Indefinite Integration 2 Question 1

1. The value of π/2π/2dx[x]+[sinx]+4, where [t] denotes the greatest integer less than or equal to t, is

(a) 112(7π5)

(b) 112(7π+5)

(c) 310(4π3)

(d) 320(4π3)

(2019 Main, 10 Jan II)

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Answer:

Correct Answer: 1. (d)

Solution:

  1. Let I=π2π2dx[x]+[sinx]+4

=π21dx[x]+[sinx]+4+10dx[x]+[sinx]+4

+01dx[x]+[sinx]+4+1π2dx[x]+[sinx]+4

[x]=2,π/2<x<11,1x<00,0x<11,1x<π/2

and [sinx]=1,π/2<x<1 1,1<x<0 0,0<x<1 0,1<x<π/2

[ For x<0,1sinx<0 and for x>0,0<sinx1]

So, I=π21dx21+4+10dx11+4+01dx0+0+4

=π21dx1+10dx2+01dx4+1π2dx5=1+π2+12(0+1)+14(10)+15π21=1+12+1415+π2+π10=20+10+5420+5π+π10=920+3π5=320(4π3)



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