Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 78

If $f''(x) = x^{-2}$, $x > 0$, $f(1) = 0$ and $f(2) = 0$, find $f$

If $f''(x) = x^{-2}$, then by applying integration...


$
\begin{equation}
\begin{aligned}

f'(x) &= \int x^{-2} dx
\\
\\
&= \frac{x^{-1}}{-1} + c_1
\\
\\
&= -\frac{1}{x} + c_1


\end{aligned}
\end{equation}
$



Again, by applying integration...


$
\begin{equation}
\begin{aligned}


f(x) &= \int \left( -\frac{1}{x} + c_1 \right) dx
\\
\\
f(x) &= - \ln x + c_1 x + c_2

\end{aligned}
\end{equation}
$


If $f(1) = 0$, then


$
\begin{equation}
\begin{aligned}


0 &= - \ln (1) + c_1(1)+c_2
\\
\\
0 &= c_1 + c_2
\\
\\
c_1 &= c_2 \qquad \to \text{(Equation 1)}

\end{aligned}
\end{equation}
$



Also, if $f(2) = 0$, then



$
\begin{equation}
\begin{aligned}

0 &= -\ln(2) + c_1(2)+c_2
\\
\\
\ln(2) &= 2c_1 + c_2 \qquad \to \text{(Equation 2)}

\end{aligned}
\end{equation}
$



By using Equations 1 and 2 simultaneously...



$
\begin{equation}
\begin{aligned}

\ln (2) &= 2c_1 - c_1
\\
\\
c_1 &= \ln 2

\end{aligned}
\end{equation}
$



Thus, $c_2 = - \ln 2$

Therefore,

$f(x) = -\ln x + x \ln (2) - \ln (2)$