1. Given integrals

We need to compare $I_1=\int_0^1 2^{x^2}\,dx,\qquad I_2=\int_0^1 2^{x^3}\,dx,$ $I_3=\int_1^2 2^{x^2}\,dx,\qquad I_4=\int_1^2 2^{x^3}\,dx.$

Since the base $2>1$, the function $2^t$ is strictly increasing in $t$. So comparison of $2^{x^2}$ and $2^{x^3}$ depends on comparison of $x^2$ and $x^3$ on the given intervals.

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2. Compare $I_1$ and $I_2$ on $[0,1]$

For $x\in[0,1]$, we have $x^2\ge x^3,$ because multiplying by $x\in[0,1]$ makes the number smaller.

Since $2^t$ is increasing, $2^{x^2}\ge 2^{x^3}\quad \text{for } x\in[0,1].$ Also, for all $x\in(0,1)$, $x^2>x^3 \implies 2^{x^2}>2^{x^3}.$ Therefore, integrating over $[0,1]$, $I_1=\int_0^1 2^{x^2}\,dx > \int_0^1 2^{x^3}\,dx=I_2.$

Hence, $I_1>I_2.$ So:

• Option A: $I_2>I_1$ is false
• Option B: $I_1>I_2$ is true

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3. Compare $I_3$ and $I_4$ on $[1,2]$

For $x\in[1,2]$, we have $x^3\ge x^2.$ In fact, for $x>1$, $x^3>x^2.$ Since $2^t$ is increasing, $2^{x^3}\ge 2^{x^2}\quad \text{for } x\in[1,2],$ and for $x\in(1,2]$, $2^{x^3}>2^{x^2}.$ Therefore, $I_4=\int_1^2 2^{x^3}\,dx > \int_1^2 2^{x^2}\,dx=I_3.$

Hence, $I_4>I_3.$ So:

• Option C: $I_3=I_4$ is false
• Option D: $I_3>I_4$ is false

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4. Final conclusion

The only correct option is $\boxed{\text{B}}.$