Calc 2 Lab 2 pg3

# Calc 2 Lab 2 pg3 - A Look again at the integral of f(x =...

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Unformatted text preview: A. Look again at the integral of f(x) = 1/sqrt(2*Pi) *exp( —x"2/2) on the interval from [0,1]. Let bnd(n) = (f(a) ~ f(b))*(b — a) /n; Then rr(1000) < integral< rr(1000)+bnd(1000) or 03412662413 03414232129 < integral < HENCE, rounded to 3 decimal places the answer must be 0.341 and , most importantly, we know this without knowing the actual value of the integral. Compute the lower and upper bounds for the estimate using n = 5000 and n = 10000. Do we know what the fourth decimal digit has to be? M0, 4‘ ”VW/ fol/"\$7 ﬂit/45 6‘ ‘Aflr‘r‘mL ””443" B. Now for integral of sin(x"2) on [0 , 1.25] do this for n = 5000. Note that this function is increasing. What can we say about the value of the integral? C. Now integral of sqrt( l + x"3 ) on [ 2, 5] do this for n = 5000. Note that this function is increasing. What can we say about the value of the integral? 1H. DETERMINING N BEFORE WE DO THE CALCULATION The value of (f(a) 4 f(b))*(b-a) /n [ for decreasing functions ] determines how many digits of the answer we know. For instance, if (f(a) — f(b))*( b — a) /n < 0.00001 we would expect to know the ﬁrst four decimal digits. For the (x) = 1/sqrt(2*Pi) *exp( —x"2/2) problem, the smallest n that would work is 15,698. Using this n we get, .3413397477 < integral < .3413497472 A. Do as above for the integral of sin(x"2) on [0,1] and for integral of sin(x"2) on [0,1]. Note that in these cases the functions are increasing so the lower bound is RiemLeﬁ and bnd(n) = (f(b) — f(a))*( b — a) /n . IV. CAN WE GET GOOD ESTHVIATES WITH SMALLER VALUES OF N? Some of the values of n in the last few problems have gotten pretty large. Can we get “good" estimates with smaller 11? The answer is yes. In particular, if we average RiemRight and Riemleft to get the number half was in between it's a better estimate. It is equivalent to using "trapezoids" to estimate the pieces of the integral. We'll talk about this is class. For now assume that the following is true > trap := n - > (b—a)/n* ( f(a)/2 + f(b)/2 + sum( evalf( f( a + i*(b—a)/n)), i = l..(n—1)); A. Use this on the problem the integral of f(x) : 1/sqrt(2*Pi) *exp( -x"2/2) on the interval from [0,1]. Do trap(n) for n = 2, 4, 8, 16, 32, etc. until you're sure the ﬁrst four decimal digits are correct. How does this compare to the n in the other problems? ”HM, n Valdez, 1‘3 M 0+ 4.5 [a :70 ...
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