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
The width of interval [1,3] is 2, not 3
I see what I did now. Thanks
To find the upper and lower bounds for the definite integral of √(x) dx with the given partition, we need to calculate the upper and lower sums.
The upper sum is the sum of the areas of the rectangles formed by the partition, where we take the maximum value of √(x) in each interval.
For this partition [1,6] with x0=1, x1=3, and x2=6, the upper sum would be:
∑(i=1 to n) max(√(xi)) * Δxi
= max(√(x0)) * Δx0 + max(√(x1)) * Δx1 + max(√(x2)) * Δx2
= max(√(1)) * (3-1) + max(√(3)) * (6-3) + max(√(6)) * (6-3)
= √(1) * 2 + √(3) * 3 + √(6) * 3
= 2 + 3√(3) + 3√(6)
On the other hand, the lower sum is the sum of the areas of the rectangles formed by the partition, where we take the minimum value of √(x) in each interval.
For the same partition [1,6], the lower sum would be:
∑(i=1 to n) min(√(xi)) * Δxi
= min(√(x0)) * Δx0 + min(√(x1)) * Δx1 + min(√(x2)) * Δx2
= min(√(1)) * (3-1) + min(√(3)) * (6-3) + min(√(6)) * (6-3)
= √(1) * 2 + √(3) * 3 + √(6) * 3
= 2 + 3√(3) + 3
It seems you have made an error in calculating the lower sum since you arrived at the same expression as the upper sum. The correct lower sum in this case should be 2 + 3√(3) + 3.
Hence, the correct upper sum is 2 + 3√(3) + 3√(6), and the correct lower sum is 2 + 3√(3) + 3.
To find the upper and lower bounds for the definite integral, you need to use the given partitions and calculate the Riemann sums.
In this case, the given partition is [1,3,6], where x0=1, x1=3, and x2=6.
To calculate the upper sum, you need to find the maximum value of sqrt(x) in each subinterval and multiply it by the width of that subinterval. Then you sum up the results for all subintervals.
For the upper sum:
In the first subinterval [1,3], the maximum value of sqrt(x) is sqrt(3). So the contribution to the upper sum from this interval is (3-1) * sqrt(3) = 2sqrt(3).
In the second subinterval [3,6], the maximum value of sqrt(x) is sqrt(6). So the contribution to the upper sum from this interval is (6-3) * sqrt(6) = 3sqrt(6).
Adding up these contributions, the upper sum is 2sqrt(3) + 3sqrt(6).
To calculate the lower sum, you need to find the minimum value of sqrt(x) in each subinterval and multiply it by the width of that subinterval. Then you sum up the results for all subintervals.
For the lower sum:
In the first subinterval [1,3], the minimum value of sqrt(x) is sqrt(1) = 1. So the contribution to the lower sum from this interval is (3-1) * 1 = 2.
In the second subinterval [3,6], the minimum value of sqrt(x) is sqrt(3). So the contribution to the lower sum from this interval is (6-3) * sqrt(3) = 3sqrt(3).
Adding up these contributions, the lower sum is 2 + 3sqrt(3).
Therefore, the correct upper sum is 2sqrt(3) + 3sqrt(6), and the correct lower sum is 2 + 3sqrt(3).