SOLUTION: Please Help, Estimate the area under the curve { { { f(x)= -x^2+4x+4.25 over 0 <= x <= 5 } } } . Use 5 intervals. I tried to graph it on a graphing calculator first, but I thi

Algebra ->  Sequences-and-series -> SOLUTION: Please Help, Estimate the area under the curve { { { f(x)= -x^2+4x+4.25 over 0 <= x <= 5 } } } . Use 5 intervals. I tried to graph it on a graphing calculator first, but I thi      Log On


   

Question 302713: Please Help,
Estimate the area under the curve { { { f(x)= -x^2+4x+4.25 over 0 <= x <= 5 } } } . Use 5 intervals.
I tried to graph it on a graphing calculator first, but I think it was wrong because I can't enter 0 <= x <= 5. Here is the link of the graphed equation: http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
Thanks for the help :)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You didn't specify a method.
Break up the area into 5 regions.
Calculate an average value of f(x) and the right and left endpoint.
Then multiply the average value of f(x) by dx in each region.
+x=0, +f%280%29=4.25
+x=1, +f%281%29=7.25
+x=2, +f%282%29=8.25
+x=3, +f%283%29=7.25
+x=4, +f%284%29=4.25
+x=5, +f%285%29=-0.75
SO then area 1 equals,
A1=%281%2F2%29%28f%280%29%2Bf%281%29%29%2A%281-0%29=5.75+
A2=%281%2F2%29%28f%281%29%2Bf%282%29%29%2A%282-1%29=7.75
A3=%281%2F2%29%28f%282%29%2Bf%283%29%29%2A%283-2%29=7.75
A4=%281%2F2%29%28f%283%29%2Bf%284%29%29%2A%284-3%29=5.75
A5=%281%2F2%29%28f%284%29%2Bf%285%29%29%2A%285-4%29=1.75
Atot=A1%2BA2%2BA3%2BA4%2BA5=28.75
The exact value is 29.583 so that's a pretty good approximation (within 3%).