Let S = (- 1, ∞) and f: S → ℝ be defined as f(x) = ∫ _ - 1 ^ x ( e...

Given,

$f\left(x\right) = \mathop{∫}_{- 1}^{x}{\left({e}^{t} - 1\right)}^{11}{\left(2t - 1\right)}^{5}{\left(t - 2\right)}^{7}{\left(t - 3\right)}^{12}{\left(2t - 10\right)}^{61}dt$

Now, differentiating the above integral using Newton Leibnitz's theorem, we get,

${f}^{'}\left(x\right) = {\left({e}^{x} - 1\right)}^{11}{\left(2x - 1\right)}^{5}{\left(x - 2\right)}^{7}{\left(x - 3\right)}^{12}{\left(2x - 10\right)}^{61}$

Now, from the above diagram using first derivative test we get,

Local minima at $x = \dfrac{1}{2}, x = 5$ 

Local maxima at $x = 0, x = 2$

So, $p = 0 + 4 = 4, q = \dfrac{1}{2} + 5 = \dfrac{11}{2}$

Hence, ${p}^{2} + 2q = 16 + 11 = 27$