# `int_3^6 1/10x(6x^2-15)dx `

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### 1 Answer

`int_3^6 1/10x(6x^2-15)dx`

First, factor out 1/10.

`=1/10int_3^6x(6x^2-15)dx`

Then, distribute x to (6x^2-15)dx.

`=1/10int_3^6(6x^3-15x)dx`

Then, integrate each term inside the parenthesis uisng the integral formula `intu^n du=u^(n+1)/(n+1)+C` .

`=1/10 ((6x^4)/4-(15x^2)/2)` `|_3^6`

`=1/10((3x^4)/2-(15x^2)/2) |_3^6`

`=1/10[((3*6^4)/2-(15*6^2)/2) -((3*3^4)/2-(15*3^2)/2)]`

`=1/10[(1944-270)-(243/2-135/2)]`

`=1/10(1674-54)`

`=1/10*1620`

`=162`

**Hence, `int_3^6 1/10x(6x^2-15)dx=162` .**