If I₁ = ₀¹ 2^{x³} dx , I₂ = ₀¹ 2^{x²} dx , I₃ = ₁² 2^{x²} dx and I₄ = ₁² 2^{x³} dx, then

Correct Answer is : I1I₁I1 > I2I₂I2

Correct Answer is : I1I₁I1 > I2I₂I2

Test Index

UPSEE Solved Model Paper 4

Question : 149 of 150

Marks: +1, -0

I_1

\int\limits_{0}^{1} 2^{x^{3}}

I_2

\int\limits_{0}^{1} 2^{x^{2}}

I_3

\int\limits_{1}^{2} 2^{x^{2}}

I_4

\int\limits_{1}^{2} 2^{x^{3}}

dx, then

Solution:

Given :

I_1

\int\limits_{0}^{1} 2^{x^{3}}

I_2

\int\limits_{0}^{1} 2^{x^{3}}

I_3

\int\limits_{1}^{2} 2^{x^{2}}

and

I_4

\int\limits_{1}^{2} 2^{x^{3}}

As we know

x^{2}

x^{3}

for x ∊ (1 , 2)

2^{x^{3}}

will also be greater than

2^{x^{2}}

for x ∊ (0 , 1) and x ∊ (1 , 2)

∴

I_1

I_2

hence

I_3

\int\limits_{1}^{2} 2^{x^{2}}

dx <

I_4

\int\limits_{1}^{2} 2^{x^{3}}

And also

2^{x^{2}}

when x ∊ (1 , 2) is greater than

2^{x^{3}}

for x ∊ (0 , 1)

hence

I_1

\int\limits_{0}^{1} 2^{x^{3}}

dx <

I_3

\int\limits_{1}^{2} 2^{x^{2}}

Hence,

I_1

I_2

is correct.

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