Question

Let $ f : [1, \infty) \to [2, \infty) $ be a differentiable function, If $ \int_{1}^{x} f(t) \, dt = 5x f(x) - x^5 - 9 $ for all $ x \geq 1 $, then the value of $ f(3) $ is :

• A. \) is :
• B. 18
• C. 32
• D. 22

Hint:

For this type of problem, use the properties of definite integrals and apply the fundamental theorem of calculus along with the chain rule. Simplify the equation step by step for easy solution.

Answer 1

The Correct Option is B

\[ 10 \frac{d}{dx} \int_1^x f(t) \, dt = \frac{d}{dx} \left( 5x f(x) - x^5 - 9 \right) \] \[ \Rightarrow 10 f(x) = 5f(x) + 5x f'(x) - 5x^4 \] \[ \Rightarrow f(x) + x^4 = x f'(x) \] \[ \Rightarrow \frac{dy}{dx} + y \left( \frac{1}{x} \right) = x^3 \] \[ \Rightarrow \frac{dy}{dx} = x^3 - y \left( \frac{1}{x} \right) \] \[ \Rightarrow y e^{-\frac{1}{x}} = \int x^3 e^{\frac{1}{x}} \, dx + c \] \[ \Rightarrow y = \frac{y}{x} \cdot \int x^3 \, dx \] \[ \Rightarrow y = \frac{x^3}{3} + c \] \[ \Rightarrow y = \frac{x^3}{3} + c \] \[ \text{Put } x = 1 \text{ in the given equation:} \] \[ 0 = 5f(1) - 9 - 9 \] \[ f(1) = 2 \Rightarrow c = \frac{5}{3} \] \[ f(3) = \frac{27}{3} + \frac{5}{3} \] \[ f(3) = 32 \]