Question

Va rog sa ma ajutati cu aceasta problema

Answer 1

a)Inlocuind f(x) in integrala si simplificand, obtinem:
$\int\limits^1_0 {x^3} \, dx = \frac{x^4}{4}|_0^1= \frac{1}{4}$
b)Inlocuind f(x) in integrala si aducand la acelasi numitor, obtinem:
[tex] \int\limits^1_0 { \frac{1}{x^2+x+1} } \, dx = \int\limits^1_0 {\frac{1}{(x+\frac{1}{2})^2+ (\frac{ \sqrt{3} }{2})^2 } } \, dx =\\ = \frac{2}{ \sqrt{3} }arctg( \frac{2}{ \sqrt{3}} (x+\frac{1}{2}))|_0^1=\\ =\frac{2}{ \sqrt{3} }( \frac{\pi}{3}- \frac{\pi}{6})=\frac{2}{ \sqrt{3} }\cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}} [/tex]
c)Fie F primitiva functiei f.
[tex]\displaystyle \lim_{t \to 0} \frac{ \int\limits^t_0 {f(x)} \, dx }{t^4} =\\ =\lim_{t \to 0} \frac{ F(t)-F(0) }{t^4} =aplicam\ L'Hospital=\\ =\lim_{t \to 0} \frac{ F'(t) }{4t^3}=\lim_{t \to 0} \frac{ f(t) }{4t^3}=\\ =\lim_{t \to 0} \frac{ t^3 }{4t^3(t^2+t+1)}=\lim_{t \to 0} \frac{1 }{4(t^2+t+1)}=\frac{1}{4}[/tex]