MATH 149LABORATORY ASSIGNMENT 8THE FUNDAMENTAL THEOREMOF CALCULUS In class we defined the definite integral of the function f (x) on the interval [ a, b ] asd⌠⌡⎮⎮ab()f xx = Lim ∑ = k 1n()f ck() − xkx − k 1 where a = x0 < x1 < x2 < ... < x − n 1 < xn = b is a partition of [ a, b ] , x − k 1 < ck < xkfor k = 1...n, and the limit is taken as → n ∞ and the lengths of the subintervals [ x − k 1 , xk ] determined by the partition go to 0. The easiest and best way to evaluate the definite integral is by using the Fundamental Theorem of Calculus, which says thatd⌠⌡⎮⎮ab()f xx = F (b) - F (a)where F is any antiderivative of f ( i.e., F '(x) = f (x) ). In this laboratory we will verify the Fundamental Theorem by first using MAPLE to evaluate the limit of the sum directly and then by finding F and computing F (b) - F (a).EXAMPLE Let f (x) = x cos(πx) , 0 < x < 2π. We will use partitions in which the xk's are equally spaced and each ck is the midpoint of [ x − k 1, xk].First we illustrate the approximating rectangles using a partition with 20 points, followed by computing the limit of the Riemann sum.> with(student):> f:=x->x*cos(Pi*x); := f → xx()cos π x> middlebox(f(x),x=0..2*Pi,20);The corresponding approximating sum is> ms20:=middlesum(f(x),x=0..2*Pi,20); := ms20110π⎛⎝⎜⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟⎟∑ = i 019⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟110⎛⎝⎜⎜⎞⎠⎟⎟ + i12π⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟cosπ2⎛⎝⎜⎜⎞⎠⎟⎟ + i1210or> mv20:=value(ms); := mv20 ms> evalf(mv20);ms If we use a partition with n points an approximating sum using middle points as sample points is> msn:=middlesum(f(x),x=0..2*Pi,n);:= msn2 π⎛⎝⎜⎜⎜⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟⎟⎟⎟∑ = i 0 − n 1⎛⎝⎜⎜⎜⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟⎟⎟⎟2⎛⎝⎜⎜⎞⎠⎟⎟ + i12π⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟cos2 π2⎛⎝⎜⎜⎞⎠⎟⎟ + i12nnnMAPLE knows a formula for this sum:> mvn:=value(msn);mvn2 π2⎛⎝⎜⎜⎞⎠⎟⎟cosπ2nπ⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟− + + 1⎛⎝⎜⎜⎞⎠⎟⎟cosπ2n2⎛⎝⎜⎜⎞⎠⎟⎟sinπ2n2()cos π22⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟− + 1⎛⎝⎜⎜⎞⎠⎟⎟cosπ2n22⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜ := 2 π⎛⎝⎜⎜⎞⎠⎟⎟sinπ2n()sin π2()cos π2− + 1⎛⎝⎜⎜⎞⎠⎟⎟cosπ2n2π⎛⎝⎜⎜⎞⎠⎟⎟cosπ2n()cos π22n⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟− + 1⎛⎝⎜⎜⎞⎠⎟⎟cosπ2n2π⎛⎝⎜⎜⎞⎠⎟⎟cosπ2nn⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟− + 1⎛⎝⎜⎜⎞⎠⎟⎟cosπ2n2 − − + ⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟n / We now take the limit as → n ∞ ; this will agree with (be equal to) the exact value of d⌠⌡⎮⎮02 πx ()cos π xx .> ml:=limit(mvn,n=infinity); := ml − + 2()cos π2224π2()sin π2()cos π2π2> evalf(ml);1.51618535331448To use the Fundamental Theorem of Calculus, we firstuse MAPLE to find an antiderivative F (x) for f (x). The command is> int(f(x),x); + ()cos π xx()sin π x ππ2(NOTE THAT MAPLE DOES NOT INCLUDE THE " + C".)> F:=x->((cos(Pi*x)+Pi*x*sin(Pi*x))/(Pi^2)); := F → x + ()cos π x π x ()sin π xπ2You should differentiate F (x) "by hand" to check that the derivative is actually f (x)---non-computer methods for finding antiderivatives of functions like f (x) = x cos(πx) will be developed in Math 152.By the Fundamental Theorem, d⌠⌡⎮⎮02 πx ()cos π xx = F (2π) - F (0) .> JFTC:=F(2*Pi)-F(0); := JFTC − + ()cos 2 π22 π2()sin 2 π2π21π2 This looks different from the answer obtained above, but the two are actually equal. Indeed, > evalf(JFTC);1.51618535331451> evalf(2*(2*sin(Pi^2)*cos(Pi^2)*Pi^2-1+cos(Pi^2)^2)/(Pi^2));1.51618535331448This apparent differance is due to round off errors. If we increase the number of digits and repeat the computations the agreement is closer:> Digits:=15; := Digits15The fifteen digit value from computing the limit of the Riemann sum is:> evalf(2*(2*sin(Pi^2)*cos(Pi^2)*Pi^2-1+cos(Pi^2)^2)/(Pi^2));1.51618535331448The fifteen digit value from the Fundamental Theorem of Calculus is:>
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