Given f''(x) = -36sin(x) and f'(0)=4, and f(0)=1. Find f(pi/5).?

Answer 1

You need to take an integral for this problem. You're give f'(0) and f(0) to account for the constant of integration after each step. Here we go:

f"(x) = -36sin(x)
f'(x) = ∫ -36sin(x) dx
f'(x) = 36cos(x) + C

Plugging in 0 to find C:
4 = 36cos(0) + C
4 = 36 + C
C = -32

So,

f'(x) = 36cos(x) - 32
f(x) = ∫ (36cos(x) - 32) dx
f(x) = 36sin(x) - 32x + C

Again, plugging in 0:

1 = 36sin(0) - 32(0) + C
1 = 0 - 0 + C
C = 1

Thus,

f(x) = 36sin(x) - 32x + 1
f(pi/5) = 36sin(pi/5) + 32pi/5 + 1

Leave it this way for an exact answer, or plug it in to a calculator if you need a decimal estimate.