Question

Subpunctele b si c de la problema 2
Integrale

va rog sa rezolvati mai detaliat si sa spuneti ce proprietati ati folosit. Multumesc anticipat!

Answer 1

$f:(-5,5)\to\mathbb{R},\,\,f(x) = \sqrt{25-x^2}\\ \\\\ \displaystyle b)\,\,\, \int_{-3}^3|xf(x)|\, dx = \int_{-3}^3 |x\sqrt{25-x^2}|\, dx =$

$\displaystyle = \int_{-3}^3 |x|\cdot |\sqrt{25-x^2}|\, dx = \int_{-3}^3 |x|\cdot \sqrt{25-x^2}\, dx =$

$\displaystyle =\int_{-3}^0 (-x)\cdot \sqrt{25-x^2}\, dx + \int_{0}^3x\cdot \sqrt{25-x^2}\, dx=$

$\displaystyle =\dfrac{1}{2}\int_{-3}^0 (25-x^2)'\cdot (25-x^2)^{\frac{1}{2}}\,dx -\dfrac{1}{2}\int_{0}^3(25-x^2)'\cdot (25-x^2)^{\frac{1}{2}}\,dx=$

$=\dfrac{1}{2}\cdot \dfrac{(25-x^2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg|_{-3}^0-\dfrac{1}{2}\cdot \dfrac{(25-x^2)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg|_{0}^3 =$

$=\dfrac{\sqrt{(25-x^2)^3}}{3}\Bigg|_{-3}^0 - \dfrac{\sqrt{(25-x^2)^3}}{3}\Bigg|_{0}^3 =$

$=\dfrac{\sqrt{(25-0^2)^3}}{3}-\dfrac{\sqrt{[25-(-3)^2]^3}}{3} - \dfrac{\sqrt{(25-3^2)^3}}{3} +\dfrac{\sqrt{(25-0)^3}}{3} =$

$=\dfrac{5^3}{3}-\dfrac{4^3}{3}-\dfrac{4^3}{3}+\dfrac{5^3}{3} = \boxed{\dfrac{122}{3}}\\ \\\\ \displaystyle c)\,\,\,I_{n+1} - I_n = \int_{0}^1\dfrac{1}{f^{n+1}(x)}\, dx - \int_{0}^1\dfrac{1}{f^n(x)}\, dx =$

$\displaystyle =\int_{0}^1\dfrac{1}{f^{n}(x)}\cdot \dfrac{1}{f(x)}\, dx - \int_{0}^1\dfrac{1}{f^n(x)}\, dx =$

$\displaystyle = \int_{0}^1\left[\dfrac{1}{f^{n}(x)}\cdot \dfrac{1}{f(x)}- \dfrac{1}{f^n(x)}\right] dx =$

$\displaystyle = \int_{0}^1\dfrac{1}{f^n(x)}\cdot \left[\dfrac{1}{f(x)}-1\right]dx=$

$\displaystyle =\int_{0}^1\dfrac{1}{{\left(\sqrt{25-x^2}\right)}^n}\cdot \left[\dfrac{1}{\sqrt{25-x^2}}-1\right]dx\\ \\\\\text{Pentru }x\in [0,1]:$

${\left(\sqrt{25-x^2}\right)}^n>0 \Rightarrow \dfrac{1}{{\left(\sqrt{25-x^2}\right)}^n} > 0\,\,\,\,\,\,\,\,\big(1\big)$

$\sqrt{25-x^2} \in [\sqrt{24},5] \Rightarrow \sqrt{25-x^2} > 1 \Rightarrow$

$\Rightarrow \dfrac{1}{\sqrt{25-x^2}} < 1 \Rightarrow \dfrac{1}{\sqrt{25-x^2}}-1<0\,\,\,\,\,\,\,\,\big(2\big)$

$\text{Din }\big(1\big)\text{ si }\big(2\big)\Rightarrow \dfrac{1}{{\left(\sqrt{25-x^2}\right)}^n}\cdot \left[\dfrac{1}{\sqrt{25-x^2}}-1\right] < 0\\ \\ \\ \textbf{Proprietate:}$

$\text{Daca } f:[a,b]\to \mathbb{R}\text{ este o functie continua si }f(x) \leq 0, \,\, x\in [a,b],\\\text{atunci:}$

$\displaystyle \int_{a}^b f(x)\, dx \leq 0\\ \\ \\ \displaystyle \Rightarrow \int_{0}^1 \dfrac{1}{{\left(\sqrt{25-x^2}\right)}^n}\cdot \left[\dfrac{1}{\sqrt{25-x^2}}-1\right] dx < 0$

$\Rightarrow I_{n+1} - I_n < 0 \Rightarrow I_{n+1} < I_n, \,\,\,\,\, \forall n\in \mathbb{N}^*$

$\Rightarrow \boxed{I_{n}-\,\text{sir monoton descrescator}}$