JEE Main Area Under The Curves Practice Questions With Solutions

The area (in square units) of the region enclosed by the ellipse $x^2+3y^2=18$x^{2} + 3 y^{2} = 18 in the first quadrant below the line $y=x$y = x is

The parabola $y^2=4x$y^{2} = 4 x divides the area of the circle $x^2+y^2=5$x^{2} + y^{2} = 5 in two parts. The area of the smaller part is equal to :

The area of the region in the first quadrant inside the circle $x^2+y^2=8$x^{2} + y^{2} = 8 and outside the parabola $y^2=2x$y^{2} = 2 x is equal to :

If the area of the region $\{(x,y):\frac{\mathrm{a}}{x^2}\le y\le \frac{1}{x},1\le x\le 2,0<\mathrm{a}<1\}$\left{\right. \left(\right. x , y \left.\right) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x} , 1 \leq x \leq 2 , 0 < a < 1 \left.\right} is $({\log }_{\mathrm{e}}2)-\frac{1}{7}$\left(\right. log_{e} ⁡ 2 \left.\right) - \frac{1}{7} then the value of $7\mathrm{a}-3$7 a - 3 is equal to :

Let the area of the region enclosed by the curves $y=3x,2y=27-3x$y = 3 x , 2 y = 27 - 3 x and $y=3x-x\sqrt{x}$y = 3 x - x \sqrt{x} be $A$A. Then $10A$10 A is equal to

The area enclosed between the curves $y=x|x|$y = x \left|\right. x \left|\right. and $y=x-|x|$y = x - \left|\right. x \left|\right. is :

The area (in sq. units) of the region described by $\{(x,y):y^2\le 2x\text{, and }y\ge 4x-1\}$\left{\right. \left(\right. x , y \left.\right) : y^{2} \leq 2 x ,\text{ and}\textrm{ } y \geq 4 x - 1 \left.\right} is

One of the points of intersection of the curves $y=1+3x-2x^2$y = 1 + 3 x - 2 x^{2} and $y=\frac{1}{x}$y = \frac{1}{x} is $(\frac{1}{2},2)$\left(\right. \frac{1}{2} , 2 \left.\right). Let the area of the region enclosed by these curves be $\frac{1}{24}(l\sqrt{5}+\mathrm{m})-\mathrm{n}{\log }_{\mathrm{e}}(1+\sqrt{5})$\frac{1}{24} \left(\right. l \sqrt{5} + m \left.\right) - n log_{e} ⁡ \left(\right. 1 + \sqrt{5} \left.\right), where $l,\mathrm{m},\mathrm{n}\in \mathbf{N}$l , \textrm{ } m , n \in \mathbf{N}. Then $l+\mathrm{m}+\mathrm{n}$l + m + n is equal to

The area of the region enclosed by the parabolas $y=4x-x^2$y = 4 x - x^{2} and $3y=(x-4{)}^2$3 y = \left(\right. x - 4 \left.\right)^{2} is equal to :