Question

Find \(f\)

\(f''\left( x \right) = 6x + \sin x\)

Short answer

Required function is \(f\left( x \right) = {x^3} - \sin x + Cx + D\).

Step-by-step solution

Find a general antiderivative of \(f''\)

Here we have,

\(f''\left( x \right) = 6x + \sin x\)

Now, the general antiderivative of \(f''\left( x \right) = 6x + \sin x\) is,

\(\begin{aligned}{c}f'\left( x \right) &= 6\frac{{{x^2}}}{2} - \cos x + C\\ &= 3{x^2} - \cos x + C\end{aligned}\)

Find a general antiderivative of \(f'\)

Now, we have, \(f'\left( x \right) = 3{x^2} - \cos x + C\)

Now, the general antiderivative of \(f'\left( x \right) = 3{x^2} - \cos x + C\) is,

\(\begin{aligned}{c}f\left( x \right) &= 3\frac{{{x^3}}}{3} - sinx + Cx + D\\ &= {x^3} - \sin x + Cx + D\end{aligned}\)

So, our required function is \(f\left( x \right) = {x^3} - \sin x + Cx + D\).