Question-173836

Question Number 173836 by mathlove last updated on 19/Jul/22

Commented by aleks041103 last updated on 21/Jul/22

I = \int_{- 2}^{- 1} (10 \sqrt[4]{x^{6}} - 2 x) d x =

= 10 \int_{- 2}^{- 1} \sqrt[4]{x^{6}} d x - \int_{- 2}^{- 1} 2 x d x =

= 10 J - {[x^{2}]}_{- 2}^{- 1} = 16 \sqrt{2} - (1 - 4) = 3 + 16 \sqrt{2}

x^{6} i s e v e n \Rightarrow J = \int_{- 2}^{- 1} \sqrt[4]{x^{6}} d x = \int_{1}^{2} \sqrt[4]{x^{6}} d x = \int_{1}^{2} x^{3 / 2} d x =

= {[\frac{2}{5} x^{5 / 2}]}_{1}^{2} = \frac{8 \sqrt{2}}{5}

Answered by CElcedricjunior last updated on 19/Jul/22

I = \int_{- 2}^{- 1} (10 \sqrt{x^{3}} - 2 x) d x = \int_{- 2}^{- 1} (10 x^{\frac{3}{2}} - 2 x) dx

o r \sqrt{x^{3}} n^{'} est definie en [- 2; - 1]

alors I n^{'} existe pas dans R

m a i s d a n s C

Answered by a.lgnaoui last updated on 19/Jul/22

I = 10 \int_{- 2}^{- 1} x^{\frac{3}{2}} d x - 2 \int x d x

I = 10 {[\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}]}_{- 2}^{- 1} - 2 {[\frac{x^{2}}{2}]}_{- 2}^{- 1} = 4 {[x^{2} \sqrt{x}]}_{- 2}^{- 1} - {[x^{2}]}_{- 2}^{+ 1}

= {[x^{2} (4 \sqrt{x} - 1)]}_{- 2}^{- 1} x < 0 \Rightarrow \sqrt{x} = \sqrt{∣ x ∣ i^{2}} = i \sqrt{∣ x ∣}

I = {[x^{2} (4 i \sqrt{∣ x ∣} - 1)]}_{- 2}^{- 1} = 4 i - 1 - 16 \sqrt{2} i + 4

I = 3 + (4 - 16 \sqrt{2}) i = 3 - 18, 56 i