It is given that, f(x)=x2e2x Differentiate the above equation. f′(x)=2e2xx2+2xe2x 0=2e2x(x2+x) x2+x=0 x=0,1 The maxima and minima is obtained at -2,-1,0,2 as the function bound. Thus, f(−2)=(−2)2e−4=4e−4 f(−1)=(1)2e−2 f(0)=0 f(2)=(2)2e4=4e4 Thus, p=4e4 and q=0 pe−4+qe4=4e4(e−)+0=4e0=4