Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
If $a \leq \int_0^1 \frac{dx}{\sqrt{4-x^2-x^3}} \leq b$, then $(a,b) =$
\left(\frac{\pi}{6}, \frac{\pi}{4}\right)
\left(\frac{\pi}{6}, \frac{\pi}{4\sqrt{2}}\right)
\left(\frac{\pi}{4\sqrt{2}}, \frac{\pi}{2\sqrt{2}}\right)
None of these
Step-by-Step Solution
Key Concept: Sandwich theorem applied to integrals by establishing inequality chains that hold pointwise over the domain.
We establish that $4 - x^2 ≥ 4 - x^2 - x^3 ≥ 4 - 2x^2$ for all $x ∈ [0,1]$. Taking square roots and reciprocals: $\frac{1}{\sqrt{4-x^2}} ≤ \frac{1}{\sqrt{4-x^2-x^3}} ≤ \frac{1}{\sqrt{4-2x^2}}$. Integrating from 0 to 1 gives bounds on the middle integral using standard forms.
Correct Answer: 2