Question

If $I = \int_{0}^{1} \frac{dx}{\sqrt{4-x^2-x^3}}$ then

• A. $I < \frac{\pi}{6}$
• B. $I > \frac{\pi}{6}$
• C. $I < \frac{\pi}{4\sqrt{2}}$
• D. $I > \frac{\pi}{4}$

Answer 1

Answer: (B), (C)

B, C

Answer 2

To evaluate the integral $I = \int_{0}^{1} \frac{dx}{\sqrt{4-x^2-x^3}}$ and compare its value with the given options, we will use the comparison properties of definite integrals.

Let $f(x) = \frac{1}{\sqrt{4-x^2-x^3}}$. We need to find suitable functions $g(x)$ and $h(x)$ such that $g(x) \le f(x) \le h(x)$ for $x \in [0, 1]$, and $\int g(x) dx$ and $\int h(x) dx$ are easy to evaluate.

1. Finding a Lower Bound for I:

For $x \in [0, 1]$, we have $x^3 \ge 0$. Therefore, $4-x^2-x^3 \le 4-x^2$. Taking the square root, $\sqrt{4-x^2-x^3} \le \sqrt{4-x^2}$. Taking the reciprocal, $\frac{1}{\sqrt{4-x^2-x^3}} \ge \frac{1}{\sqrt{4-x^2}}$. Since this inequality holds for all $x \in [0, 1]$, we can integrate both sides: $I = \int_{0}^{1} \frac{dx}{\sqrt{4-x^2-x^3}} \ge \int_{0}^{1} \frac{dx}{\sqrt{4-x^2}}$.

Let's evaluate the integral on the right side: $\int_{0}^{1} \frac{dx}{\sqrt{2^2-x^2}} = \left[\arcsin\left(\frac{x}{2}\right)\right]_{0}^{1}$ $= \arcsin\left(\frac{1}{2}\right) - \arcsin(0)$ $= \frac{\pi}{6} - 0 = \frac{\pi}{6}$.

So, we have $I \ge \frac{\pi}{6}$. Furthermore, since $x^3 > 0$ for $x \in (0, 1]$, the inequality $4-x^2-x^3 < 4-x^2$ is strict for $x \in (0, 1]$. This implies $\frac{1}{\sqrt{4-x^2-x^3}} > \frac{1}{\sqrt{4-x^2}}$ for $x \in (0, 1]$. Therefore, $I = \int_{0}^{1} \frac{dx}{\sqrt{4-x^2-x^3}} > \int_{0}^{1} \frac{dx}{\sqrt{4-x^2}} = \frac{\pi}{6}$. So, $I > \frac{\pi}{6}$.

This immediately rules out option A ($I < \frac{\pi}{6}$). Option B ($I > \frac{\pi}{6}$) is consistent with this result.

2. Finding an Upper Bound for I:

For $x \in [0, 1]$, we have $x^3 \le x^2$. Therefore, $-x^3 \ge -x^2$. Adding $4-x^2$ to both sides: $4-x^2-x^3 \ge 4-x^2-x^2 = 4-2x^2$. Taking the square root, $\sqrt{4-x^2-x^3} \ge \sqrt{4-2x^2}$. Taking the reciprocal, $\frac{1}{\sqrt{4-x^2-x^3}} \le \frac{1}{\sqrt{4-2x^2}}$. Since this inequality holds for all $x \in [0, 1]$, we can integrate both sides: $I = \int_{0}^{1} \frac{dx}{\sqrt{4-x^2-x^3}} \le \int_{0}^{1} \frac{dx}{\sqrt{4-2x^2}}$.

Let's evaluate the integral on the right side: $\int_{0}^{1} \frac{dx}{\sqrt{4-2x^2}} = \int_{0}^{1} \frac{dx}{\sqrt{2(2-x^2)}} = \frac{1}{\sqrt{2}} \int_{0}^{1} \frac{dx}{\sqrt{(\sqrt{2})^2-x^2}}$ $= \frac{1}{\sqrt{2}} \left[\arcsin\left(\frac{x}{\sqrt{2}}\right)\right]_{0}^{1}$ $= \frac{1}{\sqrt{2}} \left(\arcsin\left(\frac{1}{\sqrt{2}}\right) - \arcsin(0)\right)$ $= \frac{1}{\sqrt{2}} \left(\frac{\pi}{4} - 0\right) = \frac{\pi}{4\sqrt{2}}$.

So, we have $I \le \frac{\pi}{4\sqrt{2}}$. Furthermore, since $x^3 < x^2$ for $x \in (0, 1)$, the inequality $4-x^2-x^3 > 4-2x^2$ is strict for $x \in (0, 1)$ (except at $x=0$ and $x=1$). This implies $\frac{1}{\sqrt{4-x^2-x^3}} < \frac{1}{\sqrt{4-2x^2}}$ for $x \in (0, 1)$. Therefore, $I = \int_{0}^{1} \frac{dx}{\sqrt{4-x^2-x^3}} < \int_{0}^{1} \frac{dx}{\sqrt{4-2x^2}} = \frac{\pi}{4\sqrt{2}}$. So, $I < \frac{\pi}{4\sqrt{2}}$.

3. Combining the Bounds and Checking Options:

We have established that: $\frac{\pi}{6} < I < \frac{\pi}{4\sqrt{2}}$.

Let's approximate the values: $\frac{\pi}{6} \approx \frac{3.14159}{6} \approx 0.5236$ $\frac{\pi}{4\sqrt{2}} = \frac{\pi\sqrt{2}}{8} \approx \frac{3.14159 \times 1.41421}{8} \approx \frac{4.4428}{8} \approx 0.55535$

So, $0.5236 < I < 0.55535$.

Now let's check the given options: A. $I < \frac{\pi}{6}$. This is false because $I > \frac{\pi}{6}$. B. $I > \frac{\pi}{6}$. This is true based on our derivation. C. $I < \frac{\pi}{4\sqrt{2}}$. This is true based on our derivation. D. $I > \frac{\pi}{4}$. $\frac{\pi}{4} \approx 0.785$. This is false because $I < 0.55535$.

Since both B and C are true statements based on our derived bounds, the question has multiple correct options.