"\\int_{-2}^2 \\frac{dx}{x^2} = 2 \\int_{0}^2 \\frac{dx}{x^2} = 2 \\lim_{\\epsilon \\to 0+} \\int_{\\epsilon}^2 \\frac{dx}{x^2} \\\\= 2 \\lim_{\\epsilon \\to 0+} \\left(- \\frac{1}{x}\\right)\\vert_{\\epsilon}^2 = +\\infty"
The first equality holds because the function "\\frac{1}{x^2} ,\\; x \\in [-2;2]" is even.
The rest follows from improper integral definition and the fundamental theorem of calculus.
Answer: "+\\infty" .
2. Evaluate the definite integral
"\\int_{-\\infty}^0 2^x dx = \\int_{-\\infty}^0 e^{\\ln(2)x} dx \\\\=\n\\lim_{\\Delta \\to -\\infty} \\int_{\\Delta}^0 e^{\\ln(2)x} dx \\\\=\n\\lim_{\\Delta \\to -\\infty} \\left(\\frac{ e^{\\ln(2)x}}{\\ln(2)}\\right) \\vert_{\\Delta}^0 = \\frac{1}{\\ln(2)}"
Answer: "\\frac{1}{ \\ln(2)}"
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