1.39. \( \int 5^{x-2} d x \)

To solve the integral \( \int 5^{x-2} dx \), we can start by simplifying the expression. We know that \( 5^{x-2} = \frac{5^x}{5^2} = \frac{1}{25} \cdot 5^x \). Thus, we can rewrite the integral as: \[ \int 5^{x-2} dx = \frac{1}{25} \int 5^x dx \] Now, the integral of \( 5^x \) with respect to \( x \) is \( \frac{5^x}{\ln(5)} + C \). So, we get: \[ \int 5^x dx = \frac{5^x}{\ln(5)} + C \] Substituting this back, we find: \[ \int 5^{x-2} dx = \frac{1}{25} \left( \frac{5^x}{\ln(5)} + C \right) = \frac{5^x}{25 \ln(5)} + C \] Thus, the final answer is: \[ \int 5^{x-2} dx = \frac{5^x}{25 \ln(5)} + C \] Now you've not only tackled an integral but also explored the beauty of exponential functions and natural logs! Did you know that integral calculus was developed further by mathematicians like Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century? Their collaborative discoveries helped lay the foundation for modern calculus, connecting the dots between areas under curves and instantaneous rates of change, much like what you're doing right now! And if you're keen to delve deeper into calculus, consider reading "Calculus Made Easy" by Silvanus P. Thompson. It presents principles in a simplified manner with witty anecdotes, making it an enjoyable companion for anyone looking to demystify calculus. Happy integrating!