Calculating areas.

Question

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)).

Input

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')

Answer 1

APL (Dyalog Unicode), 9 bytes

⊃⊥∘⌽⊢÷⍳∘≢

Try it online!

A tacit function that takes C, a0, a1, ... as a single vector argument.

12 bytes if a full program is needed. Note that, while the default representation for a numeric vector is the numbers separated by spaces (no comma or anything), it accepts comma-separated ones since ⎕ evals the input and , is the "concatenate" function.

(⊃⊥∘⌽⊢÷⍳∘≢)⎕

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Uses ⎕IO←0 and ⎕DIV←1, so that the range generation is 0-based and division by zero gives 0.

Because the polynomial is nonnegative and the integral starts at 0, it suffices to evaluate the antiderivative of the given polynomial at C.

The antiderivative of \$a_0 + a_1x + a_2x^2 + \dots\$ is \$a_0x + \frac{a_1}{2} x^2 + \frac{a_2}{3} x^3 + \dots\$. Because we have C attached to the start of the vector, we simply divide the entire array by their 0-based indexes:

       C  a0   a1   a2
input: 3  0    0    2
div  : 0  1    2    3
res  : 0  0    0    2/3
       0  a0/1 a1/2 a2/3

Then we simply evaluate the resulting polynomial at C.

How it works: the code

⊃⊥∘⌽⊢÷⍳∘≢  ⍝ Input: single array of C, a0, a1, a2, ...
      ⍳∘≢  ⍝ 0-based range 0, 1, 2, 3, ...
    ⊢÷     ⍝ Divide self by above
  ∘⌽       ⍝ Reverse
⊃⊥         ⍝ Convert from base C to real number
           ⍝ i.e. evaluate the polynomial, lowest term given last

Answer 2

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.

Answer 3

J, 26 characters

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

e.g.

   f 2 0 1
2
   f 3 0 0 2
18

Answer 4

Mathematica: 48 Chars

.

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

Answer 5

Haskell, 85 characters

f(c:l)=sum.map(\(i,x)->x*c**i/i)$zip[1..]l
main=getLine>>=print.f.read.('[':).(++"]")

Answer 6

APL(NARS), 8 chars, 16 bytes

∘{⍵⊥⌽⍺}∫

The new NARS2000 function integrate, ∫.

it seems f∫b is integrate(f, 0..b)=∫_0^b f Here i don't know if can be ok, the right argument is the max of interval of integration [0, max] and the left argument is the list of numbers represent polinomy...

  f←∘{⍵⊥⌽⍺}∫
  0 1 f 2
2 
  0 0 2 f 3
18 
  1 2 3 4 5 6 f 7
137256 

Answer 7

Ruby, 65 chars

i=s=0
c=gets(q=",").to_f
$<.each(q){|a|s+=a.to_f*c**(i+=1)/i}
p s

The code reads until the end of input, not the end of the line. So you need to hit Ctrl+D to terminate input. (Pipe the input in using echo or from a file.)

Answer 8

C GCC 186 182 bytes

f(){i=0,l=1;float *d,s=0.0;d=malloc(sizeof(float)*50);scanf("%f",&d[0]);while(getchar()!='\n'){scanf("%f",&d[l]);l++;}for(i=0;i<l;i++)s+=d[i+1]*pow(d[0],(i+1))/(i+1);printf("%f",s);}

This program gives an output (area) for any curve between the curve, y=0, x=C and x=0. It can take coefficients (float as well) from a₀ to a₄₈. The first accepted input is C followed by coefficients. Press Ènter after the last coefficient.

void f()
{
  int i=0,l=1;
  float *d,s=0.0;
  const int sz=100;
  d=malloc(sizeof(float)*sz);

  scanf("%f",&d[0]);
  while(getchar()!='\n')
  {
    scanf("%f",&d[l]);
    l++;
  }

  for(i=0;i<l;i++)
    s+=d[i+1]*pow(d[0],(i+1))/(i+1);

   printf("%f",s);
}

Answer 9

Mathematica, 48 bytes

Try it online!

f[a_]:=Sum[a[[i+1]]*a[[1]]^i/i,{i,1,Length@a-1}]