To find the minimum number "n" of intervals, we need to solve the inequality
"E\\leq\\frac{(b-a)^3}{12n^2}[\\text{max}|f"(x)|],a\\leq x\\leq b." Therefore, first find the second derivative:
"\\frac{d}{dx}\\big[e^{-x^2+1}\\big]=-2xe^{1-x^2}=f'(x),\\\\ \\space\\\\\n\\frac{d}{dx}\\big[-2xe^{1-x^2}\\big]=2e^{1-x^2}(2x^2-1)=f"(x)."Assume that the integration limits are "a=-1, b=2" because they are not present in the condition. Therefore:
"f"(-1)=2\\\\\nf"(2)=0.7\\\\\nf"(0)=-5.43656...=\\text{max}[f"(x)].\\\\"
Find "n":
"0.5\\cdot10^{-4}\\leq\\frac{(2-(-1))^3}{12n^2}[5.27]=\\frac{27}{12n^2}\\cdot5.4366,\n\\\\ \\space\\\\\nn^2\\geq244647,\\\\n\\geq494.62\\approx496."
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