when y=0, t = 0 or 1, so x = 8 or 19
That makes the area
∫[8,19] y dx = ∫[0,1] (t - t^2) (33t^2 dt) = 33/20
Find the area enclosed by the x-axis and the parametric curve x=11t^3+8,y=t−t^2.Enter the exact answer as a decimal, but do NOT round. Hint: First find the intercepts of the curve with the x-axis.Show Work
2 answers
or, since t = ((x-8)/11)^(1/3)
the area is
∫[8,19] y dx = ∫[8,19] (((x-8)/11)^(1/3) - ((x-8)/11)^(2/3)) dx = 33/20
the area is
∫[8,19] y dx = ∫[8,19] (((x-8)/11)^(1/3) - ((x-8)/11)^(2/3)) dx = 33/20
1 answer