What is a if the definite integral of y=(x^3+x) is (2a-1)/4, if the limits of integration are a and a+1?
First, we'll calculate the definite integral, using Leibniz Newton identity:
Int f(x)dx = F(b) - F(a)
a = lower limit of integration
b = upper limit of integration
Let y = f(x)
Int (x^3 + x)dx = Int x^3dx + Int xdx
Int x^3dx + Int xdx = x^4/4 + x^2/2
But Int (x^3 + x)dx = F(a+1) - F(a)
F(a+1) = (a+1)^4/4 + (a+1)^2/2
F(a) = a^4/4 + a^2/2
F(a+1) - F(a) = (a+1)^4/4 + (a+1)^2/2 - a^4/4 - a^2/2
[(a+1)^4 + 2(a+1)^2 - a^4 - 2a^2]/4 = (2a-1)/4
[(a+1)^4 + 2(a+1)^2 - a^4 - 2a^2] = 2a - 1
We'll raise the binomials to powers:
(a^2 + 2a + 1)^2 + 2a^2 + 4a + 2 - a^4 - 2a^2 = 2a-1
a^4 + 4a^2 + 1 + 4a^3 + 2a^2 + 4a + 4a + 2- a^4 = 2a-1
4a^3 + 6a^2 + 6a + 4 = 0
2a^3 + 3a^2 + 3a + 2 = 0
This is an odd reciprocal equation and one of it's roots is a = -1.
We'll re-write it:
2a^3 + 3a^2 + 3a + 2 = (a+1)(ca^2 + da+ e)
2a^3 + 3a^2 + 3a + 2 = ca^3 + da^2 + ea + ca^2 + da+ e
c = 2
c + d = 3 => d = 1
d + e = 3 => e = 2
The reciprocal equation is:
(a+1)(2a^2 + a+ 2) = 0
We'll calculate the roots of 2a^2 + a+ 2 = 0
delta = 1 - 16 = -15
Since the discriminant delta is negative, then the equation 2a^2 + a+ 2 = 0 has no real roots.
The only real root of the resulted equation is a = -1.