For any real number `x ,l e t[x]` denote the largest integer less than or equal to `x ,L e tf` be a real-valued function defined on the interval `[-10 , 10]` be `f(x)={x-[x],if[x]i sod d1+[x]-x ,if[x]i se v e n` Then the value of `(pi^2)/(10)int_(-1)^(10)f(x)cospixdxi s____`

Answer

4

Explanation

`f(x)={(x-1,1le xlt2),(1-x,0lexlt1):}`

`f(x)` is periodic with period 2 and even.
`:. I=int_(-10)^(10)f(x)cos pix dx`
`=2int_(0)^(10)f(x)cospix dx`
`=2xx5int_(0)^(2)f(x)cospixdx`
`=10[int_(0)^(1)(1-x)cos pix dx+int_(1)^(2)(x-1)cospi xdx]=10(I_(1)+I_(2))`
`I_(2)=int_(1)^(2)(x_1)cospixdx ( "put"x-1=t)`
`I_(2)=-int_(0)^(1)t cos pit dt`
`:. I=10[-2int_(0)^(1)xcospixdx]=-20[x(sinpix)/(pi)+(cosxpi)/(pi^(2))]_(0)^(1)`
`=-20[-1/(pi^(2))-1/(pi^(2))]=40/(pi^(2))`
`:.(pi^(2))/10I=4`