$$fk(x)= x^3+x^2*(k-1)-k$$$$f0(x)= x^3-x^2$$$$v(x)=π*f0(x)^2=$$$$π*(x^3-x^2)^2=$$$$π*(x^6-2x^5+x^4)$$
$$V(x)=$$$$ π(1/7x^7-2/6x^6+1/5x^5)$$$$V(0)=0$$$$V(-1)=$$$$π*(-1/7-2/6-1/5)$$$$≈-2,124$$$$V=V(0)-V(-1)≈2,124E^3$$
2)
$$fk(x)= x^3+x^2*(k-1)-k$$
$$fk(0)=-k$$
$$fk(1)= 1+(k-1)-k=0$$
x≠1
(x^3+x^2*(k-1)-k)/(x-1)=x^2+k*x +k
x^3-x^2
x^2*k
x^2*k-x*k
x*k -k
x*k -k
X=-k/2+\( \sqrt{(k/2)^2-k} \)
Für
$$(k/2)^2-k<0 $$
Für k<4 wird x=1 die einzige Nullstelle, ausser k=0
$$f0(x)=x^3-x^2 $$
x=1 Nullstelle; x=0 doppelte Nullstelle.