Question

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

Answer 1

Answer:

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)

Answer 2

The value will be "$f'(\pi) =(-\frac{5}{3} )$".

Given expression is:

→ $\int x_0 (3f(t)+5t) dt = sin(x)$

or,

→ $3\times \int x_0 f(t) dt+5\times \intx_0 t\times dt = sin(x)$

then,

→ $\int x_0 t\times dt = (\frac{t^2}{2} ) x_0$

      $\frac{x^2}{2} -\frac{0^2}{2} = \frac{x^2}{2}$

thus,

→ $3\times \int x_0 f(t) +5\times \frac{x^2}{2} = sin(x)$

  $\int x_0 f(t) dt = \frac{1}{3}\times sin(x) - \frac{5}{6}\times x^2$

By applying differentiation, we get

→ $\frac{d}{dx} (\int x_0 f(t) dt) = f(x) - f(0)$

  $\frac{d}{dx}(\frac{1}{3} sin(x)) - \frac{5}{6}x^2= \frac{1}{3} cos(x) - \frac{5}{3} x$

Therefore,

→ $f(x) -f(0) = \frac{1}{3}cos (x) - \frac{5}{3} x$

  $f(x) = \frac{1}{3}cos (x) -\frac{5}{3} x+f(0)$

By differentiating again, we get

→ $f'(x) = -\frac{1}{3}sin(\pi) -\frac{5}{3}$

           $= 0-\frac{5}{3}$

           $= -\frac{5}{3}$

Thus the response above is correct.

Learn more about differentiation here:

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