JEE Main 2014 (Offline) | Definite Integrals and Applications of Integrals Question 267 | Mathematics

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)

-1

The area of the region described by
$$A = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \le 1} \right.$$ and $$\left. {{y^2} \le 1 - x} \right\}$$ is :

$${\pi \over 2} - {2 \over 3}$$

$${\pi \over 2} + {2 \over 3}$$

$${\pi \over 2} + {4 \over 3}$$

$${\pi \over 2} - {4 \over 3}$$

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)

-1

The integral $$\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx$$ equals:

$$4\sqrt 3 - 4$$

$$4\sqrt 3 - 4 - {\pi \over 3}$$

$$\pi - 4$$

$${{2\pi } \over 3} - 4 - 4\sqrt 3 $$

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)

-1

Statement-1 : The value of the integral
$$\int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$ is equal to $$\pi /6$$

Statement-2 : $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx.$$

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Statement- 1 is true; Statement-2 is False.

Statement-1 is false; Statement-2 is true.

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)

-1

The area (in square units) bounded by the curves $$y = \sqrt {x,} 2y - x + 3 = 0,$$ $$x$$-axis, and lying in the first quadrant is :

$$9$$

$$36$$

$$18$$

$${{27} \over 4}$$

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