# Find area under the graph

Using the Trapezoidal rule, Midpoint rule, and Simpson's rule. I know how to use those rule's, but I don't have a solid function to go on. Just a graph. Could someone explain how I would about do this? I'll post a picture of the graph.

-
You know what the function equals at several points, this should be sufficient for a decent approximation. – Argon Jan 20 '13 at 3:56
I assume you mean the number of blocks. If so then I think I can figure it out then. – blutuu Jan 20 '13 at 3:58
$y=f(x)$, so yes, the height of the block represents the value of the function at that given $x$. – Argon Jan 20 '13 at 3:59
Just eye balling it, the area $A$, satisfies $10.5<A<11$ – Ethan Jan 20 '13 at 4:37

You have to read off the required points from the graph. For example, the trapezoidal rule using one interval just says the area is $\frac 12(y_0+y_1)(x_1-x_0)$. The endpoints are $(0,3),(6,1)$ so the rule would give $\frac 12 (3+1)(6-0)=9$ which is a terrible approximation-just count squares. The trapezoidal approximation with six intervals would use the points $(0,3), (1,5),(2,4),(3,2),(4,3),(5,4),(6,1)$. You apply the rule to each interval and add them up. You should be able to convince yourself that this is the same as $(\frac 12 3+5+4+2+3+4+\frac 12 1)1=20$