Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x^{2}+\frac{x}{\sqrt{x^{2}+9}}$.

$5 \mathbf{p}$ a) Arătați că $\int_{0}^{1}\left(f(x)-\frac{x}{\sqrt{x^{2}+9}}\right) d x=\frac{1}{3}$.

$5 \mathbf{p}$ b) Calculatii $\int_{0}^{4}(f(x)-f(-x)) d x$.

$5 p$ c) Determinați numărul real $a, a\ \textgreater \ 4$, astfel încât $\int_{4}^{a} \frac{f(x)}{x} d x=10+\ln \frac{a+\sqrt{a^{2}+9}}{9}$.

Answer 1

$f(x)=x^{2}+\frac{x}{\sqrt{x^{2}+9}}$

a)

$\int\limits^1_0 {x^2} \, dx =\frac{x^3}{3}\ |_0^1=\frac{1}{3}$

(vezi tabelul de integrale din atasament)

b)

$\int\limits^4_0 {x^2+\frac{x}{\sqrt{x^2+9} } -x^2+\frac{x}{\sqrt{x^2+9} } \, dx =\\\\=\int\limits^4_0 {\frac{2x}{\sqrt{x^2+9} } \, dx$

$\int\limits^4_0 {\frac{2x}{\sqrt{x^2+9} } \, dx=2\sqrt{x^2+9}\ |_0^4=2\sqrt{16+9}-2\sqrt{0+9} =10-6=4$

c)

$\int\limits^a_4 {x+\frac{1}{\sqrt{x^2+9} } } \, dx$

Desfacem in doua integrale si obtinem:

$\frac{x^2}{2}\ |_4^a+ln(x+\sqrt{x^2+9}) \ |_4^a=\frac{a^2}{2}-8+ln(a+\sqrt{a^2+9} ) -ln9=\frac{a^2}{2}-8+ln\frac{a+\sqrt{a^2+9} }{9}$

$\frac{a^2}{2}-8+ln\frac{a+\sqrt{a^2+9} }{9} =10+ln\frac{a+\sqrt{a^2+9} }{9}\\\\\frac{a^2}{2}-8=10\\\\\frac{a^2}{2}=18\\\\a^2=36\\a=6\\\\a=-6 < 4\ NU$

a=6

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/8885003

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