AIEEE 2006 | Definite Integration Question 277 | 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

$$\int\limits_0^\pi {xf\left( {\sin x} \right)dx} $$ is equal to

$$\pi \int\limits_0^\pi {f\left( {\cos x} \right)dx} $$

$$\,\pi \int\limits_0^\pi {f\left( {sinx} \right)dx} $$

$${\pi \over 2}\int\limits_0^{\pi /2} {f\left( {sinx} \right)dx} $$

$$\pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx} $$

AIEEE 2005

MCQ (Single Correct Answer)

-1

Let $$f:R \to R$$ be a differentiable function having $$f\left( 2 \right) = 6$$,
$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals :

$$24$$

$$36$$

$$12$$

$$18$$

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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