_ -a ^ a x² ( e^ x³ -e^ -x³ e^ x³ +e^ -x³ ) d x=

Step-by-step solution

Let aligned f(x) &=x² [ e^ x³ -e^ -x³ e^ x³ +e^ -x³ ]=x² [ e^ x³ - 1 e^ x³ e^ x³ + 1 e^ x³ ]=x² [ (e^ x³ )²-1 (e^ x³ )²+1 ] f(-x) &=(-x)² [ e^ -x³ -e^ x³ e^ -x³ +e^ x³ ] aligned =x² [ 1 e^ x³ -e^ x³ 1 e^ x³ +e^ x³ ]=x² [ 1- (e^ x³ )² 1+ (e^ x³ )² ]=-x² [ (e^ x³ )²-1 1+ (e^ x³ ) 2 ]=-f(x) Thus f(-x)=-f(x) Given function is an odd function. I=0