∫ _0 ^1 ln(x) dx


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Question Number 93248 by i jagooll last updated on 12/May/20

$\int \underset{0}{\overset{1}{}} \ln (x) dx$
Commented by abdomathmax last updated on 12/May/20

$\int_{0}^{1} l n (x) d x = {l i m}_{a \to 0^{+}} \int_{a}^{1} l n (x) d x$ $= {l i m}_{a \to 0^{+}} {[x l n x - x]}_{a}^{1}$ $= {l i m}_{a \to 0^{+}} (- 1 - a l n a + a) = - 1$
	Answered by john santu last updated on 12/May/20
	$D.I method ∫_0 ^1 ln(x) dx = lim_(p→0^+ ) { ∫_p ^1 ln(x) dx } = lim_(p→0^+ ) { x ln(x)−x }]_p ^1 = lim_(p→0^+ ) {−1+p−pln(p)} = −1+ lim_(p→0^+ ) p(1−ln(p)) = −1+ lim_(p→0^+ ) ((1−ln(p))/(((1/p)))) =−1+ lim^(LH) _(p→0^+ ) (((−((1/p)))/(−((1/p^2 ))))) = −1 + lim_(p→0^+ ) (p) = −1+0 = −1$
	$D . I method$ $\underset{0}{\int^{1}} \ln (x) d x = \lim_{p \to 0^{+}} {\underset{p}{\int^{1}} \ln (x) d x}$ ${= \lim_{p \to 0^{+}} {x \ln (x) - x}]}_{p}^{1}$ $= \lim_{p \to 0^{+}} {- 1 + p - p \ln (p)}$ $= - 1 + \lim_{p \to 0^{+}} p (1 - \ln (p))$ $= - 1 + \lim_{p \to 0^{+}} \frac{1 - \ln (p)}{(\frac{1}{p})}$ $= - 1 + \underset{p \to 0^{+}}{li \overset{L H}{m}} (\frac{- (\frac{1}{p})}{- (\frac{1}{p^{2}})})$ $= - 1 + \lim_{p \to 0^{+}} (p) = - 1 + 0 = - 1$
	Commented by i jagooll last updated on 12/May/20

	$thank you . . .$
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