Question

$\textnormal{Se consider\u a func\c tia } g:\mathbb{R}\rightarrow \mathbb{R}, g(x)=(x+1)^{3}-3x^{2}-1. \\\textnormal{S\u a se calculeze} \int_{0}^{1}(3x^{2}+3)\cdot g^{2009}(x)dx.$

Answer 1

$g(x) = (x+1)^3-3x^2-1$

$g'(x) = 3(x+1)^2-6x = 3x^2+6x+3-6x = 3x^2+3$

$\displaystyle \int_{0}^1(3x^2+3)\cdot g^{2009}(x)\, dx =\int_{0}^1g'(x)\cdot g^{2009}(x)\, dx =$

$=\dfrac{g^{2009+1}(x)}{2009+1}\Bigg|_0^1 = \dfrac{g^{2010}(x)}{2010}\Bigg|_0^1 = \dfrac{g^{2010}(1)-g^{2010}(0)}{2010}=$

$=\dfrac{(2^3-3-1)^{2010}-(1^3-0-1)^{2010}}{2010} = \dfrac{4^{2010}}{2010} =$

$=\dfrac{2^{4020}}{2010} =\boxed{\dfrac{2^{4019}}{1005}}$