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

∫(2x−5x10x)dx\int \left( \frac{2^x-5^x}{10^x} \right) dx is equal to

Answer

2−xln⁡2−5−xln⁡5+C\frac{2^{-x}}{\ln 2} - \frac{5^{-x}}{\ln 5} + C

Explanation

Solution

The problem asks us to evaluate the indefinite integral ∫(2x−5x10x)dx\int \left( \frac{2^x-5^x}{10^x} \right) dx.

Step-by-step solution:

  1. Split the integrand: The fraction can be split into two separate terms: 2x−5x10x=2x10x−5x10x\frac{2^x-5^x}{10^x} = \frac{2^x}{10^x} - \frac{5^x}{10^x}

  2. Simplify each term using exponent rules: Recall that axbx=(ab)x\frac{a^x}{b^x} = \left(\frac{a}{b}\right)^x. 2x10x=(210)x=(15)x=5−x\frac{2^x}{10^x} = \left(\frac{2}{10}\right)^x = \left(\frac{1}{5}\right)^x = 5^{-x} 5x10x=(510)x=(12)x=2−x\frac{5^x}{10^x} = \left(\frac{5}{10}\right)^x = \left(\frac{1}{2}\right)^x = 2^{-x}

  3. Rewrite the integral: Substitute the simplified terms back into the integral: ∫(2x−5x10x)dx=∫(5−x−2−x)dx\int \left( \frac{2^x-5^x}{10^x} \right) dx = \int \left( 5^{-x} - 2^{-x} \right) dx

  4. Integrate each term: We use the standard integration formula for exponential functions: ∫akxdx=akxkln⁡a+C\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + C.

    • For the first term, ∫5−xdx\int 5^{-x} dx: Here a=5a=5 and k=−1k=-1. ∫5−xdx=5−x(−1)ln⁡5=−5−xln⁡5\int 5^{-x} dx = \frac{5^{-x}}{(-1) \ln 5} = - \frac{5^{-x}}{\ln 5}

    • For the second term, ∫2−xdx\int 2^{-x} dx: Here a=2a=2 and k=−1k=-1. ∫2−xdx=2−x(−1)ln⁡2=−2−xln⁡2\int 2^{-x} dx = \frac{2^{-x}}{(-1) \ln 2} = - \frac{2^{-x}}{\ln 2}

  5. Combine the results: ∫(5−x−2−x)dx=(−5−xln⁡5)−(−2−xln⁡2)+C\int \left( 5^{-x} - 2^{-x} \right) dx = \left( - \frac{5^{-x}}{\ln 5} \right) - \left( - \frac{2^{-x}}{\ln 2} \right) + C =−5−xln⁡5+2−xln⁡2+C= - \frac{5^{-x}}{\ln 5} + \frac{2^{-x}}{\ln 2} + C Rearranging the terms to have the positive term first: =2−xln⁡2−5−xln⁡5+C= \frac{2^{-x}}{\ln 2} - \frac{5^{-x}}{\ln 5} + C This can also be written in terms of positive exponents: =12xln⁡2−15xln⁡5+C= \frac{1}{2^x \ln 2} - \frac{1}{5^x \ln 5} + C

The final answer is 2−xln⁡2−5−xln⁡5+C\boxed{\frac{2^{-x}}{\ln 2} - \frac{5^{-x}}{\ln 5} + C}.

Explanation: The integral is simplified by splitting the fraction and using exponent rules. Each resulting term is of the form akxa^{kx}, which is integrated using the standard formula ∫akxdx=akxkln⁡a\int a^{kx} dx = \frac{a^{kx}}{k \ln a}.

Answer: The integral is equal to 2−xln⁡2−5−xln⁡5+C\frac{2^{-x}}{\ln 2} - \frac{5^{-x}}{\ln 5} + C.