*Plugging the function’s values into the equation:*

= [6.515625 + 7.640625 + 8.890625 + 10.265625 + 11.765625 + 13.390625 + 15.140625 + 17.015625]0.25

= 90.625(0.25) = 22.65625

*That’s it!*

**Tip:**The midpoint came pretty close to the actual integral (22.6666). But you can check your work by using the following Definite Integral calculator. Plug your equation into the “integral” line on the calculator (note: use the ^ key to indicate an exponent, like 2^2 for 2^{2}. Then plug your interval into the “From x=” boxes (leave “x” in the first box). In this particular example, you’d type “2” into the second box and “4” into the third box. Click “Submit” to get the integral.

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