Upper and Lower Sum - Integrals

Question

Hi, I've been trying to calculate the upper and lower sum, on a regular partition of the interval, for the below integral:

int(6 - 2x)dx (where a = 1 and b = 7)

The closest that I've gotten is from calculating the lower sum to be 24 + (36(n+1))/n by using f(xi) = 6 - 2(1 + 6*i/n)

But that is still of course wrong.

Answer 1

So you are doing a RHS sum and a LHS sum with the RHS being smaller for a decreasing function.

The width of the rectangles are (7-1)/n = 6/n x_i = 1+ 6i/n for i = 0 to n

The LHS sum of the areas is the summation (i= 1 to n) of f(x_i-1)(6/n)

The RHS sum of the areas is the summation (i=1 to n) of f(x_i)(6/n)

f(x_i) = 6 - 2(1 + 6i/n) = 4 + 12i/n

The RHS summation is 6/n times summation (i=1,n) of (4 - 12i/n)

6/n(4n - (12/n)(n)(n+1)/2 = 6/n (-2n - 6) = -12 -36/n (Approaches the correct value of integral as n→∞)

I'll leave the LHS sum for you. Remember sum of i from 1 to n is n(n+1)/2)

Please consider a tutor. Take care.