JEE MAIN Area Enclosed Between The Curves Definite Integration Previous Year Questions (PYQs) – Page 1 of 7
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
The integral $\displaystyle \int_{0}^{\pi}\sqrt{1+4\sin^{2}\frac{x}{2}-4\sin\frac{x}{2}}\,dx$ equals:
1
2
3
4
Solution
The area of the region described by
$A=\{(x,y):x^{2}+y^{2}\le 1 \text{ and } y^{2}\le 1-x\}$ is :
1
2
3
4
Solution
If the area (in sq. units) bounded by the parabola $y^{2}=4\lambda x$ and the line $y=\lambda x,\ \lambda>0$, is $\dfrac{1}{9}$, then $\lambda$ is equal to:
1
2
3
4
Solution
The area of the region given by $\{(x,y):\, xy\le 8,\ 1\le y\le x^{2}\}$ is:
1
2
3
4
Solution
If the area of the region $\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq \mathrm{a}+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}>0\right\}$ is $\frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}$, then the value of $a$ is
1
2
3
4
Solution
A line passing through the point $A(-2,0)$ touches the parabola $P: y^2=x-2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $\overline{AB}$, parabola $P$ and the $x$-axis is:
1
2
3
4
Solution
$ \text{The area bounded by the curves } y=\lvert x^{2}-1\rvert \text{ and } y=1 \text{ is :}$
1
2
3
4
Solution
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 :
1
2
3
4
Solution
The area (in sq. units) of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$ is :
1
2
3
4
Solution
The area (in sq. units) of the region
${(x,y)\in \mathbb{R}^{2} : x \ge 0,\ y \ge 0,\ y \ge x-2 \text{ and } y \le \sqrt{x}}$ is :
1
2
3
4
Solution
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
