If f(x) = int_1^x frac{1}{2+t^4} dt, then | vedclass

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(A) We have, $f(x) = \int_1^x \frac{1}{2+t^4} dt$. So, $f(2) = \int_1^2 \frac{1}{2+t^4} dt$. For $t \in [1, 2]$, the function $g(t) = \frac{1}{2+t^4}$

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$\frac{1}{18} < f(2) < \frac{1}{3}$

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(A) We have, $f(x) = \int_1^x \frac{1}{2+t^4} dt$. So, $f(2) = \int_1^2 \frac{1}{2+t^4} dt$. For $t \in [1, 2]$, the function $g(t) = \frac{1}{2+t^4}$ is a decreasing function. Therefore, the minimum value occurs at $t = 2$ and the maximum value occurs at $t = 1$. Thus, $\frac{1}{2+2^4} \leq \frac{1}{2+t^4} \leq \frac{1}{2+1^4}$. $\frac{1}{18} \leq \frac{1}{2+t^4} \leq \frac{1}{3}$. Integrating from $1$ to $2$: $\int_1^2 \frac{1}{18} dt $\frac{1}{18}(2-1) $\frac{1}{18} < f(2) < \frac{1}{3}$.

If $f(x) = \int_1^x \frac{1}{2+t^4} dt$, then

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