Let f:[1, ) [2, ) be a differentiable function.

Question

Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_1^1 f(\mathrm{t}) \mathrm{dt}=5 x f(x)-x^5-9$ for all $x \geqslant 1$, then the value of $f(3)$ is :

Accepted Answer

To solve the problem, we start with the given equation:

$ 10 \int_1^x f(t) \, dt = 5x f(x) - x^5 - 9 $

By differentiating both sides with respect to $ x $, we have:

$ \frac{d}{dx}\left(10 \int_1^x f(t) \, dtight) = \frac{d}{dx}(5x f(x) - x^5 - 9) $

The left side simplifies to $ 10 f(x) $. For the right side, using the product rule and the power rule, we get:

$ 10 f(x) = 5f(x) + 5x \frac{d}{dx} f(x) - 5x^4 $

Rearranging terms, we obtain:

$ 5 f(x) = 5x \frac{d}{dx} f(x) - 5x^4 $

Let $ y = f(x) $. Thus:

$ 5 y = 5x \frac{dy}{dx} - 5x^4 $

Dividing by 5, we have:

$ y = x \frac{dy}{dx} - x^4 $

Rewriting, we get:

$ \frac{dy}{dx} - \frac{y}{x} = x^3 $

This is a linear differential equation. The integrating factor (I.F.) is calculated as:

$ 	ext{I.F.} = e^{\int \frac{-1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} $

Multiplying through by the integrating factor, we have:

$ y \cdot \frac{1}{x} = \int x^3 \cdot \frac{1}{x} \, dx = \int x^2 \, dx = \frac{x^3}{3} + C $

Thus, we solve for $ y $:

$ y = \frac{x^4}{3} + Cx $

Substituting back, we need to find $ C $ using the condition at $ x = 1 $:

Since $ \int_1^1 f(t) \, dt = 0 $, we substitute:

$ 10 \cdot 0 = 5 \cdot 1 \cdot f(1) - 1^5 - 9 \quad \Rightarrow \quad 0 = 5f(1) - 1 - 9  $

$ 5f(1) = 10 \quad \Rightarrow \quad f(1) = 2 $

Now use $ f(1) = 2 $ in the function:

$ 2 = \frac{1}{3} + C \cdot 1 \quad \Rightarrow \quad C = \frac{5}{3} $

The function $ f(x) $ is given by:

$ f(x) = \frac{x^4}{3} + \frac{5}{3}x $

To find $ f(3) $:

$ f(3) = \frac{3^4}{3} + \frac{5}{3} \cdot 3 = 27 + 5 = 32 $

Therefore, the value of $ f(3) $ is 32.

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