EXAMPLE Le rectangliste ustimate the area and the parabola y - 4x2 from 0 to 1(the...
EXAMPLE Le rectangliste ustimate the area and the parabola y - 4x2 from 0 to 1(the parabolic region Sillustrated to the left . SOLUTION We first notice that there must be where between 0 and 4 because S is continued in a rectangle of Side 1 and 4, but we can can do bethan that. Suppose we divide the region into four strips by drawing articles 1- ts in this figure (allalaw. 2.1 21.45 3 A Vin Exampled (D) We can apor osmane each stop by a rectangle wise bases the same as the soup and whecae high is the same as the night edge of the strip las Figure (h) above). In other words, the haghts of these rectangles are the values of the function (x)=<zas the night andpants of the substarvata [ Jan Sach rectangle has with and the heights are +(+24+74) and 412. ur moet wat he be the sum of the areas of these apparating rectangkes, we out R4 + ++ ) 40 We see that the area is less than R40 Instead of using the rectangles above could use the smaller rectangles which are the value of at the end of the saintvals. (Thelmestructangle has collapse besig is o.) The Stof the areas of the approximating rectangle 10. --C We that the areas arger than 14, sowelow and uppe estimates for A We can repeat this procedure with a large number of strips. The fique show what happens when wervide the sightstries of equal width (1. + (utandpoints ang right undants By computing the sum of these areas of the smaller rectang (L) and the sum of the areas of the larger clandes (Rgli w ba better and upper sales for a 1.003750041.5937500. L. LA A Se on possible answer to the is to say that there are somewhere betin 1.02750 and 1.5927500. We could obtain bat estimates by creasing the number of strip. The table at the shows the results of similar calculations with a computer using rectangles and found with left andpoints) or night andpoints R. In particular, we see by using So strips that the relies betmen 1.2036 and 1.3736. With 1000 strips, we narrow it down on A las baten 1.3212340 and 1.3258340 A good estate is obtained by averaging these numbers: 40 101.1400000 1.5400000 201.23500001.4350000 30 1.26740741.4007407 S01.2935000 1.3736000 1001.3134000 1.3534000 1000 1.33133401.3353340