If $\mathrm{I}_{1}=\int_{0}^{1} 2^{\mathrm{x}^{2}} \mathrm{dx}, \mathrm{I}_{2}=\int_{0}^{1} 2^{\mathrm{x}^{3}} \mathrm{dx}, \mathrm{I}_{3}=\int_{1}^{2} 2^{\mathrm{x}^{2}} \mathrm{dx}$ and $\mathrm{I}_{4}=\int_{1}^{2} 2^{\mathrm{x}^{3}} \mathrm{dx}$ then

JEE Main · 2005 · 08 May 2005 · Offline · Definite Integrals

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

Answer: (B)

For

0 < x < 1

we have

x^{3} < x^{2}

, and since

2^{t}

is an increasing function,

2^{x^{3}} < 2^{x^{2}}

throughout the interval. Integrating this strict inequality over

[0, 1]

gives

\int_{0}^{1} 2^{x^{2}} d x > \int_{0}^{1} 2^{x^{3}} d x

, that is

I_{1} > I_{2}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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