∫_(−3) ^5 (√(∣x∣^3 ))dx


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Question Number 156281 by SANOGO last updated on 09/Oct/21

$\int_{- 3}^{5} \sqrt{∣ x ∣^{3}} d x$
	Answered by MathsFan last updated on 09/Oct/21

	$28.596$
	Commented by SANOGO last updated on 09/Oct/21

	$l a d e m o n s t r a t i o n s t p$
	Answered by MJS_new last updated on 10/Oct/21
	$(√(∣x∣^3 ))= { (((−x)^(3/2) ; x<0)),((x^(3/2) ; x≥0)) :} ⇒ ∫_(−3) ^5 (√(∣x∣^3 ))dx=∫_0 ^3 x^(3/2) dx+∫_0 ^5 x^(3/2) dx=i [∫x^(3/2) dx=(2/5)x^(5/2) +C] =(2/5)[x^(5/2) ]_0 ^3 +(2/5)[x^(5/2) ]_0 ^5 =10(√5)+((18(√3))/5)$
	$\sqrt{∣ x ∣^{3}} = {\begin{cases} {(- x)}^{3 / 2}; x < 0 \\ x^{3 / 2}; x ⩾ 0 \end{cases} \Rightarrow \underset{- 3}{\int^{5}} \sqrt{∣ x ∣^{3}} d x = \underset{0}{\int^{3}} x^{3 / 2} d x + \underset{0}{\int^{5}} x^{3 / 2} d x = i$ $[\int x^{3 / 2} d x = \frac{2}{5} x^{5 / 2} + C]$ $= \frac{2}{5} {[x^{5 / 2}]}_{0}^{3} + \frac{2}{5} {[x^{5 / 2}]}_{0}^{5} = 10 \sqrt{5} + \frac{18 \sqrt{3}}{5}$
	Commented by SANOGO last updated on 10/Oct/21

	$m e r c i b i e n l e d u r$
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