Question

Se consideră funcția $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x^{2} \operatorname{arctg} x$.

5p a) Arătați că $\int_{1}^{3} \frac{x f(x)}{\operatorname{arctg} x} d x=20$.

5p b) Arătați că $\int_{1}^{\sqrt{3}} \frac{f(x)}{x} d x=\frac{5 \pi}{12}-\frac{\sqrt{3}-1}{2}$.

5p c) Demonstrați că $\lim _{n \rightarrow+\infty} \int_{0}^{1} f^{n}(x) d x=0$.

Answer 1

$f(x)=x^{2} arctg} x$

a)

$.\int_{1}^{3} \frac{x f(x)}{\ arctg x} d x=20$

$\int\limits^3_1 {x^3} \, dx =\frac{x^4}{4} |_1^3=\frac{81}{4} -\frac{1}{4}=\frac{80}{4} =20$

b)

$\int\limits^{\sqrt{3}} _1 {x\cdot arctgx} \, dx$

Il scriem pe x ca fiind:

$x=\frac{(x^2+1)'}{2} \\\\2x=(x^2+1)'$

Folosim tabelul de derivate si integrale (vezi atasament)

$\int\limits^{\sqrt{3}} _1 {x\cdot arctgx} \, dx=\frac{x^2+1}{2}arctgx|_1^{\sqrt{3}}-\frac{1}{2} x |_1^{\sqrt{3}} =2arctg\sqrt{3} -arctg1-\frac{1}{2} \cdot \sqrt{3}+\frac{1}{2} =\frac{2\pi}{3}-\frac{\pi}{4}-\frac{1}{2} \cdot \sqrt{3}+\frac{1}{2} =\\\\ = \frac{5\pi}{12} -\frac{\sqrt{3}-1 }{2}$

c)

x∈[0,1]

0≤x≤1

0≤xⁿ≤1

arctg0≤arctgx≤arctg1

$0\leq arctgx\leq \frac{\pi}{4}\ \ \ |^n\\\\0\leq arctg^nx\leq (\frac{\pi}{4})^n$

Inmultim cele doua relatii si le integram:

$0\leq \int\limits^n_1 x^narctg^nx} \, dx \leq \int\limits^n_1 {(\frac{\pi}{4})^n} \, dx \\\\0\leq \int\limits^n_1 {f^n(x)} \, dx \leq (\frac{\pi}{4})^n\\\\Dar\ (\frac{\pi}{4})^n\to0\\\\Deci\ \int\limits^n_1 {f^n(x)} \, dx=0$

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9905546

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