Find the definite integral of h(x) = 3x^3 + 5x^2 - 3x + 6 for the interval [ 0, 1].

f(x) = 3x^3 + 5x^2 - 3x + 6

First,
we will find the integral of f(x).

Let F(x)
= integral f(x)

Then, the definite integral is given
by:

F = F(1) - F(0)

Let us
calculate.

F(x) = intg f(x)

        =
intg ( 3x^3 + 5x^2 - 3x + 6) dx

        = intg 3x^3 dx + intg 5x^2
dx - intg 3x dx + intg 6 dx.

         = 3x^4/4 + 5x^3/3 - 3x^2/2 +
6x + C

==> F(x) = (3/4)x^4 + (5/3)x^3 -(3/2) x^2 + 6x +
C.

Now we will substitute with x=
1.

==> F(1) = (3/4) + (5/3) - 3/2 + 6 +
C

                 = ( 9 + 20 - 18 + 72) /12 =  83/
12

==> F(1) = 83/12  + C.

Now we
will substitute with x= 0.

==> F(0) = 0 +
C

==> F = F(1) -
F(0).

==> F =
83/12