[tex]f(x)=\sqrt{x^{2}+1}[/tex]
a)
Stim ca [tex](\sqrt{u} )'=\frac{u'}{2\sqrt{u} }[/tex]
[tex](\sqrt{x^2+1})' =\frac{(x^2+1)'}{2\sqrt{x^2+1} } =\frac{2x}{2\sqrt{x^2+1} } =\frac{x}{\sqrt{x^2+1} }[/tex]
Atunci integrala noastra va fi egala cu:
[tex]\int_{0}^{\sqrt{3}} \frac{x}{f(x)} d x=\sqrt{x^2+1}\ |_0^{\sqrt{3} } = \sqrt{\sqrt{3}^2+1 } -\sqrt{0^2+1} =2-1=1[/tex]
b)
[tex]\int_{0}^{1} f(x) d x=\int_{0}^{1} \sqrt{x^2+1}\ dx=\int_{0}^{1} \frac{x^2+1}{\sqrt{x^2+1} } d x[/tex]
Am amplificat cu √(x²+1), apoi desfacem in doua integrale si obtinem:
[tex]I=\int_{0}^{1} \frac{x^2+1}{\sqrt{x^2+1} } d x=\int_{0}^{1} \frac{x^2}{\sqrt{x^2+1} } d x+\int_{0}^{1} \frac{1}{\sqrt{x^2+1} } d x[/tex]
Luam integralele pe rand si le calculam, iar apoi ne intoarcem sa calculam integrala noastra I
[tex]\int_{0}^{1} \frac{x^2}{\sqrt{x^2+1} } d x[/tex]
Calculam prin metoda integrarii prin parti, adica prin formula de mai jos:
[tex]\int\limits {f(x)\times g'(x)} \, dx =f(x)\times g(x)-\int\limits {f'(x)\times g(x)} \, dx[/tex]
[tex]f(x)=x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f'(x)=1\\\\g'(x)=\frac{x}{\sqrt{x^2+1} } \ \ \ \ \ \ \ \ \ g(x)=\sqrt{x^2+1}[/tex]
[tex]\int_{0}^{1} \frac{x^2}{\sqrt{x^2+1} } d x=x\sqrt{x^2+1} \|_0^1-\int\limits^1_0 {\sqrt{x^2+1} } \, dx=x\sqrt{x^2+1}\ |_0^1-I=\sqrt{2}-I[/tex]
[tex]\int_{0}^{1} \frac{1}{\sqrt{x^2+1} } d x=ln(x+\sqrt{x^2+1)} \ |_0^1=ln(1+\sqrt{2}) -ln1=ln(1+\sqrt{2})[/tex]
Acum calculam integrala noastra I, inlocuind ce am aflat mai sus
[tex]I=\sqrt{2} -I+ln(1+\sqrt{2})\\\\ 2I=\sqrt{2} +ln(1+\sqrt{2})\\\\I=\frac{\sqrt{2}+ln(1+\sqrt{2} )}{2}[/tex]
Sau puteam folosi direct formula din tabelul de formule cu integrale (vezi poza atasata)
c)
[tex]\int_{0}^{x} e^{f^{2}(t)} d t=x[/tex]
[tex]\int_{0}^{x} e^{(t^2+1)} d t[/tex]
[tex]\int_{0}^{x} e^{f^{2}(t)} d t-x=0[/tex]
Notam partea stanga cu g(x) si studiem monotonia
[tex]g(x)=\int_{0}^{x} e^{f^{2}(t)} d t-x\\\\g'(x)=e^{(x^2+1)}-1[/tex]
[tex]e^{(x^2+1)}-1 > 0\ ,\ \ \ g(x)\ functie\ crescatoare\[/tex]
Iar g(0)=0⇒ există un unic număr real x pentru care se indeplineste relatia de mai sus
Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/1576176
#BAC2022
