Question

ajutor urgent cât se poate de repede, vă rog​

Answer 1

2)

a)

$\it \int\limits^4_3 {(4x^3+2x-3-4x^3+3)} \, dx =\int\limits^4_3 2x dx=2\cdot\dfrac{x^2}{2}\bigg|^4_3=x^2\bigg|^4_3=4^2-3^2=7$

b)

$\it F(x)=\int {f(x) }dx=\int (\it 4x^3+2x-3) dx=4\cdot\dfrac{x^4}{4}+2\cdot\dfrac{x^2}{2}-3x+\mathcal{C}=\\ \\ \\ =x^4+x^2-3x+\mathcal{C}\\ \\ F(0)=2020 \Rightarrow \mathcal{C}=2020 \Rightarrow F(x)=x^4+x^2-3x+2020$

c)

$\it \int\limits^1_0 {xf(x)} \, dx =\int\limits^1_0(4x^4+2x^2-3x) dx=4\cdot\dfrac{x^5}{5}+2\cdot\dfrac{x^3}{3}-3\cdot\dfrac{x^2}{2}\bigg|^1_0=\\ \\ \\ =\dfrac{^{6)}4}{\ 5}+\dfrac{^{10)}2}{\ \ 3}-\dfrac{^{15)}3}{\ \ 2}=\dfrac{24+20-45}{30}=-\dfrac{1}{30}$