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please solve the following question of areab under curves:

area b/w the curves y=ex log x and y=(log x)/e^x.

pankaj negi , 14 Years ago
Grade 12
anser 1 Answers
Jitender Singh
Ans:
Area ‘A’ b/w the curves:
Points of intersections:
ex lnx = \frac{lnx}{ex}
lnx(ex - \frac{1}{ex}) = 0
x = \frac{1}{e}, 1
A = \int_{1/e}^{1}(-exlnx + \frac{lnx}{ex})dx
A = -e(\frac{lnx.x^{2}}{2}- \frac{x^{2}}{4})_{1/e}^{1} + \frac{1}{e}(\frac{ln^{2}x}{2})_{1/e}^{1}
A = \frac{e}{4} - \frac{5}{4e} = \frac{e^{2}-5}{4e}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
Last Activity: 11 Years ago
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