Question
Download Solution PDF\(\int_{\rm{0}}^{\rm{1}} {{\rm{log}}\left( {\frac{{\rm{1}}}{{\rm{x}}}{\rm{ - 1}}} \right)} \)dx बराबर है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
दिया गया समाकलन है \(\int_{\rm{0}}^{\rm{1}} {{\rm{log}}\left( {\frac{{\rm{1}}}{{\rm{x}}}{\rm{ - 1}}} \right)} dx\)
संकल्पना:
निश्चित समाकलन का गुण:- \(\int_0^a f(x)dx = \int_0^a f(a - x)dx\)
प्रयुक्त सूत्र:
log A + log B = log (A×B)
गणना:
माना I = \(\int_{\rm{0}}^{\rm{1}} {{\rm{log}}\left( {\frac{{\rm{1}}}{{\rm{x}}}{\rm{ - 1}}} \right)} dx\)
⇒ I = \(\int_0^1 log\left(\frac{1 - x}{x} \right)dx\) ----- समीकरण (1)
⇒ I = \(\int_0^1 log \left( \frac{1 - (1 - x)}{1 - x}\right) dx\)
⇒ I = \(\int_0^1 log \left( \frac{x}{1 - x} \right) dx\) ----- समीकरण (2)
समीकरण (1) और (2) को जोड़ने पर, हम प्राप्त करते हैं
⇒ 2I = \(\int_0^1 log\left(\frac{1 - x}{x} \right)dx + \int_0^1 log \left( \frac{x}{1 - x} \right) dx\)
⇒ 2I = \(\int_0^1 log \left( \frac{1 -x}{x} . \frac{x}{1 -x} \right) dx \)
⇒ 2I = \(\int_ 0^1 log1dx\)
⇒ 2I = 0
⇒ I = 0
∴ \(\int_{\rm{0}}^{\rm{1}} {{\rm{log}}\left( {\frac{{\rm{1}}}{{\rm{x}}}{\rm{ - 1}}} \right)} dx\) 0 के बराबर है
Last updated on Jul 19, 2025
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