91Ó°ÊÓ

We want to find two functions, \(f(x)\) and \(g(x)\), such that \(f(x) \leq \frac{1}{1+x^{3/2}} \leq g(x)\). We can start by noticing that \(x^{3/2} \leq x^2\) for \(0 \leq x \leq 1\). This is because raising \(x\) to a higher power results in a smaller value when \(0\leq x\leq1\). Therefore, we have the inequality: $$1+x^{3/2} \geq 1+x^2$$ Now, dividing by the terms on the left and right sides, we get: $$\frac{1}{1+x^2} \leq \frac{1}{1+x^{3/2}}$$ Thus, we found our lower bounding function \(f(x) = \frac{1}{1+x^2}\). Now let's find an upper bounding function \(g(x)\). We know that for \(0\leq x\leq 1\), \(1+x^{3/2} \leq 1 + 1\) since \(x^{3/2}\leq 1\). Dividing by 2, we get: $$\frac{1}{1+x^{3/2}} \leq \frac{1}{2}$$ Thus, we found our upper bounding function \(g(x) = \frac{1}{2}\). In summary, we have the following bounds on our integrand: $$\frac{1}{1+x^2} \leq \frac{1}{1+x^{3/2}} \leq \frac{1}{2}$$

Now that we have our bounding functions, \(f(x)\) and \(g(x)\), we need to integrate them over the interval \([0,1]\): $$\int_{0}^{1} f(x) dx = \int_{0}^{1} \frac{1}{1+x^2} dx$$ Using the substitution \(u = \arctan(x)\), we get: $$\int_{0}^{1} f(x) dx = \int_{0}^{\arctan(1)} du = \left[u\right]_{0}^{\arctan(1)} = \arctan(1) = \frac{\pi}{4}$$ Now, let's integrate the upper bounding function over the interval \([0,1]\): $$\int_{0}^{1} g(x) dx = \int_{0}^{1} \frac{1}{2} dx = \left.\frac{x}{2}\right|_{0}^1 = \frac{1}{2}$$ So, we have: $$\frac{\pi}{4} \leq \int_{0}^{1} \frac{1}{1+x^{3/2}} dx \leq \frac{1}{2}$$

Since \(\ln2 < \frac{\pi}{4} < \frac{1}{2}\), we can replace the bounds of the integral in the previous step with the desired inequality: $$\ln2 < \frac{\pi}{4} \leq \int_{0}^{1} \frac{1}{1+x^{3/2}} dx \leq \frac{1}{2} < \frac{\pi}{4}$$ From this, we can clearly see that \(\ln2 < I < \frac{\pi}{4}\), as required.