Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12
Question:
Which of the following is/are true?
$\frac{\pi}{3\sqrt{3}} < \int_0^1 \frac{dx}{1+x^2+2x^5}$
$\int_0^1 \frac{dx}{1+x^2+2x^4} = \frac{\pi}{4}$
$\int_0^{\pi/2} \sin^3 x dx \leq \frac{1}{2}(\sqrt{2} + \ln(1+\sqrt{2}))$
$1 \leq \int_0^{\pi/2} \sqrt{1-\sin^3 x} dx$
Step-by-Step Solution
Key Concept: Bounding an integral by establishing inequalities on the integrand and integrating across all bounds.
Start by establishing the inequality $1 + x^2 < 1 + x^2 + 2x^3 < 1 + x^2$ is false; the correct inequality is $1 + 3x^2 < 1 + x^2 + 2x^3 < 1 + x^2$ for $0 ≤ x ≤ 1$. Taking reciprocals reverses inequalities: $\frac{1}{1+x^2} < \frac{1}{1+x^2+2x^3} < \frac{1}{1+3x^2}$. Integrate from 0 to 1: $\int_0^1 \frac{dx}{1+x^2} < \int_0^1 \frac{dx}{1+x^2+2x^3} < \int_0^1 \frac{dx}{1+3x^2}$, yielding $\frac{π}{4} < I < \frac{π}{4\sqrt{3}}$.
Correct Answer: 1,2,3,4