3q1)\int_{e^{-1}}^{e}{-\frac{lnx}{x}}dx+\int_{e}^{e^{2}}{\frac{lnx}{x}}dx
and then take t=lnx.
1. \int_{e^-^1}^{e^2}{|lnx / x | } dx = ?
2.\int_{sinx}^{1}{t^2 f(t)dt }= ( 1 - sinx ) , then find f (1 / √3) = ?
3.\lim_{x \rightarrow infinity}1/\pi \sum_{r=1}^{n}{tan^-^1(1/2r ^2)} = ?
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1MSP, A BIT CORRECTION in the limits.
\int_{e^-^1}^{e^2}{|lnx/x| }dx = -\int_{e^-^1}^{1}{lnx/x }dx + \int_{1}^{e^2}{lnx / x}
now, substituting lnx in each , we put t= lnx i.e. dt = dx / x.
= [- 1/2 (lnx)2 ](limits e-1 to 1) + [1/2 (lnx)2 ] (limits - 1 to e2]
= 5/2
12) . using Newton-Lebnitz formula , differentiate both sides.
-sin^2x. f(sinx).cosx = -cosx
so,f(sinx) = cosx / (sin^2x.cosx)
now,if, sinx = 1/\sqrt{3},
then f( 1/√3 ) = 3
1tan-1 ( 1/2r2 ) = tan -1 ( 2/4r2)
= tan-1[ (2r+1)-(2r-1)] / [ 1+ (2r+1)(2r-1)] = tan-1 ( 2r+1) - tan-1(2r-1)
So, \lim_{n\rightarrow infinity} \sum_{r=1}^{n}{tan^-^1} = pi/2 - pi/4 = pi/4
\lim_{n\rightarrow infinity}1/\pi \sum_{r=1}^{n}{tan^-^1} = 1/4
