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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
Write the quadratic equation whose roots are real and non conjugate Ans) x^2-x+6=0 ...roots are real and non conjugate
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Determine the measure of the vertex angle of the isosceles triangle. a. 34° b. 16° c. 58° d. 112° d. Simply substitute x = 34 into the equation for the vertex angle,
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