Integral Calculus-2
Integral Calculus-2
Allen Star Batch
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: Establish inequalities for the integrand by bounding $4-x^2-x^3$ between $4-x^2$ and $4-2x^2$ on $[0,1]$, then use reciprocal inequality reversal and integrate using standard arctangent formulas $\int \frac{dx}{\sqrt{a^2-x^2}} = \arcsin(x/a)$.
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