JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(\int\limits_1^e {\left\{ {\left. {{{\left( {\frac{x}{e}} \right)}^{2x}} - {{\left( {\frac{e}{x}} \right)}^x}} \right\}{{\log }_e}\,x\,dx} \right.} \) is equal to
- A \(\frac{1}{2} - e - \frac{1}{{{e^2}}}\)
- B \( - \frac{1}{2} + \frac{1}{e} - \frac{1}{{2{e^2}}}\)
- C \(\frac{3}{2} - \frac{1}{e} - \frac{1}{{2{e^2}}}\)
- D \(\frac{3}{2} - e - \frac{1}{{2{e^2}}}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{2} - e - \frac{1}{{2{e^2}}}\)
Step-by-step Solution
Detailed explanation
\(\int_{1}^{e}\left(\frac{x}{e}\right)^{2 x} \log _{e} x \cdot d x-\int_{1}^{e}\left(\frac{e}{x}\right) \log _{e} x \cdot d x\) \(\operatorname{let}\left(\frac{x}{e}\right)^{2 x}=t,\left(\frac{e}{x}\right)^{x}=v\)…
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