Mathematical Tools for Physics - Department of Physics - University

1—Basic Stuff 10

and for convenience label the endpoints asx 0 andxN. For the sketch ,N= 8.

3. Let∆xk=xk−xk− 1. That is,

∆x 1 =x 1 −x 0 , ∆x 2 =x 2 −x 1 ,···

4. In each of theN subintervals, pick one point at which the function will be evaluated. I’ll label

these points by the Greek letterξ. (That’s the Greek version of “x.”)

xk− 1 ≤ξk≤xk

x 0 ≤ξ 1 ≤x 1 , x 1 ≤ξ 2 ≤x 2 ,···

5. Form the sum that is an approximation to the final answer.

f(ξ 1 )∆x 1 +f(ξ 2 )∆x 2 +f(ξ 3 )∆x 3 +···

6. Finally, take the limit as all the∆xk→ 0 and necessarily then, asN→∞. These six steps form

the definition

lim

∆xk→ 0

∑N

k=

f(ξk)∆xk=

∫b

a

f(x)dx (1.22)

1 2

x

1 /x

To demonstrate this numerically, pick a function and do the first five steps explicitly. Pick

f(x) = 1/xand integrate it from 1 to 2. The exact answer is the natural log of 2:ln 2 = 0. 69315 ...

(1) TakeN= 4for the number of intervals

(2) Choose to divide the distance from 1 to 2 evenly, atx 1 = 1. 25 ,x 2 = 1. 5 ,x 3 = 1. 75

a=x 0 = 1. < 1. 25 < 1. 5 < 1. 75 < 2 .=x 4 =b

(3) All the∆x’s are equal to 0.25.

(4) Choose the midpoint of each subinterval. This is the best choice when you use a finite number of
divisions without taking the limit.

ξ 1 = 1. 125 ξ 2 = 1. 375 ξ 3 = 1. 625 ξ 4 = 1. 875

(5) The sum approximating the integral is then

f(ξ 1 )∆x 1 + f(ξ 2 )∆x 2 + f(ξ 3 )∆x 3 + f(ξ 4 )∆x 4 =

1

1. 125

×.25 +

1

1. 375

×.25 +

1

1. 625

×.25 +

1

1. 875

×. 25 =.

For such a small number of divisions, this is a very good approximation — about 0.3% error.

(What do you get if you takeN= 1orN= 2orN= 10divisions?)