Question

calculus 2

Answer 1

split the time into 6 intervals :
0->1,
1->2,
2->3,
,,,
5->6

Answer 2

width of interval = 1

Answer 3

so, for the left sum simply multiply this width wid the function values :
1*[3+4+5+4+7+8]

Answer 4

similarly for the right sum :
for each interval, take the right side v-value, so u wud leave the 3 this time :
1*[4+5+4+7+8+11]

Answer 5

see if that makes more/less sense

Answer 6

31 and 39

Answer 7

what is the trap one?

Answer 8

trapezoidal rule

Answer 9

see if u can try it out..
i havent worked it recently.... .

Answer 10

for trap, try below :
1*[(3+4)/2 + (4+5)/2 + (5+4)/2 + (4+7)/2 + (7+8)/2 + (8+11)/2 ]

Answer 11

thank you, that worked

Answer 12

good, make sure u knw why they worked :)

Answer 13

lol just saying ^_^

Answer 14

i understand left and right. i dont understand trap so well

Answer 15

i sensed that :)

for trap we have used below area element for each interval :
\((b-a) * \frac{f(a) + f(b)}{2}\)

Answer 16

thats the formula for area of trapezoid of :
width = (b-a)
parallel side lenths = f(a), f(b)

Answer 17

http://en.wikipedia.org/wiki/Trapezoidal_rule

Answer 18

just look at the picture in that link,
dont go thru the bottom explanation... its lil confusing in that link..

Answer 19

thank you

Answer 20

np :)