Find the integral of the function h(x) = x^3 -3x^2 +2 for the interval [0,1].

To find  the integral of the function h(x) = x^3 -3x^2 +2 
for the interval [0,1].

Int h(x) dx = H(x)  = Int (x^3
-3x^2 +2) dx.

H(x) = Int x^3 dx- Int 3x^2 dx+ Int 2dx +C
constant C.

H(x) = (1/4)x^4 -3(1/3)x^3+ 2x+
C

H(x) = (1/4)x^4 = x^3+2x
+C

Therefore Int h(x) dx from x= 0  to x= 1 is H(1)-
H(0).

H(1) - H(0) = {1/4)1^4-1^3
+2+C}-{1/4)0^4-0^3+2*0+C.

H(1) - H(0) = 1/4-1+2 , as other
terms cancel.

H(1)- H(0) =
5/4.

Therefore {Int h(x) dx  from x= 0 to x = 1} =
5/4.