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y-x-3-y-0-and-x-a-Find-a-if-the-area-of-the-figure-bounded-by-straight-lines-is-64-




Question Number 147979 by mathdanisur last updated on 24/Jul/21
y = x^3   ;  y = 0  and  x = a  Find  a  if the area of the figure  bounded by straight lines is  64
$${y}\:=\:{x}^{\mathrm{3}} \:\:;\:\:{y}\:=\:\mathrm{0}\:\:{and}\:\:{x}\:=\:{a} \\ $$$${Find}\:\:\boldsymbol{{a}}\:\:{if}\:{the}\:{area}\:{of}\:{the}\:{figure} \\ $$$${bounded}\:{by}\:{straight}\:{lines}\:{is}\:\:\mathrm{64} \\ $$
Answered by iloveisrael last updated on 25/Jul/21
If a>0 ⇒64=∫_0 ^( a)  x^3  dx   ⇒256 = a^4  ; a=4  If a<0⇒64= ∫_a ^( 0) − x^3  dx   ⇒256 = a^4  ; a=−4  ∴ a = ± 4
$$\mathrm{If}\:\mathrm{a}>\mathrm{0}\:\Rightarrow\mathrm{64}=\int_{\mathrm{0}} ^{\:\mathrm{a}} \:\mathrm{x}^{\mathrm{3}} \:\mathrm{dx}\: \\ $$$$\Rightarrow\mathrm{256}\:=\:\mathrm{a}^{\mathrm{4}} \:;\:\mathrm{a}=\mathrm{4} \\ $$$$\mathrm{If}\:\mathrm{a}<\mathrm{0}\Rightarrow\mathrm{64}=\:\int_{\mathrm{a}} ^{\:\mathrm{0}} −\:\mathrm{x}^{\mathrm{3}} \:\mathrm{dx}\: \\ $$$$\Rightarrow\mathrm{256}\:=\:\mathrm{a}^{\mathrm{4}} \:;\:\mathrm{a}=−\mathrm{4} \\ $$$$\therefore\:\mathrm{a}\:=\:\pm\:\mathrm{4}\: \\ $$
Commented by mathdanisur last updated on 25/Jul/21
Thankyou Ser
$${Thankyou}\:{Ser} \\ $$

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