calculate-2-5-dx-x-1-x-2-

Question Number 38716 by maxmathsup by imad last updated on 28/Jun/18

c a l c u l a t e \int_{2}^{5} \frac{d x}{{(x + 1 - [x])}^{2}}

Commented by abdo mathsup 649 cc last updated on 29/Jun/18

I = \sum_{k = 2}^{4} \int_{k}^{k + 1} \frac{d x}{{(x + 1 - k)}^{2}}

= \sum_{k = 2}^{4} {[- \frac{1}{x + 1 - k}]}_{k}^{k + 1}

= \sum_{k = 2}^{4} {1 - \frac{1}{2}} = \frac{1}{2} \sum_{k = 2}^{4} (1) = \frac{1}{2} (4 - 2 + 1)

= \frac{3}{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18

\int_{2}^{3} \frac{d x}{{(x + 1 - 2)}^{2}} + \int_{3}^{4} \frac{d x}{{(x + 1 - 3)}^{2}} + \int_{4}^{5} \frac{d x}{{(x + 1 - 4)}^{2}}

=∣ \frac{{(x - 1)}^{- 1}}{- 1} ∣_{2}^{3} + ∣ \frac{{(x - 2)}^{- 1}}{- 1} ∣_{3}^{4} + ∣ \frac{{(x - 3)}^{- 1}}{- 1} ∣_{4}^{5}

= - 1 {(\frac{1}{2} - \frac{1}{1} + \frac{1}{2} - \frac{1}{1} + \frac{1}{2} - \frac{1}{1})}

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

= \frac{3}{2}