Posted by **Eric** on Wednesday, May 15, 2013 at 1:50pm.

find the upper and lower bounds for definite integral sign, a=1, b=6, sqrt(x) dx. partitions are as follow: x0=1, x1= 3, x2=6 of the interval [1,6]

upper sum: 3sqrt(3) + 3sqrt(6)

lower sum: 3sqrt(3) + 3

but this is incorrect. can someone explain what i did wrong please

- Sums-Calc -
**Steve**, Wednesday, May 15, 2013 at 3:02pm
The width of interval [1,3] is 2, not 3

- Sums-Calc -
**Eric**, Wednesday, May 15, 2013 at 4:50pm
I see what I did now. Thanks

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