Question

How do you find the Riemann sum for `f(x) = x^2 + 3x` over the interval [0, 8]?

Answer 1

The formula looks like:

`lim_(n->oo)sum_(i=1)^(n)f(x_i)Deltax = int_a^bf(x)dx`

Your integral will look like:

`int_0^8(x^2+3x)dx`

We can use this information to plug in values into our Riemann sum formula.

`Deltax=(b-a)/n`

`x_i=a+iDeltax`

Therefore:

`Deltax=(8-0)/n=8/n`

`x_i=0+i(8/n)=(8i)/n`

So, as a Riemann sum:

`lim_(n->oo)sum_(i=1)^(n)(8/n)[((8i)/n)^2+3((8i)/n)]`

`=lim_(n->oo)sum_(i=1)^(n)(8/n)[(64/n^2)i^2+(24/n)i]`

Note: Since `i` is our variable and `n` is our constant, we can pull those to the front.

`lim_(n->oo)[512/n^3sum_(i=1)^(n)i^2+192/n^2sum_(i=1)^ni]`

Recall from summation formulas:

`sum_(i=1)^(n)i=((n(n+1))/2)`
`sum_(i=1)^(n)i^2=((n(n+1)(2n+1))/6)`

So, we'll have an awesome looking function:

`lim_(n->oo)[512/n^3((n(n+1)(2n+1))/6)+192/n^2((n(n+1))/2)]`

Now, since the degree of the denominators are the same as the numerators, it will result in the sum of two ratios.

`=lim_(n->oo)[(512*2)/6+192/2]=800/3~~266.7`