AIEEE 2005 | Definite Integration Question 256 | Mathematics

If $${I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } } $$ and $${I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} $$ then

$${I_2} > {I_1}$$

$${I_1} > {I_2}$$

$${I_3} = {I_4}$$

$${I_3} > {I_4}$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,} $$ is

$$a\,\pi $$

$${\pi \over 2}$$

$${\pi \over a}$$

$${2\pi }$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx $$ is

$${1 \over 2}$$

$${3 \over 2}$$

$$2$$

$$1$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

Out of Syllabus

$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right]$$
equals

$${1 \over 2}\sec 1$$

$${1 \over 2}$$cosec 1

tan 1

$${1 \over 2}$$tan 1

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