Răspuns :
[tex]f(x)=x^{2} arctg} x[/tex]
a)
[tex].\int_{1}^{3} \frac{x f(x)}{\ arctg x} d x=20[/tex]
[tex]\int\limits^3_1 {x^3} \, dx =\frac{x^4}{4} |_1^3=\frac{81}{4} -\frac{1}{4}=\frac{80}{4} =20[/tex]
b)
[tex]\int\limits^{\sqrt{3}} _1 {x\cdot arctgx} \, dx[/tex]
Il scriem pe x ca fiind:
[tex]x=\frac{(x^2+1)'}{2} \\\\2x=(x^2+1)'[/tex]
Folosim tabelul de derivate si integrale (vezi atasament)
[tex]\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}[/tex]
c)
x∈[0,1]
0≤x≤1
0≤xⁿ≤1
arctg0≤arctgx≤arctg1
[tex]0\leq arctgx\leq \frac{\pi}{4}\ \ \ |^n\\\\0\leq arctg^nx\leq (\frac{\pi}{4})^n[/tex]
Inmultim cele doua relatii si le integram:
[tex]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[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9905546
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