Answer:
(8√2) / 15
Step-by-step explanation:
A curve bounded by the y-axis is represented by in terms of dy;
[tex]\int \:x\:dt[/tex]
When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;
0 = t^2 + 2t --- (1)
Solution(s) => t = 0, t = 2
dy = (1/2 * 1/√t)dt --- (2)
Our solutions (0, 2) are our limits. The area of the curve is in the form [tex]A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt[/tex] , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....
[tex]A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}[/tex]
Your solution is 8√2 / 15
