If the area enclosed by curve `y=f(x) and y=x^2+2`
between the abscissa `x=2 and x=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.`Find f(x)
Text Solution
Verified by Experts
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`
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