Question

Let \( f(x) \) be differentiable and 

\[ \int_{0}^{t^2} x f(x)\,dx = \frac{1}{2} t^4 \quad \text{for all } t. \]

Then the value of \( f(17) \) is ______.

• A. \(17\)
• B. \(1\)
• C. \(1/17\)
• D. \(17/2\)
• E. \(19\)

Hint:

Always multiply by derivative of upper limit when differentiating integrals.

Answer 1

The Correct Option is B

Concept: Use differentiation of integrals with variable upper limit (Leibniz rule).

Step 1: Differentiate both sides w.r.t \(t\). \[ \frac{d}{dt}\left(\int_0^{t^2} x f(x)\,dx\right) = \frac{d}{dt}\left(\frac{1}{2}t^4\right) \]

Step 2: Apply chain rule to LHS. \[ = (t^2 f(t^2)) \cdot \frac{d}{dt}(t^2) \] \[ = t^2 f(t^2) \cdot 2t \] \[ = 2t^3 f(t^2) \] RHS: \[ = 2t^3 \]

Step 3: Equate both sides. \[ 2t^3 f(t^2) = 2t^3 \]

Step 4: Simplify. \[ f(t^2) = 1 \]

Step 5: Substitute value. \[ t^2 = 17 \Rightarrow f(17) = 1 \]