Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 84

If $f''(x) = 3e^x + 5 \sin x, f(0) = 1$ and $f'(0) = 2$, find $f(x)$.


$
\begin{equation}
\begin{aligned}

\text{if } f''(x) =& 3e^x + 5 \sin x, \text{ then}
\\
\\
f'(x) =& \int (3e^x + 5 \sin x) dx
\\
\\
f'(x) =& \int 3e^x dx + \int 5 \sin x dx
\\
\\
f'(x) =& 3e^x - 5 \cos x + C_1

\end{aligned}
\end{equation}
$


when $f'(0) = 2$


$
\begin{equation}
\begin{aligned}

2 =& 3e^0 - 5 \cos (0) + C_1
\\
\\
2 =& 3 - 5(1) + C_1
\\
\\
C_1 =& 4

\end{aligned}
\end{equation}
$


Thus,

$f'(x) = 3e^x - 5 \cos x + 4$

Again, by applying integration,


$
\begin{equation}
\begin{aligned}

f(x) =& \int (3e^x - 5 \cos x + 4) dx
\\
\\
f(x) =& \int 3e^x dx - \int 5 \cos x dx + \int 4 dx
\\
\\
f(x) =& 3e^x - 5 \sin x + 4x + C_2

\end{aligned}
\end{equation}
$


when $f(0) = 1$,


$
\begin{equation}
\begin{aligned}

1 =& 3 e^0 - 5 \sin (0) + 4(0) + C_2
\\
\\
1 =& 3 + C_2
\\
\\
C_2 =& -2

\end{aligned}
\end{equation}
$


Thus,

$f(x) = 3 e^x - 5 \sin x + 4 x - 2$