Two mathematical problems about anti-missile function

F' (t) = ∫ f "(t) * dt = ∫ (4 * e t+8 * Sint) dt = 4 * e t-8 * cost+C 1

F (t) = ∫ f' (t) * dt = ∫ (4 * e t-8 * cost+c1) * dt = 4 * e t-8 * Sinter+C 1*t+C2.

When t = 0, f (0) = 4 * E0-8 * SIN0+C1* 0+C2 = 4+C2 = 0, then C2 =-4.

When t = π, f (π) = 4 * eπ-8 * sinπ+c1* π+C2 = 4 * eπ+c1* π-4 = 0, then c 1 = (4-4 * e

So f (t) = 4 * Et-8 * Sint+(4-4 * eπ) * t/π-4.

F"(x) =∫f'"(x)dx = sinx+C 1, when x = 0, f"(0) = 0+C 1=6, then C 1=6, and f "(.

F' (x) = ∫ f "(x) dx =-cosx+6x+C2, when x =0, f'(0)=- 1+6 * 0+C2 = 8, then C2 = 9.

f'(x) = -cosx + 6x + 9

f(x)=∫f '(x)dx =-sinx+3x ^ 2+9x+C3。 When x= 0, f(0)= -0+3*0+9*0+C3=9, then C3=9.

f(x) = -sinx + 3x^2 + 9x + 9