The function f is differentiable and ∫x0 (3f (t) + 5t) dt=sin (x). Determine the value of f′ (1π).

f' (π) = (-5/3)

Step-by-step explanation:

for the equation

∫₀ˣ (3f (t) + 5t) dt=sin (x)

3*∫₀ˣ f (t) dt+5*∫₀ˣ t*dt=sin (x)

then

∫₀ˣ t*dt = (t²/2) |₀ˣ = x²/2 - 0²/2 = x²/2

thus

3*∫₀ˣ f (t) dt + 5 * x²/2 = sin (x)

∫₀ˣ f (t) dt = 1/3*sin (x) - 5/6*x²

then applying differentiation

d/dx (∫₀ˣ f (t) dt) = f (x) - f (0) (from the fundamental theorem of calculus)

d/dx (1/3*sin (x) - 5/6*x²) = 1/3*cos (x) - 5/3*x

therefore

f (x) - f (0) = 1/3*cos (x) - 5/3*x

f (x) = 1/3*cos (x) - 5/3*x + f (0)

applying differentiation again (f' (x) = df (x) / dx)

f' (x) = - 1/3*sin (x) - 5/3

then

f' (π) = - 1/3*sin (π) - 5/3 = 0 - 5/3 = - 5/3

f' (π) = (-5/3)