Given f(x)=∫0x22+ett2−5t+4dt
We know that,
dxd∫u(x)v(x)f(v)dx=f(v(x))dxdv(u)−f(u(x))dxdu(x)
Using above formula we get f′(x)=(2+ex2x4−5x2+4)⋅2x
For critical points f′(x)=0
So, 2+ex2x(x2−1)(x2−4)=0
⇒x(x+1)(x−1)(x+2)(x−2)=0
Now by first derivative test we get,

Here points of maxima are −1,1 & points of minima are −2,0,2