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MATH 149
LABORATORY ASSIGNMENT 8
THE FUNDAMENTAL THEOREM
OF CALCULUS
In class we defined the definite integral of the function f (x) on the interval [ a, b ] as
d
a
b
f ( x ) x = Lim ∑
k = 1
n
f ( c (^) k )( xk − x (^) k − 1 )
where a = x 0 < x 1 < x 2 < ... < x (^) n − 1 < x (^) n = b is a partition of [ a, b ] , x (^) k − 1 < ck < xk for k =
1...n, and the limit is taken as n →∞ and the lengths of the subintervals [ x (^) k − 1 , x (^) k ]
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 that
d
a
b
f ( x ) x = 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 c (^) k 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->xcos(Pix);**
f := x → x cos( π x )
> middlebox(f(x),x=0..2Pi,20);*
The corresponding approximating sum is
> ms20:=middlesum(f(x),x=0..2Pi,20);*
ms20 :=
π
i = 0
19
i + ⎟⎟
π
cos ⎟
π
i + ⎟⎟
or
> 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..2Pi,n);*
> F:=x->((cos(Pix)+Pixsin(Pix))/(Pi^2));**
F := x →
cos( π x )+π x sin( π x )
π
2
You 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(π
x) will be developed in Math 152.
By the Fundamental Theorem, d
0
2 π
x cos( π x ) x = F (2π) - F (0).
> JFTC:=F(2Pi)-F(0);*
JFTC := −
cos 2( π )+
2 2 π
2 sin 2( π )
2
π
2
π
2
This looks different from the answer obtained above, but the two are actually
equal.
Indeed,
> evalf(JFTC);
evalf(2(2sin(Pi^2)cos(Pi^2)Pi^2-1+cos(Pi^2)^2)/(Pi^**
));
This 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;
Digits := 15
The fifteen digit value from computing the limit of the Riemann sum is:
> evalf(2(2sin(Pi^2)cos(Pi^2)Pi^2-1+cos(Pi^2)^2)/(Pi^**
));
The fifteen digit value from the Fundamental Theorem of Calculus is:
> evalf((cos(2Pi^2)+2Pi^2sin(2Pi^2))/(Pi^2)-1/(Pi^2));**
