If I_1 = _0^1 2^x^2 dx, I_2 = _0^1 2^x^3 dx, I_3 = _1^2 2^x^2 dx and I_4 = _1^2 2^x^3 dx , then: - Mathematics MCQ AIEEE 2026 | ExamTest.live

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If 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:

Solution & Step-by-step Explanation

For 0 < x < 1, we know x^{2} > x^{3} .Since 2^{t} is an increasing function: 2^{x^{2}} > 2^{x^{3}} for x \in (0, 1) .Integrating both sides from 0 to 1 : \int_{0}^{1} 2^{x^{2}} d x > \int_{0}^{1} 2^{x^{3}} d x Thus, I_{1} > I_{2} .For x > 1, x^{3} > x^{2}, so I_{4} > I_{3} .The correct option is I_{1} > I_{2} .

Try it yourself before checking the explanation above.

If 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:

I_{2} > I_{1}

I_{1} > I_{2}

I_{3} = I_{4}

I_{3} > I_{4}

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