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

Question

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

Accepted Answer

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

$10{\int }_{x}^{1}f(t)\, dt=5xf(x)-{x}^{5}-9$

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

$\frac{d}{dx}(10{\int }_{x}^{1}f(t)\, dt)=\frac{d}{dx}(5xf(x)-{x}^{5}-9)$

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

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

Rearranging terms, we obtain:

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

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

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

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:

$={e}^{\int \frac{-1}{x}\, dx}={e}^{-\ln x}=\frac{1}{x}$

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:

$10\cdot 0=5\cdot 1\cdot f(1)-{1}^{5}-9\, \Rightarrow \, 0=5f(1)-1-9$

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

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

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

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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