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
To find the upper and lower bounds for the definite integral, we can use the concept of Riemann sums. A Riemann sum is an approximation of the definite integral using partitions (subintervals) and sample points within each interval.
In this case, the partition is given by:
x0 = 1
x1 = 3
x2 = 6
To compute the upper sum and lower sum, we need to find the maximum and minimum values of the function within each subinterval.
For the subinterval [1, 3], the maximum value is obtained at x = 3, which gives us sqrt(3). The minimum value is obtained at x = 1, which gives us sqrt(1) = 1.
For the subinterval [3, 6], the maximum value is obtained at x = 6, which gives us sqrt(6). The minimum value is obtained at x = 3, which gives us sqrt(3).
Now, let's calculate the upper and lower sums:
Upper sum = (x1 - x0) * max(f(xi))
= (3 - 1) * max(sqrt(x))
= 2 * max(sqrt(x))
= 2 * sqrt(3)
Lower sum = (x1 - x0) * min(f(xi))
= (3 - 1) * min(sqrt(x))
= 2 * min(sqrt(x))
= 2 * sqrt(1)
= 2
So, the correct upper sum is 2 * sqrt(3) and the correct lower sum is 2.
Therefore, the correct upper and lower bounds for the definite integral are:
Upper bound = Upper sum = 2 * sqrt(3)
Lower bound = Lower sum = 2