# Math Help - Intergration with 6^x

1. ## Intergration with 6^x

Its a definate intergral upper limit is e lower is 1 (8^x-4^x)dx

2. $\int^{e}_{1} 8^x - 4^x dx$

Do you know that $\int a^x dx = \frac{a^x}{\ln a}$ ?
We can get that rule by seeing that $\frac{d}{dx}a^x = a^x \ln a$ and its inverse is $\frac{a^x}{\ln a}$.

So,
$\int^{e}_{1} 8^x - 4^x dx$

$\int^{e}_{1} 8^x dx - \int^{e}_{1} 4^x dx$

$\frac{8^x}{\ln 8} - \frac{4^x}{\ln 4} ~|^{e}_{1}$

$(\frac{8^e}{\ln 8}-\frac{8}{\ln 8}) - (\frac{4^e}{\ln 4} - \frac{4}{\ln 4})$

$\boxed{~\frac{8^e-8}{\ln 8}-\frac{4^e-4}{\ln 4}~}$