# Find the area bounded by y=x^2-9x+1 and the lines x=0, x=1 .Find the area bounded by y=x^2-9x+1 and the lines x=0, x=1 .

hala718 | Certified Educator

y= x^2 - 9x +1     x= 0    and  x= 1

The area under the curve y is F(x) = integral y

F(x) = intg x^2 - 9x + 1

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

Now the area is:

A = F(1) - F(0)

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

= (2- 27 + 6)/6

= -19/6

Then the area = -19/6

tonys538 | Student

The area bounded by the curve y=x^2-9x+1 and the lines x=0, x=1 is required.

The graph of the three is provided below, this gives a better idea of what has to be determined.

The roots of the equation y = x^2-9x+1 are `(9+-sqrt(77))/2` . Only the root `(9-sqrt(77))/2` lies between 0 and 1.

From the graph it can be seen that the curve y = x^2 - 9x + 1 is descending from x = 0 to x = 1. The value of y changes from positive to negative.

The required area is:

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

= `[x^3/3 - (9x^2)/2 + x]_0^1`

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

= `-19/6`

But area cannot be negative.

The required area is the absolute value of `|-19/6| = 19/6`

neela | Student

The area A bounded by a curve  f(x) by the ordinates at x= a and x = b is given by:

A = Int f(x) dx.

Here f(x) = y= x^2 -9x+1 and a= 0 and b =1.

A = In (x^2-9x+1)dx x=0 to x=1.

A =( x^3 /3 -9x^2/2 +x  , at x=1 ) - =( x^2 /3 -9x^2/2 +x  , at x=0 )

A = (1/3 -9/2 + 1) -(0)

A =  (2-27+6)/6 = -19/6 = -19/6 .

william1941 | Student

The area bounded by y=x^2-9x+1 and the lines x=0 and x=1 is the integral of the function y=x^2-9x+1, between the limits x=0 and x=1.

Here we use the property that the integral of x^n is x^(n+1)/(n+1).

Integrating y=x^2-9x+1 we get (x^3)/3- (9*x^2)/2+x.

For x=1, (x^3)/3- (9*x^2)/2+x= 1/3-9/2+1.

For x=0, (x^3)/3- (9*x^2)/2+x= 0.

Subtracting the value of the integral for x=0 from the value of the integral for x=1 we get 1/3-9/2+1-0=2/6-27/6+6/6=-19/6.

Taking the absolute value of -19/6 we get the required bounded area as 19/6.

giorgiana1976 | Student

First, the area which has to be found is 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

S = -19/6