Find the area enclosed by the curve x=t2−2t, y=t and the y−aξs

x=t2−2t, y=t, y−axis The curve will intersect y-axis when x=0 t2−2t=0 t(t−2))=0 t=0,2 t=0⇒x=0, y=0 t=2⇒x=4−4=0, y=2 Area A=∫−xdy where y=t=t12 dy=12t1/2dt dy=12tdt A=∫02(t2−2tdt)2t=−12∫02(t2t−1/2−2tt1/2)dt =−12∫02(t3/2−2t1/2)dt =−12[t32+132+1−2t12+112+1] =−12[t52+132+1−2t3232]02 =−12[25(2)52−43(2)32−0] =−12[6(2)52−20(2)3215] =−130[(2)32[6(2)−20]] =−130[22(−8)] Area

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hvacwk

Answered question

2021-12-11

Find the area enclosed by the curve

x = t^{2} - 2 t, y = \sqrt{t}

and the

y - a ξ s

Answer & Explanation

vrangett

Beginner2021-12-12Added 36 answers

Any solutions?

Elaine Verrett

Beginner2021-12-13Added 41 answers

$x = t^{2} - 2 t, y = \sqrt{t}, y - a x i s$
The curve will intersect y-axis when $x = 0$
$t^{2} - 2 t = 0$
$t (t - 2)) = 0$
$t = 0, 2$
$t = 0 \Rightarrow x = 0, y = 0$
$t = 2 \Rightarrow x = 4 - 4 = 0, y = \sqrt{2}$
Area $A = \int - x d y$
where $y = \sqrt{t} = t^{\frac{1}{2}}$
$d y = \frac{1}{2} t^{1 / 2} d t$
$d y = \frac{1}{2 \sqrt{t}} d t$
$A = \int_{0}^{2} \frac{(t^{2} - 2 t d t)}{2 \sqrt{t}} = - \frac{1}{2} \int_{0}^{2} (t^{2} t^{- 1 / 2} - 2 t t^{1 / 2}) d t$
$= - \frac{1}{2} \int_{0}^{2} (t^{3 / 2} - 2 t^{1 / 2}) d t$
$= - \frac{1}{2} [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} - 2 \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}]$
$= - \frac{1}{2} {[\frac{t^{\frac{5}{2} + 1}}{\frac{3}{2} + 1} - 2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{2}$
$= - \frac{1}{2} [\frac{2}{5} {(2)}^{\frac{5}{2}} - \frac{4}{3} {(2)}^{\frac{3}{2}} - 0]$
$= - \frac{1}{2} [\frac{6 {(2)}^{\frac{5}{2}} - 20 {(2)}^{\frac{3}{2}}}{15}]$
$= - \frac{1}{30} [{(2)}^{\frac{3}{2}} [6 (2) - 20]]$
$= - \frac{1}{30} [2 \sqrt{2} (- 8)]$
Area

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