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

Given data 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=12y12−1dt dy=12tdt A=−∫02(t2−2t)dt2t=−12∫02(t2t−12−2tt−12)dt =−12∫−02(t32−2t12)dt =−12[t32+132+1−2t12+112+1]02 =−12[t5252−2t3232]02 =−12[25(2)52−43(2)32−0] =−12[6(2)52−20(2)3215] =−130[(2)32[6(2)−20]]

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Jessica Scott

Answered question

2021-11-19

Find the area enclosed by the curve

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

and the y-axis

Answer & Explanation

Tamara Donohue

Beginner2021-11-20Added 11 answers

Given data
$x = t^{2} - 2 t, y = \sqrt{t}$ , y -axis
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 $y = \sqrt{t} = t^{\frac{1}{2}}$
$d y = \frac{1}{2} y^{\frac{1}{2} - 1} d t$
$d y = \frac{1}{2 \sqrt{t}} d t$
$A = - \int_{0}^{2} \frac{(t^{2} - 2 t) d t}{2 \sqrt{t}} = - \frac{1}{2} \int_{0}^{2} (t^{2} t^{- \frac{1}{2}} - 2 t t^{- \frac{1}{2}}) d t$
$= - \frac{1}{2} \int - 0^{2} (t^{\frac{3}{2}} - 2 t^{\frac{1}{2}}) d t$
$= - \frac{1}{2} {[\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} - \frac{2 t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}]}_{0}^{2}$
$= - \frac{1}{2} {[\frac{t^{\frac{5}{2}}}{\frac{5}{2}} - 2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{2}$
$= - \frac{1}{2} [\frac{2}{5} {(2)}^{\frac{5}{2}} - \frac{4}{3} {(2)}^{\frac{3}{2}} - 0]$
$= - \frac{1}{2} [\frac{6 {(2)}^{\frac{5}{2}} - 20 {(2)}^{\frac{3}{2}}}{15}]$
$= - \frac{1}{30} [{(2)}^{\frac{3}{2}} [6 (2) - 20]]$

James Kilian

Beginner2021-11-21Added 20 answers

I can not solve this example, please, if someone knows, then tell me the correct answer

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