Question Number 156281 by SANOGO last updated on 09/Oct/21
$$\int_{−\mathrm{3}} ^{\mathrm{5}} \:\sqrt{\mid{x}\mid^{\mathrm{3}} }{dx} \\ $$
Answered by MathsFan last updated on 09/Oct/21
$$\mathrm{28}.\mathrm{596} \\ $$
Commented by SANOGO last updated on 09/Oct/21
$${la}\:{demonstration}\:{stp} \\ $$
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)](Q156293.png)
$$\sqrt{\mid{x}\mid^{\mathrm{3}} }=\begin{cases}{\left(−{x}\right)^{\mathrm{3}/\mathrm{2}} ;\:{x}<\mathrm{0}}\\{{x}^{\mathrm{3}/\mathrm{2}} ;\:{x}\geqslant\mathrm{0}}\end{cases}\:\Rightarrow\:\underset{−\mathrm{3}} {\overset{\mathrm{5}} {\int}}\sqrt{\mid{x}\mid^{\mathrm{3}} }{dx}=\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{x}^{\mathrm{3}/\mathrm{2}} {dx}+\underset{\mathrm{0}} {\overset{\mathrm{5}} {\int}}{x}^{\mathrm{3}/\mathrm{2}} {dx}={i} \\ $$$$\:\:\:\:\:\left[\int{x}^{\mathrm{3}/\mathrm{2}} {dx}=\frac{\mathrm{2}}{\mathrm{5}}{x}^{\mathrm{5}/\mathrm{2}} +{C}\right] \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left[{x}^{\mathrm{5}/\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{5}}\left[{x}^{\mathrm{5}/\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{5}} =\mathrm{10}\sqrt{\mathrm{5}}+\frac{\mathrm{18}\sqrt{\mathrm{3}}}{\mathrm{5}} \\ $$
Commented by SANOGO last updated on 10/Oct/21
$${merci}\:{bien}\:{le}\:{dur} \\ $$