IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

nik_d_1

_{[PIE/2 ]}

^{[ 3PIE/2]} [2SINX] dx [,] gint function...

nik_d_1

REPLY FRnDS...

madhuri_goteti

pie/2 to pie the value of greates interger function of 2 sinx is zero so intergal value under this region is zero .

now pie to 7pie/6 the value is -1 so intergal one . so that value is -pie/6.

now 7pie/6 to 3pie/2 the value os step function is -2 so intergal of -2 so value is -2pie/3 so total value is -5pie/6

if iam wrong plz correct me.

nigitha_17

is the ans -pi

Oppurtunity1

integral of sin is - cos.isnt it?
= 2 [ - COS 3PIE/2 - ( - COS PIE / 2 ) ]

= 2 ( 0 )

=0

A SMALL REQUEST ..PLS REPLY WID THE CORRECT ANSWER OR JUST NOTIFY ME IF THIS IS WRONG.

aysh

Hi!!!
seems easy...
Here it goes>>>

The given integral needs to be divided into separate limits.
Now,
since there is a factor of 2 within the braces, we need to divide it into limits such that within a pair of limits the function takes values between two integers.

Clearly,

_[/2]^[3/2][2sinx]

=_[/2]^[5/6][2sinx]dx + _[5/6]^[][2sinx]dx + _[]^[7/6][2sinx]dx

+ _[7/6]^[3/2][2sinx]dx

=_[/2]^[5/6]1dx + _[5/6]^[]0dx + _[]^[7/6](-1)dx + _[7/6]^[3/2] (-2)dx

= 5

/6 -

/2 - 7

/6 +

- 3

+ 7

/3

= -/2

PLEAZZZZZZ RATE MA EFFORT PEOPLE....

nik_d_1

the ans is -pie...
but aysh i got -pie/2 s well... d same way u did.. so m quite sure of dis -pie /2 thing...
hey nigitha hw did u get -pie

nigitha_17

from pi/2 to 5pi/6 2sinx ranges from 2 to 1

so[2sinx] is 1 ([1....]=1)

from 5pi/6 to pi 2sinx ranges from 1 to 0

so [2sinx] is 0

from pi to 7pi/6 2sinx ranges from 0 to -1

so [ 2sinx] is -1

from 7pi/6 to 3pi/2 2sinx ranges from -1 to -2

so [2sinx] is -2

_{[ pi/2]}^{[ 5pi/6]}1 +_{[pi ]}^{[7pi/6 ]}-1 +_{[ 7pi/6]}^{[ 3pi/2]}-2

=-pi/2

xteremly sory yes there was some mis in cal now the ans is -pi/2

nik_d_1

thnx... frnds.. for clearin it up 4 me.... i too got -pie/2.. b ws astonished 2 see d ans... thnx....

ank_einstein

i=_{[pie/2 ]}

^{[5pie/6 ]} 1.dx +_{[ 5pie/6]}

^{[ pie]} 0.dx +_{[ pie]}

^{[ 7pie/6]} -1.dx +_{[7pie/6 ]}

^{[ 3pie/2]} (-2).dx

=

{5/6-1/2 + 0 + (-1).{7/6-1} + (-2).{3/2-7/6} }

=

{-1/2}

rate me plzzzzzzzz if i m correct ...................

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