Definite Integration
Integral Equation — Standard Parabola
nta_pyq_2026_jan
Grade 12
Question:
Let a differentiable function $f$ satisfy the equation $\displaystyle\int_0^{36}f\!\left(\frac{tx}{36}\right)dt=4\alpha f(x)$. If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4,\beta)$, then $\beta^\alpha$ is equal to _____
Step-by-Step Solution
Key Concept: Substitute $u=tx/36$: $\frac{36}{x}\int_0^x f(u)\,du=4\alpha f(x)$. Differentiate both sides w.r.t. $x$: $f(x)=\frac{\alpha}{9}[f(x)+xf'(x)]$, giving $(9-\alpha)f(x)=\alpha xf'(x)$.
$\alpha=3$, $f(x)=x^2/4$, $\beta=4$. $\beta^\alpha=64$.
Correct Answer: 64