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

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To find the value of the definite integral of h(x) = x^3 -3x^2 +2 in the interval [0, 1] we first find the integral of h(x) = x^3 -3x^2 +2.

Int [ h(x) ]

= Int[x^3 -3x^2 +2]

= Int [ x^3 ] - Int [ 3x^2] + Int [2]

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

For x = 1, x^4 / 4 - x^3 + 2x + C = 1/4 - 1 + 2 + C

=> 1.25 + C

For x = 0, x^4 / 4 - x^3 + 2x + C = C

Subtracting the two we get 1.25

The required result is 1.25

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Given the function h(x) = x^3 - 3x^2 + 2

We need to find the integral of h(x) for the interval [0,1]

Let us integrate h(x).

==>H(x) =  intg h(x) = intg ( x^3 - 3x^2 +2) dx

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

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

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

Now we will find the finite integral from x= 0 to x= 1

==> H(1) = (1/4) - 1 + 2 + C = 5/4 + C

==> H(0) = C

==> H(x) from x =0 to x= 1 = H(1) - H(0) = 5/4

 

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