6.1, #1. The left and right sums with four subdivisions for ò12x2 dx are

LS
= (1.00)2(.25)+(1.25)2(.25)+(1.50)2(.25)+(1.75)2(.25) = 1.96875
RS
= (1.25)2(.25)+(1.50)2(.25)+(1.75)2(.25)+(2.00)2(.25) = 2.71875
Since x2 is increasing on [1,2], it follows that LS £ ò12x2 dx £ RS. Moreover, RS-LS = (22-12)(.25) = 0.75. In particular, the average (1/2)(LS+RS) = 2.34375 is within (1/2)(0.75) = 0.375 of the actual value.


6.1, #2. Since f(x) is positive on [1,6], the integral ò16 f(x) dx is the area bounded between the graph of f, the x-axis, and the lines x = 1, x = 6. So the area is 8.5. The average value of f on [1,6] is

1
6-1
 ó
õ 		6

1 
 f(x) dx = 8.5
5
 = 1.7


6.1, #3. The average value of f(x) = 4x+7 on [1,3] is

1
3-1
 ó
õ 		3

1 
 (4x+7) dx = 1
2
 (2x2+7x) ê
ê 		3
1 
 = 15.


6.1, #5.

ó
õ 		2

0 
 (3x2+1) dx = (x3+x) ê
ê 		2
0 
 = 10.



6.1 #6. Since -1 £ sinx £ 1 and e-x > 0 for all x, the graph of y = e-xsinx, x ³ 0, lies between the graphs of y = e-x and y = -e-x. The effect of multiplying sinx by e-x is to "dampen" the graph of y = sinx. Now the graph of y = sinx on [0, p] is above the x-axis. The portion of the graph on [p, 2p] is the mirror image of that just described and is below the x-axis. Since y = e-x is a decreasing function we now conclude that ò02p e-xsinx dx > 0.

File translated from TEX by TTH, version 2.00.
On 14 Aug 1999, 19:50.