Question

Culegere de admitere UPT, problema AM144, va rog.

Answer 1

$f(x) = x^{1006}+2014^{x}\\ \\ \displaystyle \int x\cdot f(x^2)\, dx = \int f(x^2)\cdot x \, dx\overset{(*)}{=} \\ \\ x^2 = t \Rightarrow 2x\, dx = dt \Rightarrow xdx = \dfrac{1}{2}\, dt \\ \\ \overset{(*)}{=} \int f(t) \cdot \dfrac{1}{2}\, dt = \dfrac{1}{2}\int f(t) \, dt = \\ \\ = \dfrac{1}{2} \int \Big(t^{1006}+2014^t\Big) \, dt =$

$=\dfrac{1}{2}\cdot \left(\dfrac{t^{1006+1}}{1006+1}+\dfrac{2014^t}{\ln 2014}\right)= \dfrac{1}{2}\cdot \left(\dfrac{t^{1007}}{1007}+\dfrac{2014^t}{\ln 2014}\right) = \\ \\ = \dfrac{t^{1007}}{2014}+\dfrac{2014^t}{2\ln 2014} =\dfrac{{(x^2)}^{1007}}{2014} + \dfrac{2014^{x^2}}{2\ln 2014} = \\ \\ =\dfrac{x^{2\cdot 1007}}{2014}+\dfrac{2^{(-1)} \cdot 2^{x^2}\cdot 1007^{x^2}}{\ln 2014} = \\ \\ = \boxed{\dfrac{x^{2014}}{2014} + \dfrac{2^{x^2-1} \cdot 1007^{x^2}}{\ln 2014}}$

Answer 2

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