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If $f(x)=\int_1^x \frac{1}{2+t^4} d t$, then
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Verified Answer
The correct answer is:
$\frac{1}{18} < f(2) < \frac{1}{3}$
We have,
$$
f(x)=\int_1^x \frac{1}{2+t^4} d t
$$
So, $f(2)=\int_1^2 \frac{1}{2+t^4} d t$
Now, $f(2)>\int_1^2 \min \left(\frac{1}{2+t^4}\right) d t \quad>\int_1^2 \frac{1}{2+2^4} d t$
$$
>\int_1^2 \frac{1}{18} d t \quad \Rightarrow \quad f(2)>\frac{1}{18}
$$
and $f(2) < \int_1^2 \max \left(\frac{1}{2+t^4}\right) d t \quad < \int_1^2 \frac{1}{2+(1)^4} d t$
$$ < \int_1^2 \frac{1}{3} d t < \frac{1}{3}[t]_1^2 \quad \Rightarrow \quad f(2) < \frac{1}{3}
$$
So, $\quad \frac{1}{18} < f(2) < \frac{1}{3}$
$$
f(x)=\int_1^x \frac{1}{2+t^4} d t
$$
So, $f(2)=\int_1^2 \frac{1}{2+t^4} d t$
Now, $f(2)>\int_1^2 \min \left(\frac{1}{2+t^4}\right) d t \quad>\int_1^2 \frac{1}{2+2^4} d t$
$$
>\int_1^2 \frac{1}{18} d t \quad \Rightarrow \quad f(2)>\frac{1}{18}
$$
and $f(2) < \int_1^2 \max \left(\frac{1}{2+t^4}\right) d t \quad < \int_1^2 \frac{1}{2+(1)^4} d t$
$$ < \int_1^2 \frac{1}{3} d t < \frac{1}{3}[t]_1^2 \quad \Rightarrow \quad f(2) < \frac{1}{3}
$$
So, $\quad \frac{1}{18} < f(2) < \frac{1}{3}$
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