please help

[Tueseve728]

By following the steps below, find the exact area under the curve y=f(x)=x^2 +2 from x=0 to x=2. Use the upper sum approximation.
Step 1: Divide the interval into n pieces, be sure to write down the k^th piece.

Step 2: Write the finite approximation Sn=A1+A2+.....+Ak+......+An and simplify the expression as much as possible.

Step 3: Find the exact area under the curve y=f(x) by taking the limit as n approaches infinity of the expression in Step 2.

I'm not really sure where to start or what is going on! Thanks for your help.

 

[stapel]

I'm not really sure where to start or what is going on!

Have they not covered "Riemann sums" in your class yet? It's kind of involved, and having pictures really helps. If you haven't heard of this, we can try to find online lessons that will explain what this exercise is asking for.

Thank you.

Eliz.

 

[Tueseve728]

could somone just get me started on this?

 

[pka]

Ms Sapel has given the best advice on what you really should do
But against my better judgment, I will try to start you off
Because \(\displaystyle f(x) = x^2 + 2\) is integrable on [0,2] all Riemann Sums will converge to the same sum.
So to subdivide [0,2] into n intervals, \(\displaystyle dx = \frac{{2 - 0}}{n}\) . If n is 4, \(\displaystyle dx = \frac{1}{2}\) .
We get the subintervals this way: \(\displaystyle x_k = kdx\) . If n=4 then, \(\displaystyle x_0 = 0,\;x_1 = 1/2,\;x_2 = 1,\;x_0 = 3/2,\;x_0 = 2\) .

The left-hand sum is \(\displaystyle \sum\limits_{k = 0}^{n - 1} {f(x_k )dx}\) . If n=4 then \(\displaystyle \sum\limits_{k = 0}^3 {f(x_k )dx} = f(0)(1/2) + f(1/2)(1/2) + f(1)(1/2) + f(3/2)(1/2)\) .

In general, look at \(\displaystyle \sum\limits_{k = 0}^{n - 1} {f(x_k )dx} = dx\left( {\sum\limits_{k = 0}^{n - 1} {f(x_k )} } \right) = \frac{2}{n}\left( {\sum\limits_{k = 0}^{n - 1} {\left( {2k/n } \right)^2 + 2} } \right)\)
From that we need to find the limit as n->00.