76.
77.
78.
79.
80.
(D) Note the initial conditions: when t = 0, v = 0 and s = 0. Integrate
twice: v = 6t^2 and s = 2t^3 . Let t = 3.
(D) Since y′ = x^2 − 2, y = x^3 − 2x + C. Replacing x by 1 and y by −3
yields C = − .
(D) When t = 0, v = 3 and s = 2, so
v = 2t + 3 t^2 + 3 and s = t^2 + t^3 + 3 t + 2.
Let t = 1.
(C) Let ; then
v = at + C. (*)
Since v = 75 when t = 0, therefore C = 75. Then (*) becomes
v = at + 75
so
0 = at + 75 and a = −15.
(A) Divide to obtain . Use partial fractions to get
.
6 Definite Integrals
