# find the general form of f(x) if f''(x)=4sin+5x. The general form of (f) has two arbitrary constants.

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Given `f''(x)=4sinx+5x` :

`f'(x)=int(4sinx+5x)dx`

`=4intsinxdx+5intxdx`

`=-4cosx+5/2x^2+C_1`

Then `f(x)=int(-4cosx+5/2x^2+C_1)dx`

`=-4intcosx+5/2intx^2dx+intC_1dx`

`=-4sinx+(5/2)(1/3)x^3+C_1x+C_2`

`=-4sinx+5/6x^3+C_1x+C_2`

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**Thus if `f''(x)=4sinx+5x` then `f(x)=-4sinx+5/6x^3+C_1x+C_2` for some arbitrary real constants `C_1,C_2` **

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Check: `f'(x)=d/(dx)[-4sinx+5/6x^3+C_1x+C_2]`

`=-4cosx+5/2x^2+C_1`

while `f''(x)=d/(dx)[-4cosx+5/2x^2+C_1]`

`=4sinx+5x` as required

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**Sources:**

Sorry I wrote this in a weird way. The problem is to find the general form of f(x) if f''(x)=4sin+5x.

The general form of (f) has two arbitrary constants. Again, sorry, I'm still getting used to this webiste.