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Question Number 146784 by tabata last updated on 15/Jul/21

Commented by tabata last updated on 15/Jul/21

$? ? ? ? ? ? ?$
Commented by tabata last updated on 15/Jul/21

$h e l p m e p l e a s e$
	Answered by Cyriille last updated on 15/Jul/21
	$Q2\ y=(√x) y=2x−1, x=0 at point of intersecrion, (√x) = 2x−1 ⇒x=(2x−1)^2 ⇒x=4x^2 −4x+1 ⇒4x^2 −5x+1=0 ⇒x=1 or x=(1/4) at x=1, y=1 at x=(1/4), y=-(1/2) y=(√x) ⇒x=y^2 y=2x−1 ⇒x=((y+1)/2) A=∫_(-(1/2)) ^1 y^2 dy −∫_(-(1/2)) ^1 ((y+1)/2)dy ⇒[(y^3 /3)]_(-(1/2)) ^1 −[(y^2 /4)+(y/2)]_(-(1/2)) ^1 ⇒[(1/3)−((((−(1/2))^3 )/3))]−[((1/4)+(1/2))−((((-(1/2))^2 )/4)+((-(1/2))/2))] ⇒(3/8)−((15)/(16))=-(9/(16))$
	$Q 2 ∖$ $y = \sqrt{x} y = 2 x - 1, x = 0$ $a t p o i n t o f i n t e r s e c r i o n,$ $\sqrt{x} = 2 x - 1$ $\Rightarrow x = {(2 x - 1)}^{2}$ $\Rightarrow x = 4 x^{2} - 4 x + 1$ $\Rightarrow 4 x^{2} - 5 x + 1 = 0$ $\Rightarrow x = 1 o r x = \frac{1}{4}$ $a t x = 1, y = 1$ $a t x = \frac{1}{4}, y = - \frac{1}{2}$ $y = \sqrt{x} \Rightarrow x = y^{2}$ $y = 2 x - 1 \Rightarrow x = \frac{y + 1}{2}$ $A = \int_{- \frac{1}{2}}^{1} y^{2} d y - \int_{- \frac{1}{2}}^{1} \frac{y + 1}{2} d y$ $\Rightarrow {[\frac{y^{3}}{3}]}_{- \frac{1}{2}}^{1} - {[\frac{y^{2}}{4} + \frac{y}{2}]}_{- \frac{1}{2}}^{1}$ $\Rightarrow [\frac{1}{3} - (\frac{{(- \frac{1}{2})}^{3}}{3})] - [(\frac{1}{4} + \frac{1}{2}) - (\frac{{(- \frac{1}{2})}^{2}}{4} + \frac{- \frac{1}{2}}{2})]$ $\Rightarrow \frac{3}{8} - \frac{15}{16} = - \frac{9}{16}$
	Commented by tabata last updated on 15/Jul/21

	$s i r h o w t h e a r e a i n n e g i t i v e$
	Commented by Cyriille last updated on 01/Oct/21

	$t a k e a b o s o l u t e v a l u e$
	Answered by Cyriille last updated on 15/Jul/21
	$Q4\ y=∫_1 ^x (√(t^2 −1))dt, A=∫_a ^b ydx y=∫_1 ^x (√(t^2 −1))dt ⇒y=∫_a ^b tanθ(tanθ)secθdθ =∫_a ^b ((sin^2 θ)/(cos^3 θ))dθ =∫_a ^b sinθ.((sinθ)/(cos^3 θ))dθ I can not continue but I hope my idea was correct!$
	$Q 4 ∖$ $y = \int_{1}^{x} \sqrt{t^{2} - 1} d t, A = \int_{a}^{b} y d x$ $y = \int_{1}^{x} \sqrt{t^{2} - 1} d t$ $\Rightarrow y = \int_{a}^{b} t a n θ (t a n θ) s e c θ d θ$ $= \int_{a}^{b} \frac{{s i n}^{2} θ}{{c o s}^{3} θ} d θ$ $= \int_{a}^{b} s i n θ . \frac{s i n θ}{{c o s}^{3} θ} d θ$ $I can not continue but I hope my idea was correct!$
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