Area under curve problem again

[lionely]

Homework Statement

Find the equation of the curve which passes through the point (-1,0) and whose gradient at
any point (x,y) is 3x²-6x+4. Find the area enclosed by the curve, the axis of x and the ordinates x=1 and x=2.


. The attempt at a solution

I integrated and got the equation of the curve to be x³ -3x² +4x +8

Now to find the area I'm not sure what to do, I tried to sketch a graph but I couldn't.

I tried integrating from x=1 to x=2 and I got 2, which isn't the answer in the book :S.

Help greatly appreciated.

 

[Saitama]

Homework Statement

Find the equation of the curve which passes through the point (-1,0) and whose gradient at
any point (x,y) is 3x²-6x+4. Find the area enclosed by the curve, the axis of x and the ordinates x=1 and x=2.


. The attempt at a solution

I integrated and got the equation of the curve to be x³ -3x² +4x +8

Now to find the area I'm not sure what to do, I tried to sketch a graph but I couldn't.

I tried integrating from x=1 to x=2 and I got 2, which isn't the answer in the book :S.

Help greatly appreciated.

Which equation did you integrate? You are given the gradient of the curve at any point (x,y). You found the curve. Find the area closed by this curve and the x-axis.

 

[lionely]

x^3 - 3x^2 +4x +8

 

[skiller]

x^3 - 3x^2 +4x +8

Show the steps you made when you integrated this from x=1 to x=2 to result in the value 2.

 

[Saitama]

x^3 - 3x^2 +4x +8

Integrate it again from x=1 to x=2. If you integrate 3x^2-6x+4 from x=1 to x=2, then you get the value 2. Knowing this, I asked you which equation did you integrate.

 

[lionely]

Sorry I intregrated 3x^2-6x+4 for x=1 to x=2 and got 2