Question

How do you find the definite integral for: `[(6x^2+2) / sqrt(x)] dx` for the intervals `[1, 5]`?

Answer 1

`~~136.71`

Explanation:

`"rewrite as"`

`(6x^2)/(x^(1/2))+2/x^(1/2)=6x^(3/2)+2x^(-1/2)`

`rArrint_1^5(6x^(3/2)+2x^(-1/2))dx`

`"integrate each term using the "color(blue)"power rule"`

`• int(ax^n)=a/(n+1)x^(n+1); n!=-1`

`=[12/5x^(5/2)+4x^(1/2)]_1^5`

`=(12/5. 5^(5/2)+4. 5^(1/2))-(12/5 .1+4.1)`

`=60sqrt5+4sqrt5-32/5`

`=64sqrt5-32/5~~136.71" to 2 dec. places"`

Answer 2

`64sqrt5-6.4`

Explanation:

Let, `I=int_1^5 (6x^2+2)/sqrtx dx.`

We know that, `intx^ndx=x^(n+1)/(n+1), n!=-1.`

`:. I=int_1^5(6x^2+2)/sqrtxdx,`

`=int_1^5(6x^2+2)*x^(-1/2)dx,`

`=int_1^5{(6x^(2-1/2)+2x^(-1/2)}dx,`

`=6int_1^5x^(3/2)dx+2int_1^5x^(-1/2)dx,`

`=6[x^(3/2+1)/(3/2+1)]_1^5+2[x^(-1/2+1)/(-1/2+1)]_1^5,`

`=6*2/5[x^(5/2)]_1^5+2*2[x^(1/2)]_1^5,`

`=12/5[5^(5/2)-1^(5/2)]+4[5^(1/2)-1^(1/2)]`

`=12/5(25sqrt5-1)+4(sqrt5-1),`

`=60sqrt5-12/5+4sqrt5-4,`

`rArr I=64sqrt5-6.4`

Enjoy Maths.!