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If $\mathrm{I}_1=\int_0^1 2^{x^2} d x, I_2=\int_0^1 2^{x^3} d x, I_3=\int_1^2 2^{x^2} d x$ and $I_4=\int_1^2 2^{x^3} d x$ then
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Verified Answer
The correct answer is:
$I_1>I_2$
$I_1>I_2$
$$
\begin{aligned}
& l_1=\int_0^1 2^{x^2} d x, l_2=\int_0^1 2^{x^3} d x, I_3=\int_0^1 2^{x^2} d x, I_4=\int_0^1 2^{x^3} d x \\
& \forall 0 < x < 1, x^2>x^3 \\
& \Rightarrow \int_0^1 2^{x^2} d x>\int_0^1 2^{x^3} d x \\
& \Rightarrow I_1>I_2 .
\end{aligned}
$$
\begin{aligned}
& l_1=\int_0^1 2^{x^2} d x, l_2=\int_0^1 2^{x^3} d x, I_3=\int_0^1 2^{x^2} d x, I_4=\int_0^1 2^{x^3} d x \\
& \forall 0 < x < 1, x^2>x^3 \\
& \Rightarrow \int_0^1 2^{x^2} d x>\int_0^1 2^{x^3} d x \\
& \Rightarrow I_1>I_2 .
\end{aligned}
$$
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