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]
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Area = SQRT(2)[4/3 - 4/5]
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Area = SQRT(2)[8/15) = 0.754
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Answer:
no
Step-by-step explanation:
Answer:
(-4, 5)
Step-by-step explanation:
x will never change if reflected over it. but y will be negative. also what do you mean by y coordinate or x coordinate?
