Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
$I_1 = \int 2^x dx = p(x) + c_1$ and $I_2 = \int (\frac{1}{2})^x dx = m(x) + c_1$ then $p(x) - m(x)$ is equal to
$\{\log_e(2)\}(2^x - 2^{-x})$
$\{\log_e 2\}(2^x + 2^{-x})$
$\frac{1}{\log_e 2}(2^x + 2^{-x})$
$\log_2 e(2^x + 2^{-x})$
Step-by-Step Solution
Key Concept: The integral of $a^x$ is $\frac{a^x}{\ln a}$, and recognizing that $\frac{1}{\ln 2} = \log_2 e$ makes both expressions equivalent.
For $I_1 = \int 2^x dx$, using the formula $\int a^x dx = \frac{a^x}{\ln a}$, we get $p(x) = \frac{2^x}{\ln 2}$. For $I_2 = \int (\frac{1}{2})^x dx = \int 2^{-x} dx$, we get $m(x) = \frac{2^{-x}}{\ln(1/2)} = \frac{2^{-x}}{-\ln 2} = -\frac{2^{-x}}{\ln 2}$. Therefore, $p(x) - m(x) = \frac{2^x}{\ln 2} - (-\frac{2^{-x}}{\ln 2}) = \frac{1}{\ln 2}(2^x + 2^{-x})$. Since $\frac{1}{\ln 2} = \log_2 e$, both options 3 and 4 are equivalent and correct.
Correct Answer: 3,4