Posted by **Eric** on Wednesday, May 15, 2013 at 11:46am.

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

## Answer This Question

## Related Questions

- Sums-Calc - find the upper and lower bounds for definite integral sign, a=1, b=6...
- abstract algebra - Suppose K = {{2,3,4,5,6,7},{2,5,7,8,12},{1,2,3,5,7,9,13,20...
- abstract algebra - Suppose K = {{2,3,4,5,6,7},{2,5,7,8,12},{1,2,3,5,7,9,13,20...
- Pre-Cal - Find f X g , g X f , and f X f f(x) = 3sqrt x-1 g(x) = x^3 + 1 For f(x...
- Pre-Cal - Find f X g , g X f , and f X f f(x) = 3sqrt x-1 g(x) = x^3 + 1 For f(x...
- Pre-Cal(Urgent) - Find f X g , g X f , and f X f f(x) = 3sqrt x-1 g(x) = x^3 + ...
- Math(Please check) - Find f X g , g X f , and f X f f(x) = 3sqrt x-1 g(x) = x^3...
- algebra - Given that XY =21 and 1 < x < 2, find the sum of the upper and ...
- Calculus - We're doing a lab in my Calc II class and I'm stuck on this question...
- calculas 1 - Use upper and lower sums to approximate the area of the region ...

More Related Questions