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

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### 3 Answers

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**

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**

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.