JEE Main 2025 — Definite Integration Question with Solution

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

$10{\int }_1^{x}f(t)dt=5xf(x)-x^5-9$

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

$\frac{d}{dx}\left(10{\int }_1^{x}f(t)dt\right)=\frac{d}{dx}(5xf(x)-x^5-9)$

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

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

Rearranging terms, we obtain:

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

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

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

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:

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

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:

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

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

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.