Question

$\int \left( \frac{2^x - 5^x}{10^x} \right) dx$ is equal to

Answer 1

$\frac{1}{2^x \ln 2} - \frac{1}{5^x \ln 5} + C$

Answer 2

The problem asks us to evaluate the integral $\int \left( \frac{2^x - 5^x}{10^x} \right) dx$.

First, we simplify the integrand: $\frac{2^x - 5^x}{10^x} = \frac{2^x}{10^x} - \frac{5^x}{10^x}$ Using the property $\frac{a^x}{b^x} = \left(\frac{a}{b}\right)^x$: $= \left(\frac{2}{10}\right)^x - \left(\frac{5}{10}\right)^x$ $= \left(\frac{1}{5}\right)^x - \left(\frac{1}{2}\right)^x$ This can also be written using negative exponents: $= 5^{-x} - 2^{-x}$ Now, we need to integrate this expression: $\int (5^{-x} - 2^{-x}) dx = \int 5^{-x} dx - \int 2^{-x} dx$ We use the standard integration formula for $a^{kx}$: $\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C$ For the first term, $\int 5^{-x} dx$: Here, $a=5$ and $k=-1$. $\int 5^{-x} dx = \frac{5^{-x}}{(-1) \ln 5} + C_1 = -\frac{5^{-x}}{\ln 5} + C_1$ For the second term, $\int 2^{-x} dx$: Here, $a=2$ and $k=-1$. $\int 2^{-x} dx = \frac{2^{-x}}{(-1) \ln 2} + C_2 = -\frac{2^{-x}}{\ln 2} + C_2$ Combining these results: $\int \left( \frac{2^x - 5^x}{10^x} \right) dx = -\frac{5^{-x}}{\ln 5} - \left(-\frac{2^{-x}}{\ln 2}\right) + C$ $= -\frac{5^{-x}}{\ln 5} + \frac{2^{-x}}{\ln 2} + C$ This can be rearranged for better readability: $= \frac{2^{-x}}{\ln 2} - \frac{5^{-x}}{\ln 5} + C$ Or, expressing $a^{-x}$ as $\frac{1}{a^x}$: $= \frac{1}{2^x \ln 2} - \frac{1}{5^x \ln 5} + C$

Explanation of the solution:

1. Simplify the integrand: Rewrite $\frac{2^x - 5^x}{10^x}$ as $\left(\frac{2}{10}\right)^x - \left(\frac{5}{10}\right)^x$, which simplifies to $\left(\frac{1}{5}\right)^x - \left(\frac{1}{2}\right)^x$ or $5^{-x} - 2^{-x}$.
2. Integrate term by term: Apply the integral formula $\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C$ to each term.
3. For $\int 5^{-x} dx$, $a=5, k=-1$, yielding $-\frac{5^{-x}}{\ln 5}$.
4. For $\int 2^{-x} dx$, $a=2, k=-1$, yielding $-\frac{2^{-x}}{\ln 2}$.
5. Combine the results: Subtract the second integral from the first, remembering the minus sign from the original expression. This gives $\frac{2^{-x}}{\ln 2} - \frac{5^{-x}}{\ln 5} + C$.
6. Rewrite using positive exponents: $\frac{1}{2^x \ln 2} - \frac{1}{5^x \ln 5} + C$.