EXAMPLE 6 A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = -4 cm/s and its initial displacement is s(0) = 9 cm. Find its position function, s(t). SOLUTION Since v (t) = a(t) = 12t + 10, antidifferentiation gives V(t) = + 10t + C = + C. Note that v(0) = C. But we are given that v(0) = -4, 50 C = and V(t) = Since v(t) = s'(t), s is the antiderivative of v: 5(t) = 6 + 10 - 4t+ D = + D. This gives s(0) = D. We are given that s(0) = 9, so D = and the required position function is S(0) =EXAMPLE 4 Find fif f"(x) = 12x2 + 6x -4, f(0) = 5, and f(1) = 4. SOLUTION The general antiderivative of f"(x) = 12x~ + 6x -4 is f'(x) = 12X + 4x + C + C. Using the antidifferentiation rules once more, we find that f(x) = 4X + 3 X + CX + D. + CX + D. To determine C and D we use the given conditions that f(0) = 5 and ((1) = 4. Since f(0) = 0 + D = 5, we have D = Since f(1 ) = 1+1 -2+ C+ 4 we have C = Therefore, the required function is f( x ) =