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)

idk wbu

step-by-step explanation: