Maths integration help needed if possible

[acroviak]

hey
im really stuck on a integration question for my A level maths course.

heres the question, and hope someone can help me out.

4. The curve y=12 - 3x^2 has a tangent at the point p (2,0) on the curve.
(a) find Integral 0,2 (12-x^2)dx
(b) (i) find the gradient of the curve at point P
(ii) Point Q is where the tangent crosses the y axis, find the coordinates of this
(c) find the area of the region enclosed between the tangent, curve, and y axis.

thanks for ANY help
its much much apprechiated.

[pulse876]

(a) I assume you mean the integral of (12 - 3x^2)dx?

If so then it is [12x - x^3] from x=0 to x=2, ie (12*2-2^3) - (12*0-0^3) = 24-8=16 square units.

(b)(i) The gradient is given by the value of dy/dx at point P.

dy/dx = -6x. So if x=2, the gradient is -6*2=-12.

(ii) Assume the equation of the tangent is y = -12x + a. Since it goes through the point (2,0) then 0=-24+a, ie a = 24

So the equation of the tangent is y = 24-12x.

That line cuts the y axis when x=0, ie when y=24, ie the coordinates of Q are (0,24).

(c) The area enclosed by the tangent and the x and y axes is a triangle of size 0.5*24*2=24 square units. From part (a) the area under the curve is 16 square units. So the area of the region between the tangent, curve and y axis is 24-16=8 square units.