Answer:
The complete equation is [tex]f(x) = \sin 4x -8\cdot x -2[/tex]. [tex]f(1) = -10.756[/tex]
Step-by-step explanation:
Let be [tex]f''(x) = -16\cdot \sin 4x[/tex], we need to determine the formula of [tex]f(x)[/tex] by integrating twice:
[tex]f'(x) = \int {(-16\cdot \sin 4x)} \, dx[/tex]
[tex]f'(x) = -16\int {\sin 4x} \, dx[/tex]
We apply the following algebraic substitution in expression above:
[tex]u = 4\cdot x[/tex] and [tex]du = 4\,dx[/tex]
[tex]f'(u) = -4\int {\sin u} \, du[/tex]
[tex]f'(u) = 4\cdot \cos u + C_{1}[/tex]
[tex]f'(x) = 4\cdot \cos 4x + C_{1}[/tex]
We use the same approach to determine [tex]f(x)[/tex]:
[tex]f(x) = \int {(4\cdot \cos 4x)} \, dx + \int {C_{1}} \, dx[/tex]
[tex]f(u, x) = \int {\cos u} \, du + C_{1}\int \, dx[/tex]
[tex]f(u,x) = \sin u + C_{1}\cdot x + C_{2}[/tex]
[tex]f(x) = \sin 4x + C_{1}\cdot x + C_{2}[/tex]
If we know that [tex]f'(0) = -4[/tex] and [tex]f(0) = -2[/tex], the integration constants are obtained below:
[tex]4 + C_{1} = -4[/tex]
[tex]C_{1} = -8[/tex]
[tex]C_{2} = -2[/tex]
The complete equation is [tex]f(x) = \sin 4x -8\cdot x -2[/tex]. (Angles are measured in radians) Then:
[tex]f(1) = \sin 4 - 8- 2[/tex]
[tex]f(1) = -0.756-8-2[/tex]
[tex]f(1) = -10.756[/tex]
