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Proof of: ∫ f(x) + g(x) dx = ∫ f(x) dx + ∫g(x) dx
It is also a very easy proof. Assume that F(x) is an anti-derivative of f(x) and that G(x) is an anti-derivative of g(x). Therefore we have that F′(x) = f(x) and G′(x) = g(x).
Fundamental properties of derivatives also give us that
(F(x) + G(x))' = F'(x) + G(x) = f(x) + g(x)
and thus F(x) + G(x) is an anti-derivative of f(x) + g(x) and F(x) - G(x) is an anti- derivative of f(x)- g(x). So,
∫ f(x) + g(x) dx = F(x) + G(x) + c =∫ f(x) dx + ∫g(x) dx
Solve : 4x2+2x+3=0 Ans) x^2 + (1/2)x = -(3/4) (x+1/4)^2 = 1/16 - 3/4 = -11/16 implies x = (-1+i(11)^(1/2))/4 and its conjugate.
Provide me some Example of in-equations.
Taking 2^x=m and solving the quadratic for getting D>=0 we get range= [3/4 , infinity )
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5 2 --- - --- x-1 x+1
Let a, b, c 2 Z + . (a) Prove that if a|b, then ac|bc for all c. (b) If a|bc, can you conclude that either a|b or a|c? Justify your answer with a proof or a counter example.
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