# Consider the function f(x)= -((x^2)/2)-6 In this problem you will calculate integrate from 0 to 3 of (-((x^2)/2)-6)dx by using the definitionintegrate from a to b of f(x)dx= lim as n approaches...

Consider the function f(x)= -((x^2)/2)-6

In this problem you will calculate integrate from 0 to 3 of (-((x^2)/2)-6)dx by using the definition

integrate from a to b of f(x)dx= lim as n approaches infinity of sum_(i=1,n) of f(x_i)(delta x)

The summation inside the brackets is R_n which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

**Calculate R_n for f(x)= -((x^2)/2)-6 on the interval [0,3] **

**and write your answer as a function of n without any summation signs**.

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Evaluating the definite integral using the Riemann sum yields:

`lim_(n->oo)sum_k=0^n f(x_k)Deltax_k`

Evaluating `Delta x` yields:

`Delta x = (3-0)/n => Delta x = 3/n`

`x_k = 0 + k*Delta x => x_k = 0 + k*(3/n)`

`f(x_k) = -(x_k)^2/2 - 6 => f(x_k) = -(9k^2/n^2) - 6`

You need to evaluate the Riemann sum such that:

`lim_(n->oo)sum_k=0^n f(x_k)Deltax_k = lim_(n->oo)sum_k=0^n (-(9k^2/n^2) - 6)*(3/n)`

`lim_(n->oo)sum_k=0^n f(x_k)Deltax_k = lim_(n->oo)(3/n)sum_k=0^n (-9k^2 - 6n^2)/(n^2) `

`lim_(n->oo)sum_k=0^n f(x_k)Deltax_k = lim_(n->oo)(3/n^3)sum_k=0^n (-9k^2)`

`lim_(n->oo)sum_k=0^n f(x_k)Deltax_k = lim_(n->oo)(27/n^3)(n(n+1)(2n+1)/6)`

`lim_(n->oo)sum_k=0^n f(x_k)Deltax_k = 9`

**Hence, evaluating the definite integral using Riemann sum yields `int_0^3 (x^2/2 - 6) dx = lim_(n->oo)sum_k=0^n f(x_k)Deltax_k = 9.` **