Given the curve f(x) = x^2 and the line y= 5x-6

We need to find the area between the curve and the line.

First we need to find the intersection points.

==> x^2 = 5x-6

==> x^2 - 5x + 6 = 0

==> (x-2)(x-3) = 0

==> x = 2 ==> x = 3

Then we need to find the area between the interval [ 2, 3].

First we will find the ares under the curve f(x) = x^2.

==> Int f(x) = Int x^2 dx = x^3/3

==> A1 = F(3) - F(2) = 27/3 - 8/3 = 19/3.

Then the area under the curve f(x) is A1= 19/3.

Now we will calculate the area under the line y.

==> Int y = Int 5x -6 dx = 5x^2/2 - 6x

==> Y(3) = (5/2)*9 - 18 = (45-36)/2 = 9/2

==> Y(2) = (5/2)*4 - 12 = 10 - 12 = -2

==> y(3) - y(2) = 4.5 + 2 = 6.5 = 13/2

Then the area between the line and the curve is :

A = A2 - A1 = 13/2 - 19/3 = 0.166666

**Then the ares is given by 0.16666 square units.**

We need to find the area bounded by the curve f(x) = x^2 and the line y= 5x-6.

First let's find the point of intersection of the two

x^2 = 5x - 6

=> x^2 - 5x + 6 = 0

=> x^2 - 3x - 2x + 6 = 0

=> x(x - 3) - 2(x - 3) =0

=> x = 2 and x = 3

The area bounded is equal to

Int [x^2 - 5x + 6], x = 2 to x = 3

=> [x^3/3 - 5x^2/2 + 6x], x = 2 to x = 3

=> |(3^3 - 2^3)/3 - (5/2)(9 - 4) + 6(3 - 2)|

=> |19/3 - 25/2 + 6|

=> |-1/6|

**The area bounded by the curve f(x) = x^2 and the line y= 5x - 6 is 1/6.**