# Find {eq}f(x) {/eq} given {eq}\displaystyle f''(x) = 2 + \cos (x), f(9) = -1, f(\frac {\pi}{2}) = 0 {/eq}.

## Question:

Find {eq}f(x) {/eq} given {eq}\displaystyle f''(x) = 2 + \cos (x), f(9) = -1, f(\frac {\pi}{2}) = 0 {/eq}.

## Indefinite Integral of a Real-Valued Function:

Let {eq}\displaystyle f {/eq} be a real-valued function of {eq}\displaystyle x {/eq}. If there exists a function {eq}\displaystyle F {/eq} such that {eq}\displaystyle f\left( x \right) = F'\left( x \right),\,\,\,\forall \,\,\,x {/eq} then {eq}\displaystyle F {/eq} is said to be an antiderivative of {eq}\displaystyle f {/eq}. The general form, i.e. {eq}\displaystyle F(x)+C {/eq}, where {eq}\displaystyle C {/eq} is any integration constant (independent of {eq}\displaystyle x {/eq}), is called an indefinite integral of {eq}\displaystyle f {/eq} and it is usually denoted by {eq}\displaystyle \int_{}^{} {} f\left( x \right)dx = F\left( x \right) + C {/eq}.

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Given that {eq}\displaystyle f''(x) = 2 + \cos (x) {/eq}.

Integrating, we get

{eq}\displaystyle \begin{align} & \int_{}^{} {f''\left( x...