If I 1-01 2 x 2 dx - I 2-01 2 x 3 dx - I 3-12 2 x 2 dx and I 4-12 2 x 3 dx thenA- I 3- I 4- I 3- I 4C- I 1- I 2D- I 2- I 1

The correct option is C I_1 gt;I_2I_1 gt;I_2 (∵ 2^x^2 gt;2^x^2 if 0≤ x≤ 1) I_3 lt;I_4(∵ 2^x^3 gt; 2^x^2 if x gt;1) ...

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Property 4

If I 1=∫01 2 ...

Question

$I f I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x a n d I_{4} = \int_{1}^{2} 2 x^{3} d x t h e n$

I_{3} > I_{4}

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I_{3} = I_{4}

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I_{1} > I_{2}

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I_{2} > I_{1}

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Solution

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Similar questions

I_{1} = \int_{0}^{1} 2^{x^{3}} d x, I_{2} = \int_{0}^{1} 2^{x^{2}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

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

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

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

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

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

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

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

I_{4} = \int_{1}^{4} 2^{x^{3}} d x

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