# What is the definite integral of x^2 - 9x + 1 ; x = 0 to x = 1 ?

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We have to find the definite integral of x^2 - 9x + 1 ; x = 0 to x = 1

Let f(x) = x^2 - 9x + 1

Now we know that the integral of x^n is given by x^(n+1) / (n+1), we shall use this to find the integral of f(x).

Therefore Int [ f(x)] = Int [x^2 - 9x + 1]

= x^3 / 3 - 9x^2 /2 + x

Now as we need to find the definite integral from x=0 to x=1, it is given by the Int[f(1)] - Int[f(0)]

= 1^3 / 3 - 9*1^2 /2 + 1 - [0^3 / 3 - 9*0^2 /2 + 0]

= 1/3 - 9/2 + 1 - 0

= -19/6

**Therefore definite integral of x^2 - 9x + 1 ; x = 0 to x = 1 is given as -19/6**

The definite integral of the function represents the area located between the given curve f(x), the lines x = 0 and x = 1, also the x axis.

To calculate the area, we'll use the formula:

S = Integral (f(x) - ox)dx = Int f(x)dx = Int (x^2-9x+1)dx

Int (x^2-9x+1)dx = Int x^2dx - 9Int xdx + Int dx

Int x^2dx = x^3/3 + C

Int x dx = x^2/2 + C

Int dx = x + C

Int x^2dx - 9Int xdx + Int dx = x^3/3 -9x^2/2 + x + C

Now, we'll calculate the value of the area, using Leibnitz Newton formula:

S = F(1) - F(0), where

F(1) = 1^3/3 -9*1^2/2 + 1 = 1/3 - 9/2 + 1 = (2-27+6)/6= -19/6

F(0) = 0

S = -19/6 - 0

**The definite integral of f(x) = x^2 - 9x + 1 is the area S = -19/6 square units.**