Given,
f(x)=∫−1x(et−1)11(2t−1)5(t−2)7(t−3)12(2t−10)61dt
Now, differentiating the above integral using Newton Leibnitz's theorem, we get,
f′(x)=(ex−1)11(2x−1)5(x−2)7(x−3)12(2x−10)61

Now, from the above diagram using first derivative test we get,
Local minima at x=21,x=5
Local maxima at x=0,x=2
So, p=0+4=4,q=21+5=211
Hence, p2+2q=16+11=27