Paragraph: Let f(t)= (t+ 1+t^2 ) and g(t)= (f(t)) . Question: What is _ - ^ g(t) d t equal to?

Step-by-step solution

let f(t)= (t+ 1+t^2 ) Put t=-t in f(t) aligned & f(-t)= ( 1+t^2 -t )= ( 1 t+ 1+t^2 ) & =- (t+ 1+t^2 )=-f(t) & f(-t)=-f(t) aligned f(t) is odd function and g(l)= f(t) Put t=-t in g(t) aligned & g(-t)= (f(-t))= (-f(t)) & =- (f(t))=-g(t) aligned So, g(t) is odd function. we know that g(t) is odd function then _ - ^ g(t) d t=0