Find the area of the region enclosed by x(t)=t^2-2t, y(t)=sqrt(t), and the y-axis.

First find the the value of t where the curve intersects the Y-axis. This is when x = 0. 

x = t^2 - 2t = 0 = t(t - 2) 

So t= 0 and t = 2 

dA = (0 - x)*dy Since the curve has negative x in this region 

y = SQRT(t) and dy = [(1/2)/SQRT(t)]dt 

dA = [2t - t^2][(1/2)/SQRT(t)]dt 

dA = [t^(1/2) - (1/2)t^(3/2)]dt 

Integrate to get: A = (2/3)t^(3/2) - (1/5)t^(5/2) 

Now evaluate from t= 0 to t = 2. 

Area = [(2/3)2^(3/2) - (1/5)2^(5/2)] - [0] 

Area = SQRT(2)[4/3 - 4/5] 

Area = SQRT(2)[8/15) = 0.754

I hope my answer has come to your help. Thank you for posting your question here in . We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

(1/x + 1)^2 + 6 (1/x + 1) +5

6/ 11 = 0.545454 so your answer is55%