If {I_1} = int_0^1 {2^{x^2}} dx,{I_2} = int_0 | vedclass

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(D) For $0 x^3$. Since the base $2 > 1$,the function $f(x) = 2^x$ is strictly increasing,so $2^{x^2} > 2^{x^3}$ for $0 < x < 1

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${I_1} > {I_2}$

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(D) For $0  x^3$. Since the base $2 > 1$,the function $f(x) = 2^x$ is strictly increasing,so $2^{x^2} > 2^{x^3}$ for $0 Integrating both sides from $0$ to $1$,we get $\int_0^1 2^{x^2} dx > \int_0^1 2^{x^3} dx$,which implies ${I_1} > {I_2}$.For $1  x^2$. Similarly,$2^{x^3} > 2^{x^2}$ for $1 Integrating both sides from $1$ to $2$,we get $\int_1^2 2^{x^3} dx > \int_1^2 2^{x^2} dx$,which implies ${I_4} > {I_3}$.Thus,the correct statement is ${I_1} > {I_2}$.

If ${I_1} = \int_0^1 {2^{x^2}} dx$,${I_2} = \int_0^1 {2^{x^3}} dx$,${I_3} = \int_1^2 {2^{x^2}} dx$,and ${I_4} = \int_1^2 {2^{x^3}} dx$,then which of the following is true?

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