+129852 I don't know what they are asking for in (a) or (b), GM.....but I know the others ....
(c) With left endpoints, we have
7
Width of subinterval * ∑ f ( x)
x = 4
(d) The area with Left Endpoints is :
Width of subinterval * [ f(4) + f(5) + f(6) + f(7) ] =
(1) * [ 17 + 26 + 37 + 50 ] =
130 units^2
(e) With right endpoints, we have
Width of Interval * [ f(5) + f(6) + f(7) + f(8) ] =
(1) * [26 + 37 + 50 + 65 ] =
178 units^2
The actual area is [hide your eyes if you haven't had Calculus ]
8 8
∫ x^2 + 1 dx = [ x^3/3 + x ] = [ 8^3/3 + 8 ] - [ 4^4/3 + 4 ] = 460 / 3 ≈
4 4
153.33 units^2
(f) The Left Enpoints underestimate the area
(g) The Right Enpoints overestimate the area
(h) Trapezoidal method =
[ b - a ]
______ * [ f(4) + 2f(5) + 2f(6) + 2f(7) + f(8) ]
2n
Where b = 8 a = 4 and n = number of trapezoids = 4
So we have
[ 8 - 4]
______ * [ 17 + 52 + 74 + 100 + 65 ] =
2 * 4
(1/2) * [ 308 ] =
154 units^2
Note that this is very close to the actual area !!!
+73 Hey, CPhill thanks for the help which I understand but you made a mistake in your work and skipped something in the problem. In your work, you solved for (d) but labeled it as (c) and you solved for (e) but labeled it as (d). So you skipped (c) in the original problem.
