# SICP Goodness - Stream (10)

integral using stream

Do you think Computer Science

equalsbuilding websites and mobile apps?

Are you feeling that you are doing

repetitiveandnot so intelligentwork?

Are you feeling a bit

sickaboutreading manualsandcopy-pastingcode and keeppoking arounduntil it worksall day long?

Do you want to understand the

soulofComputer Science?

If yes,

read SICP!!!

Let’s talk about the integral example in the book.

This has always been a problem for me, whenever I get here, I either give up or pretend that I get it and move on. Not this time!

Let’s put the example here first for your reference.

And this is the mathematical formula:

And the code:

```
(define (integral integrand initial-value dt)
(define int
(cons-stream initial-value
(add-stream (scale-stream integrand dt)
int)))
int)
```

In order to fully grasp this, we need to refresh on what is integral.

In my own words, it basically represents the area between the function and x-axis. For example, the following picture shows the integral of a certain function *f(x)* between *a* and *b*.

But computational wise, how to we calculate this?

One simple way is to model this function into many many thin rectangles. Then, the sum of the area of these rectangles can be used as an approx. of the area of the function.

If you have difficulty understanding this, please watch this excellent video:

# Example 1

Let’s now play with the `integral`

procedure a little bit and see if we can actually use it to do something.

The simplest function that we can test is the function `f(t) = 1`

.

We already know its integral is function `f(t) = t`

.

Now let’s verify this using the `integral`

procedure.

We need to figure out what are the arguments `integrand`

, `initial-value`

and `dt`

.

`integrand`

is simple, it is just the stream `ones`

we have define previously.

`initial-value`

should be 0.

`dt`

let’s set it to 0.001.

Let’s say we want to calculate the integral of the function between 0 and 1. Since `dt`

is 0.001, that means we need 1000 values to reach t = 1.

```
(stream-ref (integral ones 0 0.001) 1000)
;Value: 1.0000000000000007
```

To calculate the integral of the function between 0 and 2. We need 2000 items from the input stream.

```
(stream-ref (integral ones 0 0.001) 2000)
;Value: 1.9999999999998905
```

Let’s try more,

```
(stream-ref (integral ones 0 0.001) 3000)
;Value: 2.9999999999997806
(stream-ref (integral ones 0 0.001) 4000)
;Value: 3.9999999999996705
```

These values do fit the function `f(t) = t`

nicely.

# Example 2

Now let’s try to integral the function `f(t) = t`

.

We know that the integral is `f(t) = 0.5t^2`

.

Let’s try the `integral`

procedure.

Most important thing is to figure out what is the `integrand`

.

The `integrand`

is a stream of values `f(t)`

, where `t`

is a stream of values that determined by `dt`

.

Say if we use `dt = 0.001`

, then `t`

will be a stream of values of `0, 0.001, 0.002, 0.003 ...`

and `f(t)`

will be a stream of values of `0, 0.001, 0.002, 0.003 ...`

.

Now we know that the input stream will be the `integers`

stream scaled by 0.001.

Let’s now see if the integrals are like the function `f(t) = 0.5t^2`

.

```
(stream-ref (integral (scale-stream integers 0.001) 0 0.001) 1000)
;Value: .5005000000000002
(stream-ref (integral (scale-stream integers 0.001) 0 0.001) 2000)
;Value: 2.001
(stream-ref (integral (scale-stream integers 0.001) 0 0.001) 3000)
;Value: 4.5015
```

And indeed they are.

I hope now you have a better understanding of the `integral`

procedure.

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