IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Definite integrals.. solve to get rates..

ananth_patri

Find the value of

_-1

¹ { x² - x +1}dx....plz solve it fast to get rates...

ashish_banga

IS IT 8/3

ananth_patri

no ,,, anyne else pplz solve..

akhil_o

1/6-^{[ ]}

5 /2?

ananth_patri

sorry .....try again...

akhil_o

let f(x)=x²-x+1
to find integral of g(x)={f(x)}
g(x)=f(x)-[f(x)]

so to find integral fo [f(x)]

at -1 f(x)=3
f(x) goes from 3 to 2 from -1 to a
a is root of negative root of equation
x²-x+1=2, which cgives a=(1-

5)/2

in (-1,a), [f(x)]=2
from (a,0) [f(x)]=1
from (0,1) the function is below 1 and above 3/4 except at 0 and 1
so [f(x)]=0

so _{[ -1]}

^{[1 ]} [f(x)] dx=(a-(-1))2 + (0-a)1
=3-

5+(

5-1)/2
=5/2+

5/2

also _{[ -1]}

^{[ 1]} f(x) dx= 8/3

so required integral
=8/3-5/2-

5/2
=1/6-

5/2

ananth_patri

well u got it....answer is rite....sorry dfor the last post,,,

varshavallig

ans is 2/3 which is the fractional part of the integral.
@ ashish_banga i think he is asking for the fractional part.,otherwise ur ans is right.
!!!!!!!!!!!!!!

ananth_patri

@ akhil_o got it rite...

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