# Calculating areas.

The smallest code that gives the area between the curve p(x) = a0 + a1*x + a2*x2 + ..., the line y = 0, the line x = 0 and the line x = C

(i.e. something like this:

)

You can assume that p(x) >= 0 for x < C (bonus points if your code works for negative values of p(x)).

C, a0, a1, ...

### Output

a real number - the area

Example 1:

input: 2, 0, 1
output: 2.0


Examlpe 2:

input: 3.0, 0, 0.0, 2
output: 18


UPDATE:

• C > 0 is also assumed
• the area is between the curve, y=0, x=C and x = 0
• the input can be a list of any form; not necessarily comma separated.
• the output can be a real of any form (thus, '18' is a valid output, as is '18.0')
-
Since the answer is going to be "infinite" for almost any input, I think you've misstated the problem. – Peter Taylor Feb 3 '11 at 21:37
Should the input be read from standard input as a comma-separated string? Or can we write a function which takes a list of floats as an argument? – sepp2k Feb 3 '11 at 21:47
Do you mean between x=0, x=C, y=0, and the curve? – Keith Randall Feb 3 '11 at 22:06
@Peter: I don't think so. He shows a picture of an inverse (the integral of which would diverge), but the function he specifies is a polynomial. The definite integral over [0,C) should be well defined and finite for finite coefficients. – dmckee Feb 4 '11 at 1:20
@dmckee, I had noticed that, but my point was more that he was integrating a polynomial from -\infty to C, and for any non-trivial polynomial that diverges. The question has now been amended to fix this. – Peter Taylor Feb 4 '11 at 7:11

## Mathematica: 48 Chars

.

Sum[#[[i+1]]#[[1]]^i/i,{i,Length@#-1}]&@Input[]

-

Python - 71 63 chars:

a=input()
print sum(1.*a[i]*a[0]**i/i for i in range(1,len(a)))


It's a simple integration of a polynomial function between 0 and C. And I haven't tested it, but I'm quite sure it works for negative values.

-
 Learned something new about input() today:) – st0le Feb 4 '11 at 4:39

## J, 26 characters

f=:3 :'((1}.y)&p.d._1)0{y'


eg.

   f 2 0 1
2
f 3 0 0 2
18

-
 Neat! I can't find a way to make it more tacit. That d. being a conjunction doesn't make it very easy to my novice J skills. – J B Feb 12 '11 at 15:13 @J-B: Yes, that d. is a "problem" for me too. :) – Eelvex Feb 12 '11 at 15:21

f(c:l)=sum.map(\(i,x)->x*c**i/i)$zip[1..]l main=getLine>>=print.f.read.('[':).(++"]")  - Whoever downvoted, please leave a comment. – Eelvex Feb 5 '11 at 11:42 The question isn't as strict as you treat it. You could definitely simplify the input code, and possibly do away with explicit I/O altogether. – J B Feb 12 '11 at 15:19 ## Ruby, 65 chars i=s=0 c=gets(q=",").to_f$<.each(q){|a|s+=a.to_f*c**(i+=1)/i}

i think assigning the "," to a variable will help...how about this c=gets(q=",").to_f and $<.each(q){|a|s+=a.to_f*c**(i+=1)/i}, saves one char.... – st0le Feb 4 '11 at 4:31 @st0le: Very nice. Thanks. – sepp2k Feb 4 '11 at 4:32 Assigning "," (or ?,, which is even shorter) to $/ allows you to omit the argument to $<.each. And $<.map is one character shorter than \$<.each. ;) – Ventero Feb 12 '11 at 19:58