Answer
$$\frac{16}{5 \ln 5}-\frac{1}{2 \ln 2}$$
Work Step by Step
Given
$$ y=2^x,\ \ y=5^x ,\ \ \ x=-1,\ \ x=1 $$
first, we find the intersection points
\begin{aligned}
2^x&= 5^x\\
x\ln(2)&= x\ln (5)\\
x(\ln (5)-\ln (2))&=0\\
x&=0
\end{aligned}
Since
\begin{aligned}
2^x&\geq 5^x,\ \ \ \ \ -1\leq x\leq 0\\
2^x&\leq 5^x,\ \ \ \ \ 0\leq x\leq 1
\end{aligned}
Then area given
\begin{aligned}
A &=\int_{-1}^0\left(2^x-5^x\right) d x+\int_0^1\left(5^x-2^x\right) d x\\
&=\left[\frac{2^x}{\ln 2}-\frac{5^x}{\ln 5}\right]_{-1}^0+\left[\frac{5^x}{\ln 5}-\frac{2^x}{\ln 2}\right]_0^1 \\
&=\left(\frac{1}{\ln 2}-\frac{1}{\ln 5}\right)-\left(\frac{1 / 2}{\ln 2}-\frac{1 / 5}{\ln 5}\right)+\left(\frac{5}{\ln 5}-\frac{2}{\ln 2}\right)-\left(\frac{1}{\ln 5}-\frac{1}{\ln 2}\right) \\
&=\frac{16}{5 \ln 5}-\frac{1}{2 \ln 2}
\end{aligned}