# Find the area between the parabola f(x) = x^2 - 3x +5 and the lines x= 0 and x = 2.

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### 3 Answers

f(x) = x^2 -3x + 5 and x =0 and x= 2

area = intg f(x)

= intg (x^2 -3x + 5)

= intg x^2 - intg 3x + intg 5 0< x < 2

= x^3/3 - 3x^2/2 + 5x 0< x< 2

= 8/2 - 12/2 + 10 - 0 - 0 + 0

= 4 - 6 + 10 = 8

Then the area = 8 square units.

First, the area which has to be calculated is located between the given curve f(x), the lines x = 0 and x = 2, 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 - 3x +5)dx

Int (x^2 - 3x +5)dx = Int x^2dx - 3Int xdx + 5Int dx

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

Int x dx = x^2/2 + C

Int dx = x + C

Int x^2dx - 3Int xdx + 5Int dx = x^3/3 -3x^2/2 + 5x + C

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

S = F(2) - F(0), where

F(2) = 2^3/3 -3*2^2/2 + 5*2 = 8/3 - 12/2 + 10 = 8/3 + 4 = 20/3

F(0) = 0^3/3 -3*0^2/2 + 5*0 = 0

S = 20/3 - 0

**S = 20/3**

The area of f(x) between x = a and x = b is given by:

A = Integral f(x) dx from x = a to b.

f(x) = x^2-3x+15

Intf(x) dx = Int (x^2-3x+5)dx = Int(x^2)dx -Int 3x dx + int % dx

= x^3 /3 - 3x^2 /2 -5x +C , as Int kx^n dx = k* x^n+1 /n+1.

Area = {x^3 /3 -3x^2 /2 -5 x + C) at x= 2 } - {x^3 /3 -3x^2 /2 -5 x + C) at x= 0 }

= {2^3 /3 -3*2^2/2 +5*2} - 0

= 8/3 -12/2 +10

= 8/3 +4

=20/3