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Integral Calculus

Blazing goIITian

Joined:	19 Mar 2007
Post:	585

26 Oct 2007 22:27:37 IST

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333

area : aakhri alvida

None

the area bound by curvs

y=ln x

y=ln mod.x

y=mod.in x

y=mod.ln.mod x

is :::::::::::::::::::

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Comments (5)

Priyesh

Blazing goIITian

Joined: 18 Feb 2007

Posts: 1042

26 Oct 2007 22:51:59 IST

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draw the graphs of the four functions

see fig. below (click to enlarge)

now the area needed(shaded in red) is 4_{[ 0]}

^{[ 1]} -lnx dx = [-4 (xlnx - x)]₀¹ = 4

SPARTIAN

Blazing goIITian

Joined: 19 Mar 2007

Posts: 585

26 Oct 2007 22:56:02 IST

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don't u think the graph should be open at y axis ??????

SPARTIAN

Blazing goIITian

Joined: 19 Mar 2007

Posts: 585

26 Oct 2007 23:27:18 IST

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no ans ??????????

Gaurav |spideyunlimited| Ragtah

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Joined: 16 Dec 2006

Posts: 3373

27 Oct 2007 01:56:25 IST

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yes it will extend to + infinity and - infinity
the area will be
= 4 * integral [ln x ] 0 to 1
= 4 * (xlnx + x) 0 to 1
= 4 * (1 - 0)
= 4

Priyesh

Blazing goIITian

Joined: 18 Feb 2007

Posts: 1042

27 Oct 2007 13:48:02 IST

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hi shubham sorry for replying late

see the graph does not touch y - axis agreed but the area enclosed (in approximation is the one which i have shaded) also when we integrate from x = 0 to 1 we take the limiting value of xlnx at x=0 (which is 0)since lnx is not defined at x=0

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