If the Integral of t·f(t) From 0 to x Equals x²·f(x) and f(2) = 3, Then f(6) Equals

JEE Main 2025 — 28 January, Morning Shift. Previous Year Question.

Problem

If $\displaystyle\int_0^x t\cdot f(t)\,dt = x^2\cdot f(x)$ and $f(2)=3$, then $f(6)$ equals:

Key insight. An equation with $f(x)$ trapped inside a definite integral on one side isn’t directly solvable — but differentiating both sides with respect to $x$ turns it into an ordinary differential equation in $f(x)$ alone, which is solvable by separating variables.

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Approach

Differentiate both sides of the given equation with respect to $x$. The left side becomes $x\cdot f(x)$ by the Fundamental Theorem of Calculus (the derivative of $\int_0^x g(t)\,dt$ is just $g(x)$); the right side needs the product rule. The result is a separable differential equation relating $f'(x)$ and $f(x)$, which integrates to give $f(x)$ explicitly — the given value $f(2)=3$ then fixes the constant of integration, and $f(6)$ follows by direct substitution.

Solution

Step 1 — Differentiate both sides

$$\frac{d}{dx}\int_0^x t\cdot f(t)\,dt = \frac{d}{dx}\left[x^2f(x)\right]$$

The left side, by the Fundamental Theorem of Calculus, is simply the integrand evaluated at $x$:

$$x\cdot f(x)$$

The right side, by the product rule:

$$2x\cdot f(x)+x^2f'(x)$$

So:

$$x\,f(x) = 2x\,f(x)+x^2f'(x)$$

Step 2 — Simplify into a separable equation

$$x\,f(x)-2x\,f(x) = x^2f'(x) \implies -x\,f(x) = x^2f'(x)$$

Dividing both sides by $x$ (valid since $x\ne0$):

$$-f(x) = x\,f'(x) \implies \frac{f'(x)}{f(x)} = -\frac1x$$

Step 3 — Integrate both sides

$$\int\frac{f'(x)}{f(x)}\,dx = -\int\frac1x\,dx$$

$$\ln|f(x)| = -\ln|x|+C$$

Writing the constant as $\ln K$ for some $K>0$:

$$\ln|f(x)| = \ln\left|\frac Kx\right| \implies f(x) = \frac Kx$$

Step 4 — Use f(2)=3 to find K

$$f(2) = \frac K2 = 3 \implies K=6$$

So:

$$f(x) = \frac6x$$

Step 5 — Evaluate f(6)

$$f(6) = \frac66 = 1$$

Answer

Common mistakes

• Forgetting the Fundamental Theorem of Calculus for the left side. It’s tempting to try to differentiate $\int_0^x t\cdot f(t)\,dt$ using the product rule, as if it were an ordinary product — but the integral evaluates the integrand only at the upper limit $x$, giving simply $x\cdot f(x)$, not something involving $f'(x)$.
• Forgetting the product rule on the right side. $x^2f(x)$ is a product of $x^2$ and $f(x)$, so differentiating it needs both terms $2xf(x)$ and $x^2f'(x)$ — dropping either term changes the resulting differential equation entirely.
• Losing the negative sign while separating variables. Since $-f(x)=xf'(x)$ rearranges to $\dfrac{f'(x)}{f(x)}=-\dfrac1x$, a dropped minus sign here flips $f(x)$ from $K/x$ to $Kx$, which changes every downstream value.

Practise next

• If $\displaystyle\int_0^x t\cdot f(t)\,dt = x^2\cdot f(x)$ and $f(3)=5$, find $f(9)$.