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Definite Integral interval [a, b]: Sf(x)dx = G(b) - G(a) where G'(x) = f(x) , or differential of G(x) = f(x).

Integrals appear in many practical situations.
Consider a swimming pool. If it is rectangular, then from its length, width, and depth we can easily determine the volume of water it can contain (to fill it), the area of its surface (to cover it), and the length of its edge (to rope it).

But if it is oval with a rounded bottom, all of these quantities call for integrals. Practical approximations may suffice for such trivial examples, but precision engineering (of any discipline) requires exact and rigorous values for these elements.

I create the query which calculate integral of 3 * X^2 - 2 * X in interval [2, 4]....

S(3 * X^2 - 2 * X)dx = X^3 - X^2 and exact value of integral will be
S = (4^3 - 4^2) - (2^3 - 2^2) = 48 - 4 = 44.

We know nothing in our solution about exact value and differential and suppose to calculate approximate value of integral.
Process will stop when the absolute difference between current value of integral and previous become less or equal some eps real number.

Code:
```With
Source (Xstart, Xfinish, imgFunc, eps) as
(select double(2), double(4), '3 * X^2 - 2 * X', double(1.e-3)
from sysibm.sysdummy1
)
,
Integral_calc (Xs, Xf, Xc, step, curint, prevint, eps, iterno) as
(select Xstart, Xfinish, double(Xstart - (Xfinish - Xstart) / 10.) Xc,
double((Xfinish - Xstart) / 10.) step,  double(0), double(0), eps, int(0)
from Source
union All
select Xs, Xf, Xc + step, step, curint + Fc * step, prevint, eps, iterno + 1
from Integral_calc, table
(select 3 * power(Xc + step, 2) - 2 * (Xc + step) Fc
from sysibm.sysdummy1 ) it
where  Xc + step <= Xf
Union All
select Xs, Xf, Xs - (step / 2.), step / 2., 0., curint, eps, iterno + 1
from Integral_calc
where Xc + step > Xf
and abs(curint - prevint) > eps
)
,
Integral(integral_value, integral_image) as
(select curint, 'interval: [' || varchar(Xs) || ', ' || varchar(Xf) || ']:  S' || '('
|| imgFunc || ')dx = ' || varchar(round(curint, 3))
from Integral_calc, Source
where iterno = (select max(iterno) from Integral_calc)
)
select integral_value, integral_image from Integral```
Result of calculation:

INTEGRAL_VALUE............................ INTEGRAL_IMAGE
4.40005859378289E+001............... interval: [2.0E0, 4.0E0]: S(3 * X^2 - 2 * X)dx = 4.4001E1
Could be very useful for students and engineers.

Lenny

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This query is working fast for short intervals.

Do not use it for intervals with length > 10....

Also, if you want calculate yours integral, you have to change
Code:
```table (select 3 * power(Xc + step, 2) - 2 * (Xc + step) Fc
from sysibm.sysdummy1 ) it```
and

Code:
```Source (Xstart, Xfinish, imgFunc, eps) as
(select double(1), double(12), '3 * X^2 - 2 * X', double(1.e-3)
from sysibm.sysdummy1```
Lenny

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Lenny, do you think somebody here remember what is it "definite integral" ?

Kara S.

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Posts
150

## Just better look

For me looks better if looks so:

Code:
```With
Source (Xstart, Xfinish, imgFunc, eps) as
(select double(2), double(4), '3 * X^2 - 2 * X', double(1.e-3)
from sysibm.sysdummy1
)
,
Integral_calc (Xs, Xf, Xc, step, CurInt, PrevInt, eps, iterno) as
(select Xstart, Xfinish, double(Xstart - (Xfinish - Xstart) / 10.) Xc,
double((Xfinish - Xstart) / 10.) step,  double(0), double(0), eps, int(0)
from Source
union All
select Xs, Xf, Xc + step, step, CurInt + FuncVal * step, PrevInt, eps,
iterno + 1

From Integral_calc,
table (select 3 * power(X, 2) - 2 * X  as FuncVal
From
(select (Xc + step) as X rom sysibm.sysdummy1 ) ii     ) it
where  Xc + step <= Xf

Union All
select Xs, Xf, Xs - (step / 2.), step / 2., 0., CurInt, eps, iterno + 1
from Integral_calc
where Xc + step > Xf
and abs(curint - prevint) > eps
)
,
Integral(integral_value, integral_image) as
(select curint, 'interval: [' || varchar(Xs) || ', ' || varchar(Xf) || ']:  S' || '('
|| imgFunc || ')dx = ' || varchar(round(curint, 3))
from Integral_calc, Source
where iterno = (select max(iterno) from Integral_calc)
)
select integral_value, integral_image from Integral```
Kara S.

5. :-)
Join Date
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Originally Posted by DB2Plus
Lenny, do you think somebody here remember what is it "definite integral" ?

Kara S.
No, of course not. You two guys are the only ones to carry the torch of knowledge. All the rest are just some lowly yeomen capable of no more than herding tablespaces...

6. SPAMMER
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4
wow that is so difficul for me...to bad

7. Registered User
Join Date
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Posts
150
Originally Posted by n_i
No, of course not. You two guys are the only ones to carry the torch of knowledge. All the rest are just some lowly yeomen capable of no more than herding tablespaces...
Sorry, if I did make mistake.

Kara S.

