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Penguin200
Pe: 2018-12-09T08:26:52+02:00 In: In:

Exercitii primitive si integrale

1.Fie F o primitiva a functiei f:R->R
f(x)=e^{x^{2}} care verifica egalitatea F(-1)=0.Atunci:
A.F(1)<0
B.F(1)=0
C.F(1)>2
D.F(1)=2=
2.Fie F o primitiva a functiei f:R*->R, f(x)=\frac{1}{x} care verifica egalitatile F(-1)=1 si F(1)=0.Aratati ca F(e)+F(-e)=3.
3.\int_{0}^{1}\frac{\sqrt{x}}{(x+3)\sqrt{x+3})}dx

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  1. Felixx
    veteran (III)

    3)Este o integrala binoma de forma:
    \int x^{m}\left ( ax^{n}+b \right )^{p}dxunde
    m,n,p\in \mathbb{Q},m=\frac{m_{1}}{m_{2}},n=\frac{n_{1}}{n_{2}},p=\frac{p_{1}}{p_{2}}
    Pentru calculul acestor integrale se folosesc substitutiile lui Cebasev.In cazul problemei noastre ne situam in cazul substitutiei 3):
    Daca \frac{m+1}{n}+p\in \mathbb{Z} facem substitutia:
    \frac{ax^{n}+b}{x^{n}}=t^{p_{2}}
    In cazul nostru :
    I=\int x^{\frac{1}{2}}\left ( x+3 \right )^{\frac{-3}{2}}dx
    m=\frac{1}{2},n=\frac{1}{1},p=\frac{-3}{2} si
    \frac{m+1}{n}+p=\frac{\frac{1}{2}+1}{1}+\left ( -\frac{3}{2} \right )=0\in \mathbb{Z}
    Facem substitutia:
    \frac{x+3}{x}=t^{2},x\neq 0
    x=\frac{3}{t^{2}-1},dx=\frac{-6t}{\left ( t^{2}-1 \right )^{2}}dt
    Ramane sa inlocuiesti, sa efectuezi calculele si vei obtine :
    I=\left ( -2 \right )\int \frac{1}{t^{2}\left ( t^{2}-1 \right )}dt= \left ( -2 \right )\int \left ( \frac{1}{t^{2}-1}-\frac{1}{t^{2}} \right )dt
    care se rezolva usor.
    Ajungi la :
    I=-ln\left | \frac{\sqrt{x+3}}{\sqrt{x}}-1 \right |+ln\left ( \frac{\sqrt{x+3}}{\sqrt{x}}+1 \right )-2\frac{\sqrt{x}}{\sqrt{x+3}}+c=ln\left ( \frac{\sqrt{x+3}+\sqrt{x}}{\sqrt{x+3}-\sqrt{x}} \right )-2\frac{\sqrt{x}}{\sqrt{x+3}}+c
    Rezultatul este : I=ln3-1

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