let A be non empty closed_interval Subset of REAL; :: thesis: for n being Element of NAT st A = [.((2 * n) * PI),(((2 * n) + 1) * PI).] holds
integral ((- sin),A) = - 2
let n be Element of NAT ; :: thesis: ( A = [.((2 * n) * PI),(((2 * n) + 1) * PI).] implies integral ((- sin),A) = - 2 )
assume A = [.((2 * n) * PI),(((2 * n) + 1) * PI).] ; :: thesis: integral ((- sin),A) = - 2
then ( upper_bound A = ((2 * n) + 1) * PI & lower_bound A = (2 * n) * PI ) by Th37;
then integral ((- sin),A) = (cos (0 + (((2 * n) + 1) * PI))) - (cos (0 + ((2 * n) * PI))) by Th46
.= (- (cos 0)) - (cos (0 + ((2 * n) * PI))) by Th4
.= (- (cos 0)) - (cos 0) by Th3
.= (- (cos (0 + (2 * PI)))) - (cos 0) by SIN_COS:79
.= (- 1) - 1 by SIN_COS:77, SIN_COS:79 ;
hence integral ((- sin),A) = - 2 ; :: thesis: verum