As you increase the subintervals the area will be closer and closer to the real value. In other words your approximation gets better.
As you increase the intervals, there will be more rectanagles and the added area of these rectangles are converging towards the actual area under the curve.
Given:
and 
.
To find:
The value of f(5).
Solution:
We have,
For 
,
For 
,
For 
,
For 
,
Therefore, the value of 
is 
.
Answer:
a 10 yd 2ft 7 in
b 10 yd 1 ft 2 in
c 99 yd
d. 4840 sq yd 7 sq ft 119 sq in
Step-by-step explanation:
a. 2 ft. 5 in. +
9 yd. 3 ft. 2 in.
---------------------
9 yd 5ft 7 in
but 3 ft = 1yd
9 yd 5ft 7 in
+1yd - 3ft
---------------------------
10 yd 2ft 7 in
b. 4 yd. 8 in
+ 6 yd. 6 in.
----------------------
10 yd 14 in
but 12 in = 1ft
10 yd 14 in
+ 1ft - 12 in
---------------------
10 yd 1 ft 2 in
c. 29 yd. 2 ft. 11 in.
55 yd. 1 ft. 10 in.
+ 13 yd. 1 ft. 3 in.
--------------------------
97 yd 4 ft 24 in
12 in = 1ft
24 in = 2ft
97 yd 4 ft 24 in
+2ft - 24 in
--------------------------------
97 yd 6ft
3 ft = 1yd
6f = 2yd
97 yd 6ft
+2yd - 6ft
---------------------
99 yd
d. 4,839 sq. yd. 8 sq. ft. 139 sq. in.
+ 7 sq. ft. 124 sq. in.
--------------------------------------------------
4839 sq yd 15 sq ft 263 sq in
12*12 = 144 sq in = 1 sq ft
4839 sq yd 15 sq ft 263 sq in
+ 1 sq ft - 144 sq in
--------------------------------------------------
4839 sq yd 16 sq ft 119 sq in
3ft * 3 ft = 9 sq ft = 1 sq yd
4839 sq yd 16 sq ft 119 sq in
+1 sq yd - 9 sq ft
----------------------------------------------
4840 sq yd 7 sq ft 119 sq in
Answer:
ok ok ok ok ok ok okok ok ok
