How do you evaluate the definite integral by the limit definition given #int (8-x)dx# from [0,8]?

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Answer 1 · 2017-05-26T23:02:55

#int_0^8 (8-x)dx = 32#

Divide the interval #[0,8]# using the points:

#x_k = (8k)/n# for #k=0,1,...,n#

The left Riemann sum is:

#S_n = sum_(k=0)^n f(x_k) (x_(k+1)-x_k)#

#S_n = sum_(k=0)^n (8-(8k)/n) ((8(k+1))/n-(8k)/n)#

#S_n = 64sum_(k=0)^n (1-k/n) ((k+1-k)/n)#

#S_n = 64sum_(k=0)^n (1-k/n) (1/n)#

#S_n = 64sum_(k=0)^n (1/n-k/n^2) #

#S_n = 64sum_(k=0)^n (n-k)/n^2 #

Substitute #j=(n-k)#:

#S_n = 64/n^2 sum_(j=0)^n j #

Using Gauss' formula for the sum of the first #n# integers:

#S_n = 64/n^2 ((n(n+1))/2)#

#S_n = 32 (n^2+n)/n^2#

and then:

#lim_(n->oo) S_n = 32#