Question:

Step 1: Establish bounds on the integrand using the constraint on $x$. For $x \in [0,1]$, we have $0 \leq x^8 \leq 1$. Therefore: $$1 \leq 1+x^8 \leq 2$$ Taking reciprocals (and reversing inequalities): $$\frac{1}{2} \leq \frac{1}{1+x^8} \leq 1$$ Step 2: Apply integration to obtain preliminary bounds on $L$. Integrating all parts of the inequality from $0$ to $1$: $$\int_0^1 \frac{1}{2}\,dx \leq L \leq \int_0^1 1\,dx$$ $$\frac{1}{2} \leq L \leq 1$$ Step 3: Compare $x^8$ with $x^2$ on the interval $[0,1]$. For $x \in [0,1]$, we have $x^8 = (x^2)^4 \leq x^2$ (since raising a number in $[0,1]$ to a higher power makes it smaller). Therefore: $$1 + x^8 \leq 1 + x^2$$ Taking reciprocals (and reversing inequalities): $$\frac{1}{1+x^8} \geq \frac{1}{1+x^2}$$ Step 4: Integrate the comparison to establish a lower bound for $L$. Integrating both sides from $0$ to $1$: $$L = \int_0^1 \frac{1}{1+x^8}\,dx \geq \int_0^1 \frac{1}{1+x^2}\,dx$$ The integral on the right is a standard result: $$\int_0^1 \frac{1}{1+x^2}\,dx = \arctan(x)\Big|_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4}$$ Therefore: $$L \geq \frac{\pi}{4}$$ Step 5: Establish that $L$ is strictly less than $1$. For $x \in (0,1]$, we have $x^8 > 0$, which means $1 + x^8 > 1$. Therefore: $$\frac{1}{1+x^8} < 1 \text{ for } x \in (0,1]$$ Since the integrand is strictly less than $1$ on the interior of the interval: $$L = \int_0^1 \frac{1}{1+x^8}\,dx < \int_0^1 1\,dx = 1$$ Step 6: Verify that the lower bound is strict. Since $x^8 < x^2$ for $x \in (0,1)$ (strict inequality on the open interval), we have: $$\frac{1}{1+x^8} > \frac{1}{1+x^2} \text{ for } x \in (0,1)$$ Therefore: $$L > \int_0^1 \frac{1}{1+x^2}\,dx = \frac{\pi}{4}$$ Step 7: Conclude with the final answer. Combining the results from Steps 5 and 6: $$\frac{\pi}{4} < L < 1$$ The answer is **Option 1: $\dfrac{\pi}{4} < L < 1$**