evaluate-0-2-x-2-dx-

Question Number 66561 by Rio Michael last updated on 17/Aug/19

e v a l u a t e

\int_{0}^{2} ∣ x + 2 ∣ d x .

Commented by kaivan.ahmadi last updated on 17/Aug/19

0 < x < 2 \Rightarrow x + 2 > 0 \Rightarrow∣ x + 2 ∣= x + 2

\int_{0}^{2} (x + 2) d x = \frac{x^{2}}{2} + 2 x ∣_{0}^{2} = (2 + 4) - 0 = 6

Commented by Rio Michael last updated on 17/Aug/19

s i r i d o n t u n d e r s t a n d w h a t y o u d i d a b o v e

Answered by Tanmay chaudhury last updated on 17/Aug/19

∣ x + 2 ∣

= x + 2 s i n c h e r e i n t e r v a l f r o m 0 t o 2

s o x i s p o s i t i v e

\int_{0}^{2} (x + 2) d x

∣ \frac{x^{2}}{2} + 2 x ∣_{0}^{2}

= \frac{4}{2} + 2 \times 2

= 6

Commented by Rio Michael last updated on 17/Aug/19

t h a n k s s o h o w d o i s o l v e t h i s t h e n

\int_{2}^{3} ∣ x^{2} + x + 1 ∣ d x ?

Commented by mr W last updated on 17/Aug/19

\int_{2}^{3} \dots . m e a n s t h e r a n g e f o r x i s [2, 3] .

i n t h e r a n g e 2 ⩽ x ⩽ 3 {i t}^{'} s c l e a r t h a t

x^{2} + x + 1 > 0, h e n c e ∣ x^{2} + x + 1 ∣= x^{2} + x + 1,

\int_{2}^{3} ∣ x^{2} + x + 1 ∣ d x

= \int_{2}^{3} (x^{2} + x + 1) d x

= {[\frac{x^{3}}{3} + \frac{x^{2}}{2} + x]}_{2}^{3}

= [\frac{3^{3}}{3} + \frac{3^{2}}{2} + 3] - [\frac{2^{3}}{3} + \frac{2^{2}}{2} + 2]

= \frac{27 - 8}{3} + \frac{9 - 4}{2} + 3 - 2

= \frac{19}{3} + \frac{5}{2} + 1

= \frac{19 \times 2 + 5 \times 3 + 6}{6}

= \frac{59}{6}

Commented by Rio Michael last updated on 17/Aug/19

t h a n k y o u