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Lab Assignment 8 - The Fundamental Theorem of Calculus | MATH 149, Lab Reports of Pre-Calculus

Material Type: Lab; Professor: McCray; Class: Calculus/Precalculus II; Subject: Mathematics; University: Illinois Institute of Technology; Term: Unknown 1989;

Typology: Lab Reports

Pre 2010

Uploaded on 08/18/2009

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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
()fxx = Lim
=
k1
n
()fck()xkx
k1
where a = x0 < x1 < x2 < ... < xn1 < xn = b is a partition of [ a, b ] , xk1 < ck < xkfor k =
1...n, and the limit is taken as n and the lengths of the subintervals [ xk1 , 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 that
d
a
b
()fxx = 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 o
f
[ xk1, 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);
:= fxx()cos πx
> middlebox(f(x),x=0..2*Pi,20);
pf3
pf4

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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 )( xkx (^) 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 := xx 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));**