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Question Number 137226 by SLVR last updated on 31/Mar/21

	Answered by MJS_new last updated on 31/Mar/21

	$f (x) = 2 \ln x$ $x^{3} - 6 x^{2} + 11 x - 6 = (x - 1) (x - 2) (x - 3)$ $the curves intersect at x = 1$ $\underset{0}{\int^{1}} 2 \ln x d x = {[2 x \ln x - 2 x]}_{0}^{1} = - 2$ $\underset{0}{\int^{1}} (x^{3} - 6 x^{2} + 11 x - 6) d x = {[\frac{x^{4}}{4} - 2 x^{3} + \frac{11 x^{2}}{2} - 6 x]}_{0}^{1} = - \frac{9}{4}$ $area is - 2 + \frac{9}{4} = \frac{1}{4}$
	Commented by SLVR last updated on 31/Mar/21

	$g r e a r t s i r . . t h a n k s$
	Commented by mr W last updated on 31/Mar/21

	$t h e a r e a b o u n d e d b y y = f (x) a n d$ $y = x^{3} - 6 x^{2} + 11 x - 6 i s \frac{1}{4} . b u t t h e$ $q u e s t i o n a s k s t h e a r e a b o u n d e d b y$ $y = f (x) a n d y =∣ x^{3} - 6 x^{2} + 11 x - 6 ∣,$ $w h i c h i s t h e n 2 + \frac{9}{4} = \frac{17}{4} .$ $s o i t h i n k t h e q u e s t i o n i s w r o n g .$
	Commented by MJS_new last updated on 31/Mar/21

	$you are right$
	Commented by Dwaipayan Shikari last updated on 31/Mar/21

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