# Consider the function f(x)=-((x^2)/2)-9. In this problem you will calculate integrate from 0 to 3 of ((-x^2)/2)-9)dx by using the definitionintegrate from a to b of (f(x))dx= sum_(i=1)^n of...

Consider the function f(x)=-((x^2)/2)-9.

In this problem you will calculate integrate from 0 to 3 of ((-x^2)/2)-9)dx by using the definition

integrate from a to b of (f(x))dx= 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)-9 on the interval [0,3] and write your answer as a function of n without any summation signs.

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You need to remember that you can always evaluate the definite integral using the Riemann sum such that:

`int_a^b f(x)dx - lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k`

Notice that the problem provides the following informations for a and b such that:

`a = 0, b = 3`

You need to evaluate `Delta x` such that:

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

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

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

You need to evaluate the Riemann sum such that:

`lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = lim_(n->oo)sum_(k=1)^n (9k^2/(2n^2) - 9)(3/n)`

`lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = lim_(n->oo)(3/n)sum_(k=1)^n (9k^2 - 18n^2)/(2n^2)`

`lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = lim_(n->oo)(3/n)sum_(k=1)^n 9k^2/(2n^2) -lim_(n->oo)(3/n)* (18)/(2)`

`lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = lim_(n->oo)(6/n^3)sum_(k=1)^n 9k^2 - lim_(n->oo)(27/n)`

`lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = lim_(n->oo)(54/n^3)(n(n+1)(2n+1))/6 - lim_(n->oo)27/n`

`lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = lim_(n->oo) 9(n(n+1)(2n+1))/n^3 -0`

`lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = 18`

**Hence, evaluating the definite integral using Riemann sum yields `int_0^3 (x^2/2 - 9)dx = lim_(n->oo)sum_(k=1)^n f(x_k)*Delta x_k = 18.` **