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Question

Evaluate : π0x tanxsec x+tan xdx. OR Evaluate : 41 {|x-1|+|x-2|+|x-4|}dx.

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Solution

Let I= π0x tanxsec x+tan xdx....(i) π0(πx)tan(πx)sec(πx)tan(πx)dx=π0(πx)tanxsecx+tanxdx...(ii)

Adding (i) and (ii),2I=ππ0tan xsecx+tanxdx 2I=ππ0tanxsec x+tan x×secxtanxsecxtan xdx

2I=ππ0(secx tanxtan2x)dx 2I=ππ0(secx tanxsec2x+1)dx

2I=π[secxtanx+x]π0 2I=π{[secπtanπ+π][sec0tan0+0]}

2I=π{[10+π][10+0]} I=π(π2)2.

OR Let I=41{|x1|+|x2|+|x4|dx}

I=41|x1|dx+=41|x2|dx+41|x4|dx

=41(x1)dx+21|x2|dx+42|x2|dx41(x4)dx

I=12[(x1)2]4121(x2)dx+42(x2)dx12[(x4)2]41

I=12[90]12[(x2)2]21+12[(x2)2]4212[09]

I=9212[01]+12[40]+92=11+12=232


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