3         4
If  ∫c(x)dx = ∫c(x)dx, then what is c(3)?
    1          2

My calculus is a bit rusty (haven’t used it for real since college and I usually don’t do the calculus problems here) so I won’t be surprised if there are mistakes. But here goes…

Since we know c(x) is a cubic function, let’s say c(x) = ax^3 + bx^2 + cx + d

From the three points given, we know the following three equations:
c(1) = a + b + c + d = 2
c(2) = 8a + 4b + 2c + d = 7
c(4) = 64a + 16b + 4c + d = 14

If we subtract systems of equations, we find the following relationships:
R1: c(2) – c(1): 7a + 3b + c = 5
R2: c(4) – c(2): 56a + 12b + 2c = 7

And if we subtract these systems of equations we find this relationship:
R3: R2 – 2*R1: 42a + 6b = -3 or 14a + 2b = -1

I believe the integral of c(x) by dx from point s to point t =
a/4(t^4 – s^4) + b/3(t^3 – s^3) + c/2(t^2 – s^2) + d(t – s) + e(t^0 – s^0)

(Notice that e(t^0 – s^0) = e(1-1) = 0. So you won’t see the e coefficient mentioned again).

Since we are told that the integral from 1 to 3 is equal to the integral from 2 to 4, we know that:
a/4(81-1) + b/3(27-1) + c/2(9-1) + d(3-1) = a/4(256-16) + b/3(64-8) + c/2(16-4) + d(4-2)
20a + 26b/3 + 4c + 2d = 60a + 56b/3 + 6c + 2d
40a + 10b + 2c = 0
20a +5b + c = 0

Let’s call this R4.

R5: R4 – R1: 13a + 2b = -5

R3 – R5: a = 4
So now for R3 we have 14*4 + 2b = -1. So b = -28.5
So now for R1 we have 7*4 – 3*28.5 + c = 5. So c = 62.5
So now for c(1) we have 4 – 28.5 + 62.5 + d = 2. So d = -36

Therefore c(x) = 4x^3 – 28.5x^2 + 62.5x – 36
So c(3) = 3