evaluate ∫_0 ^2 ∣ x+ 2∣ dx.


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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

	$\int_{2}^{3} . . . . m e a n s t h e r a n g e f o r x i s [2, 3] .$ $i n t h e r a n g e 2 ⩽ x ⩽ 3 {i t}^{'} s c l e a r t h a t$ $x^{2} + x + 1 > 0, h e n c e ∣ x^{2} + x + 1 ∣= x^{2} + x + 1,$ $\int_{2}^{3} ∣ x^{2} + x + 1 ∣ d x$ $= \int_{2}^{3} (x^{2} + x + 1) d x$ $= {[\frac{x^{3}}{3} + \frac{x^{2}}{2} + x]}_{2}^{3}$ $= [\frac{3^{3}}{3} + \frac{3^{2}}{2} + 3] - [\frac{2^{3}}{3} + \frac{2^{2}}{2} + 2]$ $= \frac{27 - 8}{3} + \frac{9 - 4}{2} + 3 - 2$ $= \frac{19}{3} + \frac{5}{2} + 1$ $= \frac{19 \times 2 + 5 \times 3 + 6}{6}$ $= \frac{59}{6}$
	Commented by Rio Michael last updated on 17/Aug/19

	$t h a n k y o u$
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