Question

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Answer 1

1) Facem substitutia 2x+1 =t
                                  2 dx = dt 

  x1=-1  -> t1 = 2*(-1) +1 =-1
  x2=1   -> t2 = 2*1+1 = 3

Integrala devine 

$\int\limits^b_a {t ^{3} } \ dt = \frac{ t^{4} }{4} \ \ |_{-1} ^{3} = \frac{3^{4} }{4} - \frac{ (-1)^{4} }{4} = \frac{81}{4} - \frac{1}{4} = \frac{80}{4} =20 \\b =3 \ si \ a=-1$

2)$\int\limits^b_a { \frac{1}{ x^{5} } } \, dx = \frac{-1}{4 x^{4} } \ | _ {1} ^{2} = \frac{-1}{4*16} + \frac{1}{4} = \frac{-1}{64} + \frac{16}{64}= \frac{15}{64}$

 b=2 si a =1

3)[tex] \int\limits^a_b {2x \frac{1}{2} } \, dx = \int\limits^a_b {x} \, dx = \frac{ x^{2} }{2} \ |_{1} ^{4} = \frac{16}{2} - \frac{1}{2} = \frac{15}{2}=7.5 \\b=4 \ si \ a=1[/tex]