If the area enclosed by curve `y=f(x)a n dy=x^2+2` between the abscissa `x=2a n dx=alpha,alpha>2,` is `(alpha^3-4alpha^2+8)s qdot` unit. It is known that curve `y=f(x)` lies below the parabola `y=x^2+2.`

Explanation

According to the question,
`alpha^(3)-4alpha^(2)+8=overset(alpha)underset(2)int(x^(2)+2-f(x))dx`
Differentiating w.r.t. `alpha` on both sides, we get
`3alpha^(2)-8alpha=alpha^(2)+2-f(alpha)`
`therefore" "f(x)=-2x^(2)+8x+2`