KCET · Maths · Indefinite Integration
\(\int e^{-x \log 2} \cdot 2^x\,dx = \)
- A \(\log x + C\)
- B \(x + C\)
- C \(\dfrac{1}{x} + C\)
- D \(\dfrac{x^2}{2} + C\)
Answer & Solution
Correct Answer
(B) \(x + C\)
Step-by-step Solution
Detailed explanation
The given integral is \(\int e^{-x \log 2} \cdot 2^x \, dx\)
Using the property \(e^{a \log b} = b^a\), we can write \(e^{-x \log 2} = e^{\log(2^{-x})} = 2^{-x}\)
Substituting this into the integral:
\(\int 2^{-x} \cdot 2^x \, dx\)
\(\int 2^{-x + x} \, dx\)
\(\int 2^0 \, dx\)
\(\int 1 \, dx\)
\(x + C\)
Answer: \(x + C\)
Using the property \(e^{a \log b} = b^a\), we can write \(e^{-x \log 2} = e^{\log(2^{-x})} = 2^{-x}\)
Substituting this into the integral:
\(\int 2^{-x} \cdot 2^x \, dx\)
\(\int 2^{-x + x} \, dx\)
\(\int 2^0 \, dx\)
\(\int 1 \, dx\)
\(x + C\)
Answer: \(x + C\)
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