Question

Am nevoie de ajutor la punctul b). Mie imi da rezultatul (-2√2+1)/3.. iar in barem e alt rezultat

Answer 1

fie 2-x²=t
     -2xdx=dt
x=-dt/2

∫x√2-x² dx=-1/2·∫√t dt=-1/2·2/3·t√t=-1/3t√t

revenim la substitutie
∫x√2-x² dx=-1/3(2-x²)√(2-x²)
F(1)-F(0)=-1/3+2/3·√2=(-1+2√2)/3

Answer 2

$f:[0,1] \rightarrow \mathbb{R},\quad f(x) = x\sqrt{2-x^2}\\ \\ \\ \displaystyle \int_0^1 f(x) \, dx = \int_0^1 x\sqrt{2-x^2}\, dx =\int_0^1 \sqrt{2-x^2}\cdot x\, dx \overset{(*)}{=} \\ \\ \sqrt{2-x^2} = t \Rightarrow 2-x^2 = t^2 \Rightarrow -2xdx = 2tdt\Big|:-2 \Rightarrow \\ \\ \Rightarrow x dx = -tdt\\ \\\bullet x = 0 \Rightarrow t = \sqrt{2};\quad x= 1 \Rightarrow t = 1$

[tex] \overset{(*)}{=} \int_{\sqrt 2} ^{1}t\cdot (-t) \,dt = \int_{\sqrt 2} ^{1}(-t^2)dt = \int_{1} ^{\sqrt 2} t^2 \, dt = \dfrac{t^3}{3} \Big|_{1}^{\sqrt 2} = \\ \\ = \dfrac{\sqrt{2}^3}{3} - \dfrac{1^3}{3} = \dfrac{2\sqrt2}{3} - \dfrac{1}{3} = \boxed{\dfrac{2\sqrt2 -1}{3}}[/tex]




Varianta in care notam doar pe 2-x^2 cu t.

[tex] \displaystyle \\ \int_{0}^1 f(x) \, dx = \int_0^1 x\sqrt{2-x^2}\, dx \overset{(*)}{=} \\ \\ 2-x^2 = t \Rightarrow -2x = 1\cdot dt \Big|:-2 \Rightarrow x = \dfrac{-1}{2}dt \\ \\ \bullet x = 0 \Rightarrow t = 2, \quad x = 1 \Rightarrow t = 1\\ \\ \overset{(*)}{=} \int_2^1 \sqrt{t}\, \cdot \Big(-\dfrac{1}{2}\Big)\, dt = -\dfrac{1}{2}\int_2^1 \sqrt t\, dt = [/tex]

$\displaystyle = -\left(-\dfrac{1}{2}\int_1^2\sqrt{t}\, dt\right) = \dfrac{1}{2}\int_1^2\sqrt{t}\, dt = \dfrac{1}{2}\cdot \dfrac{t^{\Big(\dfrac{1}{2}\big +\big 1\Big)}}{\dfrac{1}{2}+1}\Big|_{1}^2 = \\ \\ = \dfrac{1}{2}\cdot \dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}}\Big|_{1}^2 = \dfrac{\sqrt{t}^3}{3}\Big|_{1}^2 = \dfrac{\sqrt 2^3}{3} - \dfrac{\sqrt 1^3}{3} = \boxed{\dfrac{2\sqrt2 -1}{3}}$