# Given that f''(x) = \cos{x}, f'(\frac{\pi}{2}) = 12 and f(\frac{\pi}{2}) = 8 find f'(x) and f(x)

## Question:

Given that {eq}\displaystyle f''(x) = \cos{x} {/eq}, {eq}\displaystyle \ \ \ f'(\frac{\pi}{2}) = 12 {/eq} and {eq}\displaystyle \ \ \ f(\frac{\pi}{2}) = 8 {/eq} find {eq}\displaystyle f'(x) {/eq} and {eq}\displaystyle f(x) {/eq}

## Integration:

• We deal with the concepts of differential equations and integration. Let F(x) be the anti-derivative of a function a function f(x) .Now, indefinite integral is given as {eq}\displaystyle \int f(x)dx = F(x) + c {/eq}, where c is the constant of integration.

Essential formulae:

{eq}\displaystyle *\int \cos ax = \frac{\sin ax}{a} + c \\ *\int \sin ax = -\frac{\cos ax}{a} + c {/eq}

Given:

{eq}\displaystyle f''(x) = \cos{x} \\ f'(\frac{\pi}{2}) = 12 \\ f(\frac{\pi}{2}) = 8 {/eq}

On integration,

{eq}\displaystyle \begin{align} \int f''(x) dx &= \int \cos x dx \\ f'(x) &= \sin x + c &&\left[ \int \cos ax = \frac{\sin ax}{a} + c \right] \\ \text{But}, f'(\frac{\pi}{2}) = 12 : \\ f'(\frac{\pi}{2}) &= \sin \frac{\pi}{2} + c \\ 12 &= 1 + c \\ c &= 11 \\ \end{align} {/eq}

Now,

{eq}\displaystyle f'(x) = \sin x + 11 {/eq}

On integration,

{eq}\displaystyle \begin{align} \int f'(x) dx &=\int \sin x dx +\int 11 dx \\ f(x) &= -\cos x + 11x + c &&\left[ \int \sin ax = -\frac{\cos ax}{a} + c \right] \\ \text{But}, f(\frac{\pi}{2}) = 8 : \\ f(\frac{\pi}{2}) &= -\cos \frac{\pi}{2} + \frac{11\pi}{2} + c \\ 8 &= -0+\frac{11\pi}{2} + c \\ c &=8 - \frac{11\pi}{2} \\ \end{align} {/eq}

Now,

{eq}\displaystyle f(x) = -\cos x + 11x + 8 - \frac{11\pi}{2} {/eq}

Therefore,

{eq}\displaystyle \boxed{\mathbf{ f'(x) = \sin x + 11 \\ f(x) = -\cos x + 11x + 8 - \frac{11\pi}{2} }} {/eq}