The area under a curve. Today we (begin to) ask questions of the type: How much area sits under the graph of f(x) = x 2 over the interval [ 1, 2]? Before we work on How we will figure out Why velocity The distance problem. time Suppose that you are studying a particle traveling with velocity equal to v(t) = t 2. How much distance does the particle cover from t = 1 to t = 2? Now if the particle had a constant velocity of 3 then from t = 1 to t = 2 the particle would have covered (distance) = (speed) (elapsed time) = ( ) ( ) =. Now our particle does not have constant velocity, but we can approximate this velocity function by a bunch of constant functions: velocity time These constant functions (depicted to the right) are (1) velocity = 1 from t = 1 to t =.5, (2) velocity = 0.25 from t =.5 to t = 0, (3) velocity = 0 from t = 0 to t = 0.5, (4) velocity = 0.25 from t = 0.5 to t = 1, (5) velocity = 1 from t = 1 to t = 1.5, (6) velocity = 2.25 from t = 1.5 to t = 2, Adding up these distances we get an approximation for the distance the particle covers from t = 1 to t = 2. Some leading questions: (1) How accurate do you think that our estimate is? (2) How could we improve its accuracy? (3) Did we basically just compute an area? 1
2 The payoffs of these questions: (3) Did we basically compute an area when we computed distance covered? As a consequence we see that two quantities are the same: = This will be super important later on. As a consequence for the rest of today we will think about computing areas under curves. (1) and (2) How accurate do you think that our estimate is? How could we improve its accuracy? Philosophy of computing the area under a curve The area under the curve y = f(x) over the interval [a, b] can be approximated by (1) breaking [a, b] up into lots of subintervals, (2) approximating the area under f(x) over each of these pieces by a rectangle (3) adding up the areas of these rectangles. By using more subintervals (and so more rectangles) we get better approximations. Example By breaking the interval [0, 4] at 1, 2, 3 and using left endpoints, approximate the area under the curve. Underestimate Over [0, 1] use Area = Over [1, 2] use Area = Over [2, 3] use Area = Over [3, 4] use Area = What is the total area of these rectangles? The actual area should be close to this Area How could we get a better estimate?
use more rectangles! By breaking the interval [0, 4] at 1 2, 1, 3 2, 2, 5 2, 3, 7, draw eight rectangles which and using 2 right endpoints approximate the area under the curve y = x 2 over [0, 4]. 3 What is the total area of these rectangles? The result of the previous problem gives us a better estimate AREA If we use more rectangles (Say, 10,000) we can get better estimates. A computer will be necessary. In order to allow us to think of these, we will need to make the problem more abstract so that we can get a computer to work for us. Working towards Riemann sums We are about to build up to the right way to get good area estimates. Recall how to approximate area: (1) (2) (3) Step 1 in abstractifying Understanding the subintervals: If we break up [a, b] into n subintervals all of the same length then each subinterval has length x =. Breakin [a, b] up into n subintervals each of length x = gives a sequence of endpoints a...... b x 0 x 1 x 2 x 3 x k x k+1 x n 2 x n 1 x n x 0 =, x 1 =, x 2 =, x 3 =,... x k =,... x n = The most important is x k = The k th subinterval is: Example: When you break [0, 4] into 8 subintervals you get: x = and x k =
4 Step 2: Pick an element in each subinterval to build a rectangle. The notation x k is used to denote an element of the k th subinterval. (The k th subinterval is =.) There are infinitely many such points here are some particular plans of attack: The left endpoint of the interval The right endpoint of the interval The midpoint of the interval Evaluate the function f(x) at x k to approximate the function by a rectangle. Draw a picture of the situation here The area of this rectangle is Area of k th rectangle = (base) (height) = example If you break the interval [0, 4] into 8 subintervals then x =, x k = The left endpoint x k = =. if f(x) = x 2 then height = f(x k ) = So that the area is Step 3: Add these areas up We now see that (Area of 1 st rect) + (area of 2 nd rec) +... (area of k th rect) +... (area of n th rect) substitute in our formula for the area of the k th rectangle ( ) + ( ) +... ( ) +... ( ) Example: If you break [0, 4] up into 8 subintervals, and approximate the area under f(x) = x 2 using the subintervals and left endpoints then you get
Summation notation If we are to make computations using a computer then we need to have a way of writing (Area of 1 st rect) + (area of 2 nd rec) +... (area of k th rect) +... (area of n th rect) which is precise enough for a computer to figure out what we mean. Summation notation n The notation a k means: The sum of a k where you let k take values 1, 2, 3,..., n. n a k = Example 4 k 2 = Thus in summation notation: ( ) + ( ) +... ( ) +... ( ) becomes the stuff on the right is called a Riemann sum Example: Write down a Riemann sum that approximates the area under the curve f(x) = x 2 over [0, 4] using n = 8 right endpoints, x =, x k =, x k =, f(x k ) =. So that Use a computer to compute this. ( A sample link to wolfram alpha can be found on the webpage. Example: Write down a Riemann sum that approximates the area under the curve f(x) = x 2 over [0, 4] using n = 4, 000 right endpoints, x =, x k =, x k =, f(x k ) =. So that Use a computer to compute this. ( A sample link to wolfram alpha can be found on the webpage.) 5