Question

dau 50 puncte ma ajuta cineva va rog

Answer 1

[tex]\it a)\ \int^1_0 \dfrac{1}{x+1} dx=ln(x+1) \Big{|}^1_0 =ln2-ln1 =ln2-0=ln2 \\ \\ \\ b)\ \int f^2(x) dx = \int\dfrac{1}{(x+1)^2} dx = \int(x+1)^{-2}dx =-\dfrac{1}{x+1} +\mathcal{C}[/tex]

Relația din enunț devine:

$\it -\dfrac{1}{x+1} \Big{|}^a _0 = \dfrac{1}{2} \Rightarrow -\dfrac{1}{a+1} + 1 =\dfrac{1}{2} \Rightarrow -\dfrac{1}{a+1} = -\dfrac{1}{2} \Rightarrow a=1$

$\it c)\ \Big{\int} _0 ^1 \dfrac{1}{x^2+1} dx = arctgx \Big{|}_0^1 = arctg1-arctg 0 = \dfrac{\pi}{4} -0=\dfrac{\pi}{4}$