The basic problem we want to solve is how to find the area between the graph of a function and the interval [a,b] on the x-axis. In order to approximate this we will subdivide the interval [a, b] into subintervals. Although, it is not absolutely necessary, this activity will divide the interval into n equal size subintervals. Each of these has width [math]\Delta x=\frac{b-a}{n}[/math]. We will then draw rectangles with these subintervals as bases and (signed) heights equal to some y-value on the curve. We will then add up the (signed) areas as an approximation of the (net signed) area between the curve and x-axis over this interval. These approximations are called Riemann Sums, after Bernhard Rieamann.[br][br]In the App:[br] Enter the formula for a function you want to integrate in the input box for f(x).[br] Adjust the values of a and b via the sliders or the input boxes to set the interval.[br] Adjust the value of n via its slider or input box to determine the number of subintervals.[br] Hide or show various elements by using the check boxes.
For the Left Riemann Sum (L) we take the signed heights of the rectangles to be the y-values at the left end of each subinterval.
For the Right Riemann Sum (R) we take the signed heights of the rectangles to be the y-values at the right end of each subinterval.
When do the Left and Right Riemann Sums perfectly find the actual exact area? When are they definitely too large, and when are they definitely too small?
If the function is constant (or at least constant on each subinterval) then both the Left and Right Riemann Sums are perfect estimates. If the function is strictly increasing, then the Right Riemann Sum is an over estimate and the Left Riemann Sum is an under estimate. This is reversed if the function is decreasing.
Why is the answer to the last question good news? How can we combine the Left and Right Riemann Sum estimates to get a better estimate?
Since one is an over estimate when the other is under estimate (at least for monotonic functions) then the average of the two should usually give a better estimate.
Since there is no reason to believe that one of the L or R estimate is better than the other we will just average them to find a better estimate. Averaging the L and R estimate yields the Trapezoidal Rule Estimate (T). If you click on the Trapezoidal Rule Rectangles, Left Riemann Sum, and Right Riemann Sum, then you will see that the Trapezoidal Rule is represented by rectangles whose height is exactly half-way between the heights of the corresponding Left and Right Riemann Sum rectangles.
Why is it called the Trapezoidal Rule?
Click on only the Trapezoidal Rule Rectangle and Trapezoidal Rule [br]Trapezoids check boxes. Can you prove that each of these Trapezoids [br]have the same area as the corresponding rectangles? Hint: prove that [br]there are some congruent triangles present. [br][br]The Trapezoidal [br]Rule is equivalent to approximating the curve with secant line segments [br]with the ends of the line segments being points on the curve above each [br]end of the subinterval. The areas we then sum are the areas of right [br]trapezoids.
The Midpoint Rule uses a Riemann Sum where the (signed) heights of the rectangle are y-values on the graph of the function above the midpoint of each of the subintervals.
Can we visualize the Midpoint Rule with trapezoids?
Yes, Both the Trapezoidal Rule and Midpoint Rule can be thought of summing the areas of rectangle or as summing the areas of right trapezoids.[br][br]Check both the Midpoint Rule Rectangles and Midpoint Rule Trapezoids check boxes. Again, you should be able to show that each right trapezoid has the same area as the corresponding rectangle. This time the top of the right trapezoids are line segments that are tangent to the graph of the function above the midpoint of the subinterval.
Show only the Midpoint Rule Trapezoids and the Trapezoid Rule Trapezoids.[br]When are these estimates perfect?[br]When are they too large, and when are they too small?
If the function is linear (or at least linear on each subinterval) then both the Midpoint and Trapezoid estimates are perfect. If the function is concave down, then the Midpoint Rule gives an over estimate and the Trapezoidal Rule gives an under estimate. This is reversed if the function is concave up. Once again, this is good news, because averaging the two will produce an even better estimate.
Which of the two gives a better estimate, the Trapezoidal Rule or the Midpoint Rule. Look carefully at the graphs with the Trapezoid Rule Trapezoids and Midpoint Rule Trapezoids turned on.
The Midpoint Rule usually gives a better estimate than the Trapezoidal Rule, with the error being around one-half as much on the Midpoint Rule. (It is exactly half as much for quadratic functions.)
We will find a weighted average the M and T estimates with the M estimate counting twice as much to find the Simpson's Rule estimate (S). [math]S_{2n}=\frac{T_n+2M_n}{3}[/math]. You can think of this as taking n intervals and using function values from both ends and the midpoint. An alternate way to think of this is to think of it as 2n intervals with differences in how some intervals are treated. For this reason, most people will call this Simpson's 2n. For example, the Simpson's 6 estimate is a weighted average of the Trapezoid 3 and Midpoint 3 estimates. [br][br]Graphically, Simpson's Rule estimate turns out to be the same as approximating the curve with a quadratic function (parabola) going through the three points on the curve above both ends and the midpoint of each interval. This is illustrated (without shading) when you check the Simpson's Rule checkbox. Simpson's Rule is very good, so we will not try to find a better approximate now. It may be hard to distinguish the parabolic pieces from the graph of the function visually.
What happens to our estimates as we increase the value of n?
Regardless of whether we use the L, R, T, M, or S estimates, they all get better and better the larger and larger we make n (and consequently the shorter and shorter we make the subintervals).
What Calculus concept are we talking about here?
We are taking the LIMIT of a Riemann Sum as n approaches infinity (and delta x approaches 0).
The definite integral is the limit of any of these estimates as n approaches infinity, and it is exactly the area we are trying to find. For simplicity sake, here we define the definite integral as the limit of the Right Riemann Sum as n approaches infinity.[br][br]Activate the checkbox for values to see the value of the integral and the various estimates accurate to five decimal places. Activate the checkbox for definitions to see the formulas that define the estimates and the integral.
We have been talking about lengths and areas. This is accurate when the graph of the function is above the x-axis, since then the function values are positive. However, lengths and areas must be positive, and if the graph of the function is below the x-axis, then the function values are negative (the opposite of the heights) and the products with the widths are also negative (the opposite of the areas). So, in the approximations and in the definite integral areas above the x-axis are added and areas below the x-axis are subtracted. This results in a nest signed area between the x-axis and the graph of the function over some interval.
When you select the Average Value checkbox it shows a Riemann Sum illustrated with rectangles. However, this one is chosen so that the top of the rectangle is at the average value of the function. This makes the signed area of each of the rectangles the same as the integral of the function over the subinterval. This Riemann Sum is a perfect estimate of the definite integral.