THE DEFINITE INTEGRAL RECTANGULAR APPROXIMATION, RIEMANN SUM, AND INTEGRTION RULES

time velocity After 4 seconds, the object has gone 12 feet. Consider an object moving at a constant rate of 3 ft/sec. Since rate. time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

Heck – it’s a triangle! That’s easy! Consider this same object moving at a varying rate defined by the function f(x) = x Distance is still found using rate. time = distance: That means the area under the curve STILL represents the distance traveled, but since we don’t have a constant rate, how could we find that area now (after 4 seconds)?

What happens, though, if the velocity function is not constant, nor does it form a regular geometric shape (with an easy area formula) when we shade the area underneath it? What’s a student to do?? How can we find the area under that kind of curve?? Example:

VIDEO TUTORIAL INITIATE SLIDE TO VIEW Links to videos--> Definite Integral Limit of Riemann Sum Geometry I CU5L1a LRAM MRAM RRAM Rectangular Approximations left hand riemann calculus right CU5L1b pt II LRAM MRAM RRAM Rectangular Approximations Left right hand riemann calculus CU5L1c pt III LRAM MRAM RRAM Rectangular Approximations hand Riemann left right calculus CU5L3 Rules for Definite Integrals CU5l4 Using Antiderivatives to Find Definite Integrals CU5l5 fundamental theorem of calculus I and II Cu5l6 trapezoidal rule approximation calculus I and II

When we find the area under a curve by adding rectangles, the answer is called a Rieman sum. subinterval partition The width of a rectangle is called a subinterval. The entire interval is called the partition. Subintervals do not all have to be the same size.

subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by. As gets smaller, the approximation for the area gets better. if P is a partition of the interval

What if: We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. It seems reasonable that the distance will equal the area under the curve.

The area under the curve We can use anti-derivatives to find the area under a curve!

is called the definite integral of over. If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:

Leibnitz introduced a simpler notation for the definite integral: Note that the very small change in x becomes dx.

Integration Symbol lower limit of integration upper limit of integration integrand variable of integration (dummy variable) It is called a dummy variable because the answer does not depend on the variable chosen.

We have the notation for integration, but we still need to learn how to evaluate the integral.

Rules for definite integrals

Using the Properties of the Definite Integral Given:

Riemann Sums There are three types of Riemann Sums Right Riemann: Left Riemann: Midpoint Riemann:

If the velocity is not constant, we might guess that the distance traveled is still equal to the area under the curve. (The units work out.) Example: We could estimate the area under the curve by drawing rectangles touching at their left corners. This is called the Left-hand Rectangular Approximation Method (LRAM). Approximate area:

A particle starts at x=0 and moves along the x- axis with velocity v(t)=t 2 for time t >0. Where is the particle at t=3? (use 6 intervals) LRAM = y x + y x + y x +... = v t + v t + v t +... = 0 2 ( 1 / 2 ) + ( 1 / 2 ) 2 ( 1 / 2 ) + (1) 2 ( 1 / 2 ) + (1½) 2 ( 1 / 2 ) + (2) 2 ( 1 / 2 ) + (2 1 / 2 ) 2 ( 1 / 2 ) = 6.875

We could also use a Right-hand Rectangular Approximation Method (RRAM). Approximate area:

A particle starts at x=0 and moves along the x- axis with velocity v(t)=t 2 for time t >0. Where is the particle at t=3? (use 6 intervals) RRAM = y x + y x + y x +... = v t + v t + v t +... = ( 1 / 2 ) 2 ( 1 / 2 ) + (1) 2 ( 1 / 2 ) + (1½) 2 ( 1 / 2 ) + (2) 2 ( 1 / 2 ) + (2 1 / 2 ) 2 ( 1 / 2 ) + (3) 2 ( 1 / 2 ) =

A particle starts at x=0 and moves along the x- axis with velocity v(t)=t 2 for time t >0. Where is the particle at t=3? (use 6 intervals) MRAM = y x + y x + y x +... = v t + v t + v t +... = ( 1 / 4 ) 2 ( 1 / 2 ) + ( 3 / 4) 2 ( 1 / 2 ) + (1 1 / 4 ) 2 ( 1 / 2 ) + (1 3 / 4 ) 2 ( 1 / 2 ) + (2 1 / 4 ) 2 ( 1 / 2 ) + (2 3 / 4 ) 2 ( 1 / 2 ) = The actual area is 9.

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM). Approximate area: In this example there are four subintervals. As the number of subintervals increases, so does the accuracy.

Approximate area: width of subinterval With 8 subintervals: The exact answer for this problem is.

Circumscribed rectangles are all above the curve: Inscribed rectangles are all below the curve:

Area

Area from x=0 to x=1 Example: Find the area under the curve from x = 1 to x = 2. Area from x=0 to x=2 Area under the curve from x = 1 to x = 2.

Example: Find the area under the curve from x = 1 to x = 2. To do the same problem on the TI-89: ENTER 7 2nd

Example: Find the area between the x-axis and the curve from to. On the TI-89: If you use the absolute value function, you don’t need to find the roots. pos. neg. 