AIEEE 2006 | Definite Integration Question 253 | Mathematics

$$\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx$$ is equal to

$${{{\pi ^4}} \over {32}}$$

$${{{\pi ^4}} \over {32}} + {\pi \over 2}$$

$${\pi \over 2}$$

$${\pi \over 4} - 1$$

AIEEE 2006

MCQ (Single Correct Answer)

-1

The value of $$\int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1$$ where $${\left[ x \right]}$$ denotes the greatest integer not exceeding $$x$$ is

$$af\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( {\left[ a \right]} \right)} \right\}$$

$$\left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( {\left[ a \right]} \right)} \right\}$$

$$\left[ a \right]f\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + ...........f\left( a \right)} \right\}$$

$$af\left( {\left[ a \right]} \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + .............f\left( a \right)} \right\}$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

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

$${I_2} > {I_1}$$

$${I_1} > {I_2}$$

$${I_3} = {I_4}$$

$${I_3} > {I_4}$$

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