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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate definite integral such that:

`int_0^1 (x^2 - 9x + 1)dx`

You may use the property of linearity of integral, such that:

`int_a^b (f(x) + g(x))dx = int_a^b f(x) dx + int_a^b g(x) dx`

Reasoning by analogy yields:

`int_0^1 (x^2 - 9x + 1)dx = int_0^1 (x^2)dx - int_0^1 9x dx + int_0^1 dx`

`int_0^1 (x^2 - 9x + 1)dx =(x^3/3 - 9x^2/2 + x)|_0^1`

You need to use the fundamental theorem of calculus, such that:

`int_0^1 (x^2 - 9x + 1)dx = (1^3/3 - 9*1^2/2 + 1) - (0^3/3 - 9*0^2/2 + 0)`

`int_0^1 (x^2 - 9x + 1)dx =1/3 - 9/2 + 1 => int_0^1 (x^2 - 9x + 1)dx = (2 - 27 + 6)/6 => int_0^1 (x^2 - 9x + 1)dx = -19/6`

Hence, evaluating the given definite integral, using the property of linearity, yields `int_0^1 (x^2 - 9x + 1)dx = -19/6.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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