Question

Salut, am nevoie de ajutor la ex . 1059(explicatii pas cu pas, va rog).

Answer 1

$\displaystyle f =\int_{-1}^1(x^2-a-bx)^2\, dx\\ \\ \\ f'_a =\dfrac{d}{da}\left(\int_{-1}^1(x^2-a-bx)^2 dx\right) = \\ \\ = \int_{-1}^1\dfrac{d}{da}\Big[(x^2-a-bx)^2\Big]\, dx = \int_{-1}^12(x^2-a-bx)\cdot (-1)\, dx = \\ \\ =-2\int_{-1}^1(x^2-a-bx)\, dx = -2\cdot\Big( \dfrac{x^3}{3}-ax-\dfrac{bx^2}{2}\Big)\Big|_{-1}^1 = \\ \\ = -2\cdot \Big(\dfrac{2}{3}-2a\Big)\\ \\ \\ f_b' = \dfrac{d}{db}\int_{-1}^1(x^2-a-bx)^2\, dx = \int_{-1}^1 2(x^2-a-bx)\cdot (-x)\, dx =$

$\displaystyle = -2\int_{-1}^1x(x^2-a-bx)\, dx = -2\cdot \Big(\dfrac{x^4}{4}-\dfrac{ax^2}{2}-\dfrac{bx^3}{3}\Big)\Big|_{-1}^1 = \\ \\ = -2\cdot \Big(\dfrac{1}{4}-\dfrac{a}{2}-\dfrac{b}{3} - \dfrac{1}{4}+\dfrac{a}{2}-\dfrac{b}{3}\Big) = \dfrac{4b}{3}\\ \\ \\ \text{Facem }f_a ' \text{ si }f_b' = 0\\ \Rightarrow \dfrac{2}{3}-2a = 0 \text{ si }\dfrac{4b}{3} = 0 \Rightarrow a = \dfrac{1}{3} \text{ si }b = 0$

$\Rightarrow \displaystyle \underset{a,b\in \mathbb{R}}{\min}\int_{-1}^1(x^2-a-bx)^2\, dx= \int_{-1}^1\Big(x^2-\frac{1}{3}\Big)^2\, dx = \\ \\ = \int_{-1}^1\Big(x^4-\dfrac{2x^2}{3}+\dfrac{1}{9}\Big)\, dx = \Big(\dfrac{x^5}{5}-\dfrac{2x^3}{9}+\dfrac{x}{9}\Big)\Big|_{-1}^1 = \\ \\ =\dfrac{1}{5}-\dfrac{2}{9}+\dfrac{1}{9}+\dfrac{1}{5}-\dfrac{2}{9}+\dfrac{1}{9} = \boxed{\dfrac{8}{45}}$