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In my opinion when interval includes special numbers pi or e, we have to use something special,
because we don't know exact values of these two numbers and have to use scalar functions, e.g.
pi = 2 * asin(1.0) , e = exp(1.)

Kara S.
Last edited by DB2Plus; 09-27-09 at 16:49.

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Posts
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## Example with trigonometric function

For example: interval [0, 3/4pi] S sin(x)dx:

Code:
```With
Source (Xstart, Xfinish, imgFunc, eps) as
(select double(0), 1.5 * asin(1.), 'sin(x)', double(1.e-4)
from sysibm.sysdummy1
)
,
Integral_calc (Xs, Xf, Xc, step, CurInt, PrevInt, eps, iterno) as
(select Xstart, Xfinish, double(Xstart - (Xfinish - Xstart) / 10.) Xc,
double((Xfinish - Xstart) / 10.) step,  double(0), double(0), eps, int(0)
from Source
union All
select Xs, Xf, Xc + step, step, CurInt + FuncVal * step, PrevInt, eps,
iterno + 1

From Integral_calc,
table (select sin(x) as FuncVal
From
(select (Xc + step) as X rom sysibm.sysdummy1 ) ii     ) it
where  Xc + step <= Xf

Union All
select Xs, Xf, Xs - (step / 2.), step / 2., 0., CurInt, eps, iterno + 1
from Integral_calc
where Xc + step > Xf
and abs(curint - prevint) > eps
)
,
Integral(integral_value, integral_image) as
(select curint, 'interval: [' || varchar(Xs) || ', ' || varchar(Xf) || ']:  S' || '('
|| imgFunc || ')dx = ' || varchar(round(curint, 4))
from Integral_calc, Source
where iterno = (select max(iterno) from Integral_calc)
)
select integral_value, integral_image from Integral```
Kara S.

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## Some notes

Originally Posted by Lenny77
This query is working fast for short intervals.

Do not use it for intervals with length > 10....

Also, if you want calculate yours integral, you have to change
Code:
```table (select 3 * power(Xc + step, 2) - 2 * (Xc + step) Fc
from sysibm.sysdummy1 ) it```
and

Code:
```Source (Xstart, Xfinish, imgFunc, eps) as
(select double(1), double(12), '3 * X^2 - 2 * X', double(1.e-3)
from sysibm.sysdummy1```
Lenny
Hi, Lenny !
We have to change
Code:
```Integral(integral_value, integral_image) as
(select curint, 'interval: [' || varchar(Xs) || ', ' || varchar(Xf) || ']:  S'
|| '('  || imgFunc || ')dx = '
|| varchar(round(curint, 3))
from Integral_calc, Source
where iterno = (select max(iterno) from Integral_calc)```
Because of
round(curint, 3)
depends on EPS

Kara S.

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Location
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Posts
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## Small correction to trigonometric integral

Thank you Kara for yours trigonometric example !

One small correction:

Code:
```With
Source (Xstart, Xfinish, imgFunc, eps) as
(select double(0), 1.5 * asin(1.), 'sin(x)', double(1.e-4)
from sysibm.sysdummy1
)
,
Integral_calc (Xs, Xf, Xc, step, CurInt, PrevInt, eps, iterno) as
(select Xstart, Xfinish, double(Xstart - (Xfinish - Xstart) / 10.) Xc,
double((Xfinish - Xstart) / 10.) step,  double(0), double(0), eps, int(0)
from Source
union All
select Xs, Xf, Xc + step, step, CurInt + FuncVal * step, PrevInt, eps,
iterno + 1

From Integral_calc,
table (select sin(x) as FuncVal
From
(select (Xc + step) as X from sysibm.sysdummy1 ) ii     ) it
where  Xc + step <= Xf

Union All
select Xs, Xf, Xs - (step / 2.), step / 2., 0., CurInt, eps, iterno + 1
from Integral_calc
where Xc + step > Xf
and abs(curint - prevint) > eps
)
,
Integral(integral_value, integral_image) as
(select curint, 'interval: [' || varchar(Xs) || ', ' || varchar(Xf) || ']:  S' || '('
|| imgFunc || ')dx = ' || varchar(round(curint, 4))
from Integral_calc, Source
where iterno = (select max(iterno) from Integral_calc)
)
select integral_value, integral_image from Integral```
and it works !

Result:

interval: [0E0, 2.35619449019234E0]: S(sin(x))dx = 1.7072E0
Lenny

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Location
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Posts
963

## Check the result

We can check the result of calculation (less then 1 second).

We know, or you can find in the smart books:
S(six(x))dx = -cos(x)

So, on interval: [0E0, 2.35619449019234E0] exact value of integral will be:

Code:
```select -cos(2.35619449019234E0) - (-cos(0E0))
from sysibm.sysdummy1```
E.g.
S(six(x))dx = 1.70710678118654E+000
Lenny

13. Registered User
Join Date
Jul 2009
Posts
150
Thank you, Lenny.

But where is my mistake ?

Kara S.
Last edited by DB2Plus; 09-29-09 at 22:21.

14. Registered User
Join Date
Jul 2009
Posts
150
Lenny, thanks.

Now I found my mistake:

(select (Xc + step) as X rom sysibm.sysdummy1 ) ii ....

where has to be:

Code:
`(select (Xc + step) as X From sysibm.sysdummy1 ) ii ....`
Sorry, everybody!
I have made this query without test, because I don't have DB2 at my home.

Kara S.

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