Question

P 4.00 Figure 1 x9. Two rectangles have been formed: CDQP (known as the upper rectangle) and ABQP (known as the lower rectangle). ve * Choose any value for k. 'にA+ Find the area of each rectangie. a)

Please Answer to these all parts, Thank you so much

Answer 1

2 VE dr 4 � (2/3) � (93/2-43/2) = (8/3) x(27-8) = 50.67 \r\n"",""error"":null,""modified"":1556178226,""source"":""automated""}"

Area of a rectangle = width x height

(a) Area(CDQP) = (9-4) x y(9) = 5 x 4 x 3 = 60

Area(ABQP) = (9-4) x y(4) = 5 x 4 x 2 = 40

(b) Area(LMRP) = (6.5-4) x y(6.5) = 2.5 x 4 x 2.55 = 25.50

Area(TUQR) = (9-6.5) x y(9) = 2.5 x 4 x 3 = 30

Sum of areas of upper rectangles = Area(LMRP) + Area(TUQR) = 25.50 + 30 = 55.50

(c) Area(STRP) = (6.5-4) x y(4) = 2.5 x 4 x 2 = 20

Area(MVQR) = (9-6.5) x y(6.5) = 2.5 x 4 x 2.55 = 25.50

Sum of areas of lower rectangles = Area(STRP) + Area(MVQR) = 20 + 25.50 = 45.50

(d) In general, if you divide the interval into equal sections, then, the sum of the upper rectangles is given by,

and the sum of the lower rectangles is given by

$\sum_{i=1}^n h \times y(4+(i-1)h)$

See the table in case when the interval is divided into 25 equal sections

i	Section	Area (Upper)	Area (lower)
1	4 - 4.2	1.64	1.60
2	4.2 - 4.4	1.68	1.64
3	4.4 - 4.6	1.72	1.68
4	4.6 - 4.8	1.75	1.72
5	4.8 - 5	1.79	1.75
6	5 - 5.2	1.82	1.79
7	5.2 - 5.4	1.86	1.82
8	5.4 - 5.6	1.89	1.86
9	5.6 - 5.8	1.93	1.89
10	5.8 - 6	1.96	1.93
11	6 - 6.2	1.99	1.96
12	6.2 - 6.4	2.02	1.99
13	6.4 - 6.6	2.06	2.02
14	6.6 - 6.8	2.09	2.06
15	6.8 - 7	2.12	2.09
16	7 - 7.2	2.15	2.12
17	7.2 - 7.4	2.18	2.15
18	7.4 - 7.6	2.21	2.18
19	7.6 - 7.8	2.23	2.21
20	7.8 - 8	2.26	2.23
21	8 - 8.2	2.29	2.26
22	8.2 - 8.4	2.32	2.29
23	8.4 - 8.6	2.35	2.32
24	8.6 - 8.8	2.37	2.35
25	8.8 - 9	2.40	2.37
Sum ==>		51.07	50.27

(e)

(f, g and h)

The sums of upper and lower rectangles in (d) become equal when the interval is divided into a large number of sections, $n \to \infty$ . Both the sums converge to the integral value in (e).