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koshka99
@koshka99
July 2022
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помогите пж решить интегралы, 30 б
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yugolovin
Verified answer
А) = (- 2√x+sin x)|_0^(π/2)=-2√(π/2)+1=1 - √(2π)
б) = ∫_1^2(2x+7-3/x-5x^(-2))dx=
(x^2+7x-3ln|x|+5x^(-1)|_1^2=
4+14-3ln2+5/2-1-7-5=15/2-3ln 2
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Answers & Comments
Verified answer
А) = (- 2√x+sin x)|_0^(π/2)=-2√(π/2)+1=1 - √(2π)б) = ∫_1^2(2x+7-3/x-5x^(-2))dx=
(x^2+7x-3ln|x|+5x^(-1)|_1^2=
4+14-3ln2+5/2-1-7-5=15/2-3ln 2