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Question Number 174420 by ali009 last updated on 31/Jul/22

d e t e r m i n e w h e t h e r t h e f o l l o w i n g i n t e g r a l

a r e c o n v e r g e n c e o r d i v e r g e n c e

1) \int_{- \infty}^{1} \frac{3}{4 - 2 x} d x

2) \int_{1}^{2} \frac{x^{2}}{\sqrt{8 - x^{3}}} d x

Answered by floor(10²Eta[1]) last updated on 31/Jul/22

1) - \frac{3}{2} \int_{- \infty}^{1} \frac{dx}{x - 2} = - \frac{3}{2} \ln ∣ x - 2 ∣_{- \infty}^{1}

- \frac{3}{2} (\ln ∣ - 1 ∣ - \ln ∣ - \infty ∣)

- \frac{3}{2} (- \ln (\infty)) = \infty

Answered by floor(10²Eta[1]) last updated on 31/Jul/22

2) u = 8 - x^{3} \Rightarrow du = - {3 x}^{2} dx

- \frac{1}{3} \int_{1}^{8} \frac{dx}{\sqrt{u}} = - \frac{1}{3} \int_{1}^{8} u^{- 1 / 2} = - \frac{2}{3} {[\sqrt{u}]}_{1}^{8}

= - \frac{2}{3} (2 \sqrt{2} - 1) = \frac{2 - 4 \sqrt{2}}{3}