For the graph of the function using 6 rectangles:Example #1Example #2Signed AreaWhat is the area of the region R that is enclosed between the x-axis and the graph of 2( ) 22xf x x  for x between 1 and 4?One way we can estimate the area under a curve is to use rectangles. In particular, we will use a midpoint sum formula. In this example, we are going to estimate this area by using the area of these 6 rectangles. What do I need to find the area of these rectangles? First, I will divide the interval from 1 to 4 into 6 equal intervals each of width 0.5 which gives me 7 x values (to have 6 intervals, I need 7 x values). What information does this give me about each of my rectangles?Next, I will find the midpoints of each of these intervals (there willbe 6). Evaluating the function at each of the 6 midpoints gives me the heights of each of these rectangles. Question: Did we have to choose the midpoint to determine the height? Lastly, we have the widths (5 = x) and the heights, so we can calculate the area of each rectangle and add up all the areas to get the approximate area under the curve. 1 4211.522.533.5In the table below are listed the x values used, the interval number (i), the midpoint of interval i (mi ), and the value of the function at the midpoint of interval i (f(mi)). Also depicted is a generalized graph with proper notation. For the graph of the function 2( ) 22xf x x using 6 rectangles:x0 = 1 i mif(mi) area of rectangle i = xmfi*x1 = 1.51 (1+1.5)/2 = 1.252*(1.25) -(1.25)2/2=1.718750.5*1.71875 = 0.859375x2 = 2 2 (1.5+2)/2 = 1.751.96875 0.984375x3 = 2.53 (2+2.5)/2 = 2.251.96875 0.984375x4 = 3 4 2.75 1.71875 0.859375x5 = 3.55 3.25 1.21875 0.609375x6 = 4 6 3.75 0.46875 0.234375Approximation of area: 4.531250We just found 616)5)(.(])4,1[(iimffS1421x0x6x2x1x3x4x5m1m6m2m3m4m5fm1( ){x531250.4)(areasofSum61xmfiiThe formula for the midpoint sum of the areas of n rectangles is]),[,( bafSn, where n is the number of intervals, f is the function and[a,b] is the interval of integration.Example #1Let .501015)(23 xxxxf Find ]).2,1[,(6fSExample #2Let .25)(2xxf Find ]).3,5[,(4fSHow do we get a better approximation? As n gets larger and larger (closer and closer to infinity), we get closer and closer to the actual area underneath the curve between 1 and 4.Note: If you have n intervals of equal length, you will need (n + 1) x-values. To find the x-values, start with a, a +x, a + 2x, …..b.1 421Looking at infinitely the area of infinitely many rectangles, we are really looking at nibaindxxfxmf1.)()(lim This is read “the integral of f over [a,b]”. For example, the area between the horizontal axis and the graph501015)(23 xxxxfover the interval [-1,2] is given by2123.)501015( dxxxx Graphically what does this look like?Signed AreaFor areas that are bound by a function under the x-axis, this method gives you a “negative” area. In other words, if your function is below the horizontal axis, the height of your rectangle will be negative. Why? So, then the product is negative [0*)( xmfi].This means that the area of your rectangle will result in a negative number. This is called the “signed area”. So, if you have a curve that is both above and below the x-axis, the resulting area under thecurve will be the difference of the area above the axis and the area below the