NIMCET 2012 Question Paper with solutions

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Question : 8 of 120

Marks: +1, -0

If

I_1

=

∫↖{1}↙{0} 2^{x^2}

dx ,

I_2

=

∫↖{1}↙{0} 2^{x^3}

dx ,

I_3

=

∫↖{2}↙{1} 2^{x^2}

dx and

I_4

=

∫↖{2}↙{1} 2^{x^3}

dx , then

$I_1$ = $I_2$
$I_2$ > $I_1$
$I_3$ > $I_1$
$I_4$ > $I_3$

Solution:

Since

x^2

>

x^3

∀ x ∊ (0 , 1)
⇒

2^{x^2}

>

2^{x^3}

∀ x ∊ (0 , 1)
⇒

∫↖{1}↙{0} 2^{x^2}

dx >

∫↖{1}↙{0} 2^{x^3}

dx
⇒

I_2

>

I_2

Now ,

x^2

<

x^3

, ∀ x ∊ (1 , 2)
⇒

2^{x^2}

<

2^{x^3}

, ∀ x ∊ (1 , 2)
⇒

∫↖{2}↙{1} 2^{x^2}

dx <

∫↖{2}↙{1} 2^{x^3}

dx
⇒

I_3

<

I_4

or

I_4

>

I_3

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