# `int_a^bf(x)dx` `` `a= 3, b= 6``` `f(x) = 1/10x(6x^2-15)dx` `` `Evaluate ` `` `` `` `` `` `` `` `` `` `` `` `` ``

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You need to simplify the function removing the round brackets, such that:

`f(x) = (x/10)(6x^2 - 15) => f(x) = (6x^3)/10 - (15x)/10`

Reducing duplicate factors yields:

`f(x) = (3x^3)/5 - (3x)/2`

You need to evaluate the definite integral, such that:

`int_3^6 f(x) dx = int_3^6 ((3x^3)/5 - (3x)/2) dx`

Using the property of linearity of integral, you need to split the integral in two simpler integrals, such that:

`int_3^6 f(x) dx = int_3^6 ((3x^3)/5) dx -int_3^6 ((3x)/2) dx`

`int_3^6 f(x) dx =(3/5)(x^4/4)|_3^6 - (3/2)(x^2/2)|_3^6`

Using the fundamental theorem of calculus yields:

`int_3^6 f(x) dx = (3/20)(6^4 - 3^4) - (3/4)(6^2 - 3^2)`

`int_3^6 f(x) dx = (3/20)(1296 - 81) - (3/4)(36 - 9)`

`int_3^6 f(x) dx = (3*1215/20) - 81/4`

`int_3^6 f(x) dx = (729 - 81)/4`

`int_3^6 f(x) dx = 162`

**Hence, evaluating the definite integral, under the given conditions, yields **`int_3^6 f(x) dx = 162.`