(B) , so v = 3t^3 + c.
Since v = 1 when t = 0, c = 1.
Now , so s = t^3 + t + c.
Since s = 3 when t = 0, c = 3; then s = t^3 + t + 3.
(A) Let u = x^2 . Then
(C)
To find a, the point of intersection of y = x^2 and y = cos (x), use your
calculator to solve the equation x^2 − cos (x) = 0. (Store the value for later
use; a ≈ 0.8241.)
As shown in the diagram above, ΔA = (cos (x) − x^2 )Δx.
Evaluate the area: .
(D) If x = 2t + 1, then , so . When t = 0, x = 1; when t = 3,
x = 7.