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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

First let's integrate h(x) = 3x^3 + 5x^2 - 3x + 6.

Int [ 3x^3 + 5x^2 - 3x + 6 ]

=> Int [ 3x^3 ]+ Int [5x^2] - Int [ 3x ] + Int [6] + C

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

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

For the interval [ 0, 1 ], the value of the integral is

(3/4)*1^4 + (5/3)*1^3 - (3/2)*1^2 + 6*1 + C - (3/4)*0^4 + (5/3)*0^3 - (3/2)*0^2 + 6*0 + C

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

=> 83/12

Therefore the value of the definite integral of h(x) = 3x^3 + 5x^2 - 3x + 6 for the interval [ 0, 1] is 83/12.

